Array Reverse - Complete Tutorial
Last Updated : 25 Sep, 2024
Given an array arr[], the task is to reverse the array. Reversing an array means rearranging the elements such that the first element becomes the last, the second element becomes second last and so on.
Examples:
Input: arr[] = {1, 4, 3, 2, 6, 5}
Output: {5, 6, 2, 3, 4, 1}
Explanation: The first element 1 moves to last position, the second element 4 moves to second-last and so on.
Input: arr[] = {4, 5, 1, 2}
Output: {2, 1, 5, 4}
Explanation: The first element 4 moves to last position, the second element 5 moves to second last and so on.
[Naive Approach] Using a temporary array - O(n) Time and O(n) Space
The idea is to use a temporary array to store the reverse of the array.
- Create a temporary array of same size as the original array.
- Now, copy all elements from original array to the temporary array in reverse order.
- Finally, copy all the elements from temporary array back to the original array.
Working:
Below is the implementation of the algorithm:
C++ // C++ Program to reverse an array using temporary array #include <iostream> #include <vector> using namespace std; // function to reverse an array void reverseArray(vector<int> &arr) { int n = arr.size(); // Temporary array to store elements in reversed order vector<int> temp(n); // Copy elements from original array to temp in reverse order for(int i = 0; i < n; i++) temp[i] = arr[n - i - 1]; // Copy elements back to original array for(int i = 0; i < n; i++) arr[i] = temp[i]; } int main() { vector<int> arr = { 1, 4, 3, 2, 6, 5 }; reverseArray(arr); for(int i = 0; i < arr.size(); i++) cout << arr[i] << " "; return 0; } // C Program to reverse an array using temporary array #include <stdio.h> #include <stdlib.h> // function to reverse an array void reverseArray(int arr[], int n) { // Temporary array to store elements in reversed order int temp[n]; // Copy elements from original array to temp in reverse order for(int i = 0; i < n; i++) temp[i] = arr[n - i - 1]; // Copy elements back to original array for(int i = 0; i < n; i++) arr[i] = temp[i]; } int main() { int arr[] = { 1, 4, 3, 2, 6, 5 }; int n = sizeof(arr) / sizeof(arr[0]); reverseArray(arr, n); for(int i = 0; i < n; i++) printf("%d ", arr[i]); return 0; } // Java Program to reverse an array using temporary array import java.util.Arrays; class GfG { // function to reverse an array static void reverseArray(int[] arr) { int n = arr.length; // Temporary array to store elements in reversed order int[] temp = new int[n]; // Copy elements from original array to temp in reverse order for (int i = 0; i < n; i++) temp[i] = arr[n - i - 1]; // Copy elements back to original array for (int i = 0; i < n; i++) arr[i] = temp[i]; } public static void main(String[] args) { int[] arr = { 1, 4, 3, 2, 6, 5 }; reverseArray(arr); for (int i = 0; i < arr.length; i++) System.out.print(arr[i] + " "); } } # Python Program to reverse an array using temporary array # function to reverse an array def reverseArray(arr): n = len(arr) # Temporary array to store elements in reversed order temp = [0] * n # Copy elements from original array to temp in reverse order for i in range(n): temp[i] = arr[n - i - 1] # Copy elements back to original array for i in range(n): arr[i] = temp[i] if __name__ == "__main__": arr = [1, 4, 3, 2, 6, 5] reverseArray(arr) for i in range(len(arr)): print(arr[i], end=" ")
// C# Program to reverse an array using temporary array using System; class GfG { // function to reverse an array static void reverseArray(int[] arr) { int n = arr.Length; // Temporary array to store elements in reversed order int[] temp = new int[n]; // Copy elements from original array to temp in reverse order for (int i = 0; i < n; i++) temp[i] = arr[n - i - 1]; // Copy elements back to original array for (int i = 0; i < n; i++) arr[i] = temp[i]; } static void Main() { int[] arr = { 1, 4, 3, 2, 6, 5 }; reverseArray(arr); for (int i = 0; i < arr.Length; i++) Console.Write(arr[i] + " "); } } // JavaScript Program to reverse an array using temporary array // function to reverse an array function reverseArray(arr) { let n = arr.length; // Temporary array to store elements in reversed order let temp = new Array(n); // Copy elements from original array to temp in reverse order for (let i = 0; i < n; i++) temp[i] = arr[n - i - 1]; // Copy elements back to original array for (let i = 0; i < n; i++) arr[i] = temp[i]; } const arr = [1, 4, 3, 2, 6, 5]; reverseArray(arr); console.log(arr.join(" ")); Time Complexity: O(n), Copying elements to a new array is a linear operation.
