Print left rotation of array in O(n) time and O(1) space
Last Updated : 05 Dec, 2024
Given 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 = 1
k2 = 3
k3 = 4
k4 = 6
Output :
3 5 7 9 1
7 9 1 3 5
9 1 3 5 7
3 5 7 9 1
Input :
arr[] = {1, 3, 5, 7, 9}
k1 = 14
Output :
9 1 3 5 7
Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.
We have discussed a solution in the below post.
Quickly find multiple left rotations of an array | Set 1
Method I: The solution discussed above requires extra space. In this post, an optimized solution is discussed that doesn't require extra space.
Implementation:
C++ // C++ implementation of left rotation of // an array K number of times #include <bits/stdc++.h> using namespace std; // Function to leftRotate array multiple times void leftRotate(int arr[], int n, int k) { /* To get the starting point of rotated array */ int mod = k % n; // Prints the rotated array from start position for (int i = 0; i < n; i++) cout << (arr[(mod + i) % n]) << " "; cout << "\n"; } // Driver Code int main() { int arr[] = { 1, 3, 5, 7, 9 }; int n = sizeof(arr) / sizeof(arr[0]); int k = 2; // Function Call leftRotate(arr, n, k); k = 3; // Function Call leftRotate(arr, n, k); k = 4; // Function Call leftRotate(arr, n, k); return 0; } #include <stdio.h> // Function to leftRotate array multiple times void leftRotate(int arr[], int n, int k) { /* To get the starting point of rotated array */ int mod = k % n; // Prints the rotated array from start position for (int i = 0; i < n; i++) printf("%d ", arr[(mod + i) % n]); printf("\n"); } int main() { int arr[] = { 1, 3, 5, 7, 9 }; int n = sizeof(arr) / sizeof(arr[0]); int k = 2; // Function Call leftRotate(arr, n, k); k = 3; // Function Call leftRotate(arr, n, k); k = 4; // Function Call leftRotate(arr, n, k); return 0; } // JAVA implementation of left rotation // of an array K number of times import java.util.*; import java.lang.*; import java.io.*; class arr_rot { // Function to leftRotate array multiple // times static void leftRotate(int arr[], int n, int k) { /* To get the starting point of rotated array */ int mod = k % n; // Prints the rotated array from // start position for (int i = 0; i < n; ++i) System.out.print(arr[(i + mod) % n] + " "); System.out.println(); } // Driver code public static void main(String[] args) { int arr[] = { 1, 3, 5, 7, 9 }; int n = arr.length; int k = 2; // Function Call leftRotate(arr, n, k); k = 3; // Function Call leftRotate(arr, n, k); k = 4; // Function Call leftRotate(arr, n, k); } } // This code is contributed by Sanjal # Python implementation of left rotation of # an array K number of times # Function to leftRotate array multiple times def leftRotate(arr, n, k): # To get the starting point of rotated array mod = k % n s = "" # Prints the rotated array from start position for i in range(n): print str(arr[(mod + i) % n]), print return # Driver code arr = [1, 3, 5, 7, 9] n = len(arr) k = 2 # Function Call leftRotate(arr, n, k) k = 3 # Function Call leftRotate(arr, n, k) k = 4 # Function Call leftRotate(arr, n, k) # This code is contributed by Sachin Bisht
// C# implementation of left // rotation of an array K // number of times using System; class GFG { // Function to leftRotate // array multiple times static void leftRotate(int[] arr, int n, int k) { // To get the starting // point of rotated array int mod = k % n; // Prints the rotated array // from start position for (int i = 0; i < n; ++i) Console.Write(arr[(i + mod) % n] + " "); Console.WriteLine(); } // Driver Code static public void Main() { int[] arr = { 1, 3, 5, 7, 9 }; int n = arr.Length; int k = 2; // Function Call leftRotate(arr, n, k); k = 3; // Function Call leftRotate(arr, n, k); k = 4; // Function Call leftRotate(arr, n, k); } } // This code is contributed by m_kit <script> // JavaScript implementation of left rotation of // an array K number of times // Function to leftRotate array multiple times function leftRotate(arr, n, k){ /* To get the starting point of rotated array */ let mod = k % n; // Prints the rotated array from start position for (let i = 0; i < n; i++) document.write((arr[(mod + i) % n]) + " "); document.write("\n"); } // Driver Code let arr = [ 1, 3, 5, 7, 9 ]; let n = arr.length; let k = 2; // Function Call leftRotate(arr, n, k); document.write("<br>"); k = 3; // Function Call leftRotate(arr, n, k); document.write("<br>"); k = 4; // Function Call leftRotate(arr, n, k); </script> <?php // PHP implementation of // left rotation of an // array K number of times // Function to leftRotate // array multiple times function leftRotate($arr, $n, $k) { // To get the starting // point of rotated array $mod = $k % $n; // Prints the rotated array // from start position for ($i = 0; $i < $n; $i++) echo ($arr[($mod + $i) % $n]) , " "; echo "\n"; } // Driver Code $arr = array(1, 3, 5, 7, 9); $n = sizeof($arr); $k = 2; // Function Call leftRotate($arr, $n, $k); $k = 3; // Function Call leftRotate($arr, $n, $k); $k = 4; // Function Call leftRotate($arr, $n, $k); // This code is contributed by m_kit ?> Output5 7 9 1 3 7 9 1 3 5 9 1 3 5 7
Time Complexity: O(N), as we are using a loop to traverse N times.
Auxiliary Space: O(1), as we are not using any extra space.
Method II: In the below implementation we will use Standard Template Library (STL) which will be making the solution more optimize and easy to Implement.
Implementation:
C++ // C++ Implementation For Print Left Rotation Of Any Array K // Times #include <bits/stdc++.h> #include <iostream> using namespace std; // Function For The k Times Left Rotation void leftRotate(int arr[], int k, int n) { // Stl function rotates takes three parameters - the // beginning,the position by which it should be rotated // ,the end address of the array // The below function will be rotating the array left // in linear time (k%arraySize) times rotate(arr, arr + (k % n), arr + n); // Print the rotated array from start position for (int i = 0; i < n; i++) cout << arr[i] << " "; cout << "\n"; } // Driver program int main() { int arr[] = { 1, 3, 5, 7, 9 }; int n = sizeof(arr) / sizeof(arr[0]); int k = 2; // Function Call leftRotate(arr, k, n); return 0; } #include <stdio.h> // Function For k Times Left Rotation void leftRotate(int arr[], int k, int n) { int i, temp; // Perform k left rotations for (i = 0; i < k; i++) { // Store the first element of the array temp = arr[0]; // Shift all elements one position