Auxiliary Space: O(n), as we are using an extra array to store the reversed array.
[Expected Approach - 1] Using Two Pointers - O(n) Time and O(1) Space
The idea is to maintain two pointers: left and right, such that left points at the beginning of the array and right points to the end of the array.
While left pointer is less than the right pointer, swap the elements at these two positions. After each swap, increment the left pointer and decrement the right pointer to move towards the center of array. This will swap all the elements in the first half with their corresponding element in the second half.
Working:
Below is the implementation of the algorithm:
C++ // C++ Program to reverse an array using Two Pointers #include <iostream> #include <vector> using namespace std; // function to reverse an array void reverseArray(vector<int> &arr) { // Initialize left to the beginning and right to the end int left = 0, right = arr.size() - 1; // Iterate till left is less than right while(left < right) { // Swap the elements at left and right position swap(arr[left], arr[right]); // Increment the left pointer left++; // Decrement the right pointer right--; } } int main() { vector<int> arr = { 1, 4, 3, 2, 6, 5 }; reverseArray(arr); for(int i = 0; i < arr.size(); i++) cout << arr[i] << " "; return 0; } // C Program to reverse an array using Two Pointers #include <stdio.h> // Function to swap two numbers void swap(int *a, int *b) { int temp = *a; *a = *b; *b = temp; } // function to reverse an array void reverseArray(int arr[], int n) { // Initialize left to the beginning and right to the end int left = 0, right = n - 1; // Iterate till left is less than right while (left < right) { // Swap the elements at left and right position swap(&arr[left], &arr[right]); // Increment the left pointer left++; // Decrement the right pointer right--; } } int main() { int arr[] = { 1, 4, 3, 2, 6, 5 }; int n = sizeof(arr) / sizeof(arr[0]); reverseArray(arr, n); for (int i = 0; i < n; i++) printf("%d ", arr[i]); return 0; } // Java Program to reverse an array using Two Pointers import java.util.Arrays; class GfG { // function to reverse an array static void reverseArray(int[] arr) { // Initialize left to the beginning and right to the end int left = 0, right = arr.length - 1; // Iterate till left is less than right while (left < right) { // Swap the elements at left and right position int temp = arr[left]; arr[left] = arr[right]; arr[right] = temp; // Increment the left pointer left++; // Decrement the right pointer right--; } } public static void main(String[] args) { int[] arr = { 1, 4, 3, 2, 6, 5 }; reverseArray(arr); for (int i = 0; i < arr.length; i++) System.out.print(arr[i] + " "); } } # Python Program to reverse an array using Two Pointers # function to reverse an array def reverseArray(arr): # Initialize left to the beginning and right to the end left = 0 right = len(arr) - 1 # Iterate till left is less than right while left < right: # Swap the elements at left and right position arr[left], arr[right] = arr[right], arr[left] # Increment the left pointer left += 1 # Decrement the right pointer right -= 1 if __name__ == "__main__": arr = [1, 4, 3, 2, 6, 5] reverseArray(arr) for i in range(len(arr)): print(arr[i], end=" ")
// C# Program to reverse an array using Two Pointers using System; class GfG { // function to reverse an array static void reverseArray(int[] arr) { // Initialize left to the beginning and right to the end int left = 0, right = arr.Length - 1; // Iterate till left is less than right while (left < right) { // Swap the elements at left and right position int temp = arr[left]; arr[left] = arr[right]; arr[right] = temp; // Increment the left pointer left++; // Decrement the right pointer right--; } } static void Main() { int[] arr = { 1, 4, 3, 2, 6, 5 }; reverseArray(arr); for (int i = 0; i < arr.Length; i++) Console.Write(arr[i] + " "); } } // JavaScript Program to reverse an array using Two Pointers // function to reverse an array function reverseArray(arr) { // Initialize left to the beginning and right to the end let left = 0, right = arr.length - 1; // Iterate till left is less than right while (left < right) { // Swap the elements at left and right position [arr[left], arr[right]] = [arr[right], arr[left]]; // Increment the left pointer left++; // Decrement the right pointer right--; } } const arr = [1, 4, 3, 2, 6, 5]; reverseArray(arr); console.log(arr.join(" ")); Time Complexity: O(n), as we are visiting each element exactly once.
Auxiliary Space: O(1)
[Expected Approach - 2] By Swapping Elements - O(n) Time and O(1) Space
The idea is to iterate over the first half of the array and swap each element with its corresponding element from the end. So, while iterating over the first half, any element at index i is swapped with the element at index (n - i - 1).
Working:
Below is the implementation of the algorithm:
C++ // C++ Program to reverse an array by swapping elements #include <iostream> #include <vector> using namespace std; // function to reverse an array void reverseArray(vector<int> &arr) { int n = arr.size(); // Iterate over the first half and for every index i, // swap arr[i] with arr[n - i - 1] for(int i = 0; i < n/2; i++) { swap(arr[i], arr[n - i - 1]); } } int main() { vector<int> arr = { 1, 4, 3, 2, 6, 5 }; reverseArray(arr); for(int i = 0; i < arr.size(); i++) cout << arr[i] << " "; return 0; } // C Program to reverse an array by swapping elements #include <stdio.h> // function to reverse an array void reverseArray(int arr[], int n) { // Iterate over the first half and for every index i, // swap arr[i] with arr[n - i - 1] for (int i = 0; i < n / 2; i++) { int temp = arr[i]; arr[i] = arr[n - i - 1]; arr[n - i - 1] = temp; } } int main() { int arr[] = { 1, 4, 3, 2, 6, 5 }; int n = sizeof(arr) / sizeof(arr[0]); reverseArray(arr, n); for (int i = 0; i < n; i++) printf("%d ", arr[i]); return 0; } // Java Program to reverse an array by swapping elements import java.util.Arrays; class GfG { // function to reverse an array static void reverseArray(int[] arr) { int n = arr.length; // Iterate over the first half and for every index i, // swap arr[i] with arr[n - i - 1] for (int i = 0; i < n / 2; i++) { int temp = arr[i]; arr[i] = arr[n - i - 1]; arr[n - i - 1] = temp; } } public static void main(String[] args) { int[] arr = { 1, 4, 3, 2, 6, 5 }; reverseArray(arr); for (int i = 0; i < arr.length; i++) System.out.print(arr[i] + " "); } } # Python Program to reverse an array by swapping elements def reverseArray(arr): n = len(arr) # Iterate over the first half and for every index i, # swap arr[i] with arr[n - i - 1] for i in range(n // 2): temp = arr[i] arr[i] = arr[n - i - 1] arr[n - i - 1] = temp if __name__ == "__main__": arr = [1, 4, 3, 2, 6, 5] reverseArray(arr) for i in range(len(arr)): print(arr[i], end=" ")
// C# Program to reverse an array by swapping elements using System; class GfG { // function to reverse an array static void reverseArray(int[] arr) { int n = arr.Length; // Iterate over the first half and for every index i, // swap arr[i] with arr[n - i - 1] for (int i = 0; i < n / 2; i++) { int temp = arr[i]; arr[i] = arr[n - i - 1]; arr[n - i - 1] = temp; } } static void Main() { int[] arr = { 1, 4, 3, 2, 6, 5 }; reverseArray(arr); for (int i = 0; i < arr.Length; i++) Console.Write(arr[i] + " "); } } // JavaScript Program to reverse an array by swapping elements // function to reverse an array function reverseArray(arr) { let n = arr.length; // Iterate over the first half and for every index i, // swap arr[i] with arr[n - i - 1] for (let i = 0; i < n / 2; i++) { let temp = arr[i]; arr[i] = arr[n - i - 1]; arr[n - i - 1] = temp; } } const arr = [1, 4, 3, 2, 6, 5]; reverseArray(arr); console.log(arr.join(" ")); Time Complexity: O(n), the loop runs through half of the array, so it's linear with respect to the array size.
Auxiliary Space: O(1), no extra space is required, therefore we are reversing the array in-place.
[Alternate Approach] Using Recursion - O(n) Time and O(n) Space
The idea is to use recursion and define a recursive function that takes a range of array elements as input and reverses it. Inside the recursive function,
- Swap the first and last element.
- Recursively call the function with the remaining subarray.
C++ // C++ Program to reverse an array using Recursion #include <iostream> #include <vector> using namespace std; // recursive function to reverse an array from l to r void reverseArrayRec(vector<int> &arr, int l, int r) { if(l >= r) return; // Swap the elements at the ends swap(arr[l], arr[r]); // Recur for the remaining array reverseArrayRec(arr, l + 1, r - 1); } // function to reverse an array void reverseArray(vector<int> &arr) { int n = arr.size(); reverseArrayRec(arr, 0, n - 1); } int main() { vector<int> arr = { 1, 4, 3, 2, 6, 5 }; reverseArray(arr); for(int i = 0; i < arr.size(); i++) cout << arr[i] << " "; return 0; } // C Program to reverse an array using Recursion #include <stdio.h> // recursive function to reverse an array from l to r void reverseArrayRec(int arr[], int l, int r) { if(l >= r) return; // Swap the elements at the ends int temp = arr[l]; arr[l] = arr[r]; arr[r] = temp; // Recur for the remaining array reverseArrayRec(arr, l + 1, r - 1); } // function to reverse an array void reverseArray(int arr[], int n) { reverseArrayRec(arr, 0, n - 1); } int main() { int arr[] = { 1, 4, 3, 2, 6, 5 }; int n = sizeof(arr) / sizeof(arr[0]); reverseArray(arr, n); for(int i = 0; i < n; i++) printf("%d ", arr[i]); return 0; } // Java Program to reverse an array using Recursion import java.util.Arrays; class GfG { // recursive function to reverse an array from l to r static void reverseArrayRec(int[] arr, int l, int r) { if (l >= r) return; // Swap the elements at the ends int temp = arr[l]; arr[l] = arr[r]; arr[r] = temp; // Recur for the remaining array reverseArrayRec(arr, l + 1, r - 1); } // function to reverse an array static void reverseArray(int[] arr) { int n = arr.length; reverseArrayRec(arr, 0, n - 1); } public static void main(String[] args) { int[] arr = { 1, 4, 3, 2, 6, 5 }; reverseArray(arr); for (int i = 0; i < arr.length; i++) System.out.print(arr[i] + " "); } } # Python Program to reverse an array using Recursion # recursive function to reverse an array from l to r def reverseArrayRec(arr, l, r): if l >= r: return # Swap the elements at the ends arr[l], arr[r] = arr[r], arr[l] # Recur for the remaining array reverseArrayRec(arr, l + 1, r - 1) # function to reverse an array def reverseArray(arr): n = len(arr) reverseArrayRec(arr, 0, n - 1) if __name__ == "__main__": arr = [1, 4, 3, 2, 6, 5] reverseArray(arr) for i in range(len(arr)): print(arr[i], end=" ")
// C# Program to reverse an array using Recursion using System; class GfG { // recursive function to reverse an array from l to r static void reverseArrayRec(int[] arr, int l, int r) { if (l >= r) return; // Swap the elements at the ends int temp = arr[l]; arr[l] = arr[r]; arr[r] = temp; // Recur for the remaining array reverseArrayRec(arr, l + 1, r - 1); } // function to reverse an array static void reverseArray(int[] arr) { int n = arr.Length; reverseArrayRec(arr, 0, n - 1); } static void Main(string[] args) { int[] arr = { 1, 4, 3, 2, 6, 5 }; reverseArray(arr); for (int i = 0; i < arr.Length; i++) Console.Write(arr[i] + " "); } } // JavaScript Program to reverse an array using Recursion // recursive function to reverse an array from l to r function reverseArrayRec(arr, l, r) { if (l >= r) return; // Swap the elements at the ends [arr[l], arr[r]] = [arr[r], arr[l]]; // Recur for the remaining array reverseArrayRec(arr, l + 1, r - 1); } // function to reverse an array function reverseArray(arr) { let n = arr.length; reverseArrayRec(arr, 0, n - 1); } let arr = [1, 4, 3, 2, 6, 5]; reverseArray(arr); console.log(arr.join(" ")); Time Complexity: O(n), the recurrence relation will be T(n) = T(n - 2) + O(1), which can be simplified to O(n).
Auxiliary Space: O(n), as we are using recursion stack.
Using Inbuilt Methods - O(n) Time and O(1) Space
The idea is to use inbuilt reverse methods available across different languages.
C++ // C++ Program to reverse an array using inbuilt methods #include <iostream> #include <vector> #include <algorithm> using namespace std; // function to reverse an array void reverseArray(vector<int> &arr) { reverse(arr.begin(), arr.end()); } int main() { vector<int> arr = { 1, 4, 3, 2, 6, 5 }; reverseArray(arr); for(int i = 0; i < arr.size(); i++) cout << arr[i] << " "; return 0; } // Java Program to reverse an array using inbuilt methods import java.util.*; class GfG { // function to reverse an array static void reverseArray(List<Integer> arr) { Collections.reverse(arr); } public static void main(String[] args) { List<Integer> arr = new ArrayList<>(Arrays.asList(1, 4, 3, 2, 6, 5)); reverseArray(arr); for (int i = 0; i < arr.size(); i++) System.out.print(arr.get(i) + " "); } } # Python Program to reverse an array using inbuilt methods # function to reverse an array def reverse_array(arr): arr.reverse() if __name__ == "__main__": arr = [1, 4, 3, 2, 6, 5] reverse_array(arr) print(" ".join(map(str, arr))) // C# Program to reverse an array using inbuilt methods using System; class GfG { // function to reverse an array static void reverseArray(int[] arr) { Array.Reverse(arr); } static void Main() { int[] arr = { 1, 4, 3, 2, 6, 5 }; reverseArray(arr); for (int i = 0; i < arr.Length; i++) Console.Write(arr[i] + " "); } } // JavaScript Program to reverse an array using inbuilt methods // function to reverse an array function reverseArray(arr) { arr.reverse(); } const arr = [1, 4, 3, 2, 6, 5]; reverseArray(arr); console.log(arr.join(" ")); Time Complexity: O(n), the reverse method has linear time complexity.
Auxiliary Space: O(1) Additional space is not used to store the reversed array, as the in-built array method swaps the values in-place.
Similar Reads
Array Data Structure Complete Guide to ArraysLearn more about Array in DSA Self Paced CoursePractice Problems on ArraysTop Quizzes on Arrays What is Array?An array is a collection of items stored at contiguous memory locations. The idea is to store multiple items of the same type together. This makes it easier to calcul
3 min read
What is Array? Array is a linear data structure where all elements are arranged sequentially. It is a collection of elements of same data type stored at contiguous memory locations. For simplicity, we can think of an array as a flight of stairs where on each step is placed a value (let's say one of your friends).
2 min read
Getting Started with Array Data Structure Array is a collection of items of the same variable type that are stored at contiguous memory locations. It is one of the most popular and simple data structures used in programming. Basic terminologies of ArrayArray Index: In an array, elements are identified by their indexes. Array index starts fr
14 min read
Applications, Advantages and Disadvantages of Array Array is a linear data structure that is a collection of data elements of same types. Arrays are stored in contiguous memory locations. It is a static data structure with a fixed size. Applications of Array Data Structure:Arrays mainly have advantages like random access and cache friendliness over o
2 min read
Subarrays, Subsequences, and Subsets in Array What is a Subarray?A subarray is a contiguous part of array, i.e., Subarray is an array that is inside another array. In general, for an array of size n, there are n*(n+1)/2 non-empty subarrays. For example, Consider the array [1, 2, 3, 4], There are 10 non-empty sub-arrays. The subarrays are: (1),
10 min read
Basic operations in Array
Searching in ArraySearching is one of the most common operations performed in an array. Array searching can be defined as the operation of finding a particular element or a group of elements in the array. There are several searching algorithms. The most commonly used among them are: Linear Search Binary Search Ternar
4 min read
Array Reverse - Complete TutorialGiven an array arr[], the task is to reverse the array. Reversing an array means rearranging the elements such that the first element becomes the last, the second element becomes second last and so on.Examples: Input: arr[] = {1, 4, 3, 2, 6, 5} Output: {5, 6, 2, 3, 4, 1}Explanation: The first elemen
15+ min read
Rotate an Array by d - Counterclockwise or LeftGiven an array of integers arr[] of size n, the task is to rotate the array elements to the left by d positions.Examples:Input: arr[] = {1, 2, 3, 4, 5, 6}, d = 2Output: {3, 4, 5, 6, 1, 2}Explanation: After first left rotation, arr[] becomes {2, 3, 4, 5, 6, 1} and after the second rotation, arr[] bec
15+ min read
Array after K RotationsGiven an array arr[] and an integer k, rotate the array in place k times to the right (clockwise). In each rotation, the last element moves to the front, and all other elements shift one position to the right. Modify the array in place, do not return anything.Examples : Input: arr[] = [1, 2, 3, 4, 5
12 min read
Search, Insert, and Delete in an Unsorted Array | Array OperationsIn this post, a program to search, insert, and delete operations in an unsorted array is discussed.Search Operation:In an unsorted array, the search operation can be performed by linear traversal from the first element to the last element. Coding implementation of the search operation:C++// C++ prog
15+ min read
Search, Insert, and Delete in an Sorted Array | Array OperationsHow to Search in a Sorted Array?In a sorted array, the search operation can be performed by using binary search.Below is the implementation of the above approach:C++// C++ program to implement binary search in sorted array #include <bits/stdc++.h> using namespace std; int binarySearch(int arr[
15+ min read
Array Sorting - Practice ProblemsSorting an array means arranging the elements of the array in a certain order. Generally sorting in an array is done to arrange the elements in increasing or decreasing order. Problem statement: Given an array of integers arr, the task is to sort the array in ascending order and return it, without u
9 min read
Generating All SubarraysGiven an array arr[], the task is to generate all the possible subarrays of the given array.Examples: Input: arr[] = [1, 2, 3]Output: [ [1], [1, 2], [2], [1, 2, 3], [2, 3], [3] ]Input: arr[] = [1, 2]Output: [ [1], [1, 2], [2] ]Iterative ApproachTo generate a subarray, we need a starting index from t
8 min read
Easy problems on Array
Largest three distinct elements in an arrayGiven an array arr[], the task is to find the top three largest distinct integers present in the array.Note: If there are less than three distinct elements in the array, then return the available distinct numbers in descending order.Examples :Input: arr[] = [10, 4, 3, 50, 23, 90]Output: [90, 50, 23]
6 min read
Second Largest Element in an ArrayGiven an array of positive integers arr[] of size n, the task is to find second largest distinct element in the array.Note: If the second largest element does not exist, return -1. Examples:Input: arr[] = [12, 35, 1, 10, 34, 1]Output: 34Explanation: The largest element of the array is 35 and the sec
14 min read
Move all zeros to end of arrayGiven an array of integers arr[], the task is to move all the zeros to the end of the array while maintaining the relative order of all non-zero elements.Examples: Input: arr[] = [1, 2, 0, 4, 3, 0, 5, 0]Output: arr[] = [1, 2, 4, 3, 5, 0, 0, 0]Explanation: There are three 0s that are moved to the end
15 min read
Rearrange array such that even positioned are greater than oddGiven an array arr[], sort the array according to the following relations: arr[i] >= arr[i - 1], if i is even, ∀ 1 <= i < narr[i] <= arr[i - 1], if i is odd, ∀ 1 <= i < nFind the resultant array.[consider 1-based indexing]Examples: Input: arr[] = [1, 2, 2, 1]Output: [1 2 1 2] Expla
9 min read
Rearrange an array in maximum minimum form using Two Pointer TechniqueGiven a sorted array of positive integers, rearrange the array alternately i.e first element should be a maximum value, at second position minimum value, at third position second max, at fourth position second min, and so on. Examples: Input: arr[] = {1, 2, 3, 4, 5, 6, 7} Output: arr[] = {7, 1, 6, 2
6 min read
Segregate even and odd numbers using Lomuto’s Partition SchemeGiven an array arr[] of integers, segregate even and odd numbers in the array such that all the even numbers should be present first, and then the odd numbers.Examples: Input: arr[] = {7, 2, 9, 4, 6, 1, 3, 8, 5}Output: 2 4 6 8 7 9 1 3 5Input: arr[] = {1, 3, 2, 4, 7, 6, 9, 10}Output: 2 4 6 10 7 1 9 3
6 min read
Reversal algorithm for Array rotationGiven an array arr[] of size N, the task is to rotate the array by d position to the left. Examples: Input: arr[] = {1, 2, 3, 4, 5, 6, 7}, d = 2Output: 3, 4, 5, 6, 7, 1, 2Explanation: If the array is rotated by 1 position to the left, it becomes {2, 3, 4, 5, 6, 7, 1}.When it is rotated further by 1
15 min read
Print left rotation of array in O(n) time and O(1) spaceGiven an array of size n and multiple values around which we need to left rotate the array. How to quickly print multiple left rotations?Examples : Input : arr[] = {1, 3, 5, 7, 9}k1 = 1k2 = 3k3 = 4k4 = 6Output : 3 5 7 9 17 9 1 3 59 1 3 5 73 5 7 9 1Input : arr[] = {1, 3, 5, 7, 9}k1 = 14 Output : 9 1
15+ min read
Sort an array which contain 1 to n valuesWe are given an array that contains 1 to n elements, our task is to sort this array in an efficient way. We are not allowed to simply copy the numbers from 1 to n.Examples : Input : arr[] = {2, 1, 3};Output : {1, 2, 3}Input : arr[] = {2, 1, 4, 3};Output : {1, 2, 3, 4} Native approach - O(n Log n) Ti
7 min read
Count Possible TrianglesGiven an unsorted array of positive integers, the task is to find the number of triangles that can be formed with three different array elements as three sides of triangles. For a triangle to be possible from 3 values as sides, the sum of the two values (or sides) must always be greater than the thi
15+ min read
Find all distinct elements in a given arrayGiven an integer array arr[], find all the distinct elements of the given array. The given array may contain duplicates and the output should contain every element only once.Examples: Input: arr[] = [2, 2, 3, 3, 7, 5]Output: [2, 3, 7, 5]Explanation : After removing the duplicates 2 and 3 we get 2 3
9 min read
Unique Number IGiven an array of integers, every element in the array appears twice except for one element which appears only once. The task is to identify and return the element that occurs only once.Examples: Input: arr[] = [2, 3, 5, 4, 5, 3, 4]Output: 2 Explanation: Since 2 occurs once, while other numbers occu
8 min read
Leaders in an arrayGiven an array arr[] of size n, the task is to find all the Leaders in the array. An element is a Leader if it is greater than or equal to all the elements to its right side. Note: The rightmost element is always a leader. Examples: Input: arr[] = [16, 17, 4, 3, 5, 2]Output: [17 5 2]Explanation: 17
10 min read
Subarray with Given SumGiven a 1-based indexing array arr[] of non-negative integers and an integer sum. You mainly need to return the left and right indexes(1-based indexing) of that subarray. In case of multiple subarrays, return the subarray indexes which come first on moving from left to right. If no such subarray exi
10 min read
Intermediate problems on Array
Rearrange an array such that arr[i] = iGiven an array of elements of length n, ranging from 0 to n - 1. All elements may not be present in the array. If the element is not present then there will be -1 present in the array. Rearrange the array such that arr[i] = i and if i is not present, display -1 at that place.Examples: Input: arr[] =
13 min read
Alternate Rearrangement of Positives and NegativesAn array contains both positive and negative numbers in random order. Rearrange the array elements so that positive and negative numbers are placed alternatively. A number of positive and negative numbers need not be equal. If there are more positive numbers they appear at the end of the array. If t
11 min read
Reorder an array according to given indexesGiven two integer arrays of the same length, arr[] and index[], the task is to reorder the elements in arr[] such that after reordering, each element from arr[i] moves to the position index[i]. The new arrangement reflects the values being placed at their target indices, as described by index[] arra
15+ min read
Find the smallest missing numberGiven a sorted array of n distinct integers where each integer is in the range from 0 to m-1 and m > n. Find the smallest number that is missing from the array. Examples: Input: {0, 1, 2, 6, 9}, n = 5, m = 10 Output: 3 Input: {4, 5, 10, 11}, n = 4, m = 12 Output: 0 Input: {0, 1, 2, 3}, n = 4, m =
15 min read
Difference Array | Range update query in O(1)Given an array arr[] and a 2D array opr[][], where each row represents an operation in the form [l, r, v]. For each operation, add v to all elements from index l to r in arr. Return the updated array after applying all operations.Examples : Input: arr[] = [2, 3, 5, 6, 7], opr[][] = [[2, 4, 2], [3, 4
15+ min read
Stock Buy and Sell – Max 2 Transactions AllowedIn the stock market, a person buys a stock and sells it on some future date. Given the stock prices of n days in an array prices[ ]. Find out the maximum profit a person can make in at most 2 transactions. A transaction is equivalent to (buying + selling) of a stock and a new transaction can start o
15+ min read
Smallest subarray with sum greater than a given valueGiven an array arr[] of integers and a number x, the task is to find the smallest subarray with a sum strictly greater than x.Examples:Input: x = 51, arr[] = [1, 4, 45, 6, 0, 19]Output: 3Explanation: Minimum length subarray is [4, 45, 6]Input: x = 100, arr[] = [1, 10, 5, 2, 7]Output: 0Explanation: N
15+ min read
Count Inversions of an ArrayGiven an integer array arr[] of size n, find the inversion count in the array. Two array elements arr[i] and arr[j] form an inversion if arr[i] > arr[j] and i < j.Note: Inversion Count for an array indicates that how far (or close) the array is from being sorted. If the array is already sorted
15+ min read
Merge Two Sorted Arrays Without Extra SpaceGiven two sorted arrays a[] and b[] of size n and m respectively, the task is to merge both the arrays and rearrange the elements such that the smallest n elements are in a[] and the remaining m elements are in b[]. All elements in a[] and b[] should be in sorted order.Examples: Input: a[] = [2, 4,
15+ min read
Majority ElementGiven an array arr[], return the element that appears more than n / 2 times, where n is the array size. If no such element exists, return -1.Examples:Input: arr[] = [1, 1, 2, 1, 3, 5, 1]Output: 1Explanation: Element 1 appears 4 times, which is more than {\scriptsize \frac{7}{2} = 3.5} so it is the m
15 min read
Two Pointers TechniqueTwo pointers is really an easy and effective technique that is typically used for Two Sum in Sorted Arrays, Closest Two Sum, Three Sum, Four Sum, Trapping Rain Water and many other popular interview questions. Given a sorted array arr (sorted in ascending order) and a target, find if there exists an
11 min read
3 Sum - Triplet Sum in ArrayGiven an array arr[] of size n and an integer sum, the task is to check if there is a triplet in the array which sums up to the given target sum.Examples: Input: arr[] = [1, 4, 45, 6, 10, 8], target = 13Output: true Explanation: The triplet [1, 4, 8] sums up to 13Input: arr[] = [1, 2, 4, 3, 6, 7], t
15 min read
Equilibrium IndexGiven an array arr[] of size n, the task is to return an equilibrium index (if any) or -1 if no equilibrium index exists. The equilibrium index of an array is an index such that the sum of all elements at lower indexes equals the sum of all elements at higher indexes. Note: When the index is at the
15 min read
Hard problems on Array
MO's Algorithm (Query Square Root Decomposition) | Set 1 (Introduction)Let us consider the following problem to understand MO's Algorithm. We are given an array and a set of query ranges, we are required to find the sum of every query range.Example: Input: arr[] = {1, 1, 2, 1, 3, 4, 5, 2, 8}; query[] = [0, 4], [1, 3] [2, 4]Output: Sum of arr[] elements in range [0, 4]
15+ min read
Square Root (Sqrt) Decomposition AlgorithmSquare Root Decomposition Technique is one of the most common query optimization techniques used by competitive programmers. This technique helps us to reduce Time Complexity by a factor of sqrt(N) The key concept of this technique is to decompose a given array into small chunks specifically of size
15+ min read
Sparse TableSparse table concept is used for fast queries on a set of static data (elements do not change). It does preprocessing so that the queries can be answered efficiently.Range Minimum Query Using Sparse TableYou are given an integer array arr of length n and an integer q denoting the number of queries.
15+ min read
Range sum query using Sparse TableWe have an array arr[]. We need to find the sum of all the elements in the range L and R where 0 <= L <= R <= n-1. Consider a situation when there are many range queries. Examples: Input : 3 7 2 5 8 9 query(0, 5) query(3, 5) query(2, 4) Output : 34 22 15Note : array is 0 based indexed and q
8 min read
Range LCM QueriesGiven an array arr[] of integers of size N and an array of Q queries, query[], where each query is of type [L, R] denoting the range from index L to index R, the task is to find the LCM of all the numbers of the range for all the queries.Examples: Input: arr[] = {5, 7, 5, 2, 10, 12 ,11, 17, 14, 1, 4
15+ min read
Jump Game - Minimum Jumps to Reach EndGiven an array arr[] of non-negative numbers. Each number tells you the maximum number of steps you can jump forward from that position.For example:If arr[i] = 3, you can jump to index i + 1, i + 2, or i + 3 from position i.If arr[i] = 0, you cannot jump forward from that position.Your task is to fi
15+ min read
Space optimization using bit manipulationsThere are many situations where we use integer values as index in array to see presence or absence, we can use bit manipulations to optimize space in such problems.Let us consider below problem as an example.Given two numbers say a and b, mark the multiples of 2 and 5 between a and b using less than
12 min read
Maximum value of Sum(i*arr[i]) with array rotations allowedGiven an array arr[], the task is to determine the maximum possible value of the expression i*arr[i] after rotating the array any number of times (including zero).Note: In each rotation, every element of the array shifts one position to the right, and the last element moves to the front.Examples : I
12 min read
Construct an array from its pair-sum arrayGiven a pair-sum array arr[], construct the original array res[] where each element in arr represents the sum of a unique pair from res in a specific order, starting with res[0] + res[1], then res[0] + res[2], and so on through all combinations where the first index is less than the second. Note: Th
5 min read
Maximum equilibrium sum in an arrayGiven an array arr[]. Find the maximum value of prefix sum which is also suffix sum for index i in arr[].Examples : Input : arr[] = {-1, 2, 3, 0, 3, 2, -1}Output : 4Explanation : Prefix sum of arr[0..3] = Suffix sum of arr[3..6]Input : arr[] = {-3, 5, 3, 1, 2, 6, -4, 2}Output : 7Explanation : Prefix
11 min read
Smallest Difference Triplet from Three arraysThree arrays of same size are given. Find a triplet such that maximum - minimum in that triplet is minimum of all the triplets. A triplet should be selected in a way such that it should have one number from each of the three given arrays. If there are 2 or more smallest difference triplets, then the
9 min read
Top 50 Array Coding Problems for Interviews Array is one of the most widely used data structure and is frequently asked in coding interviews to the problem solving skills. The following list of 50 array coding problems covers a range of difficulty levels, from easy to hard, to help candidates prepare for interviews.Easy ProblemsSecond Largest
2 min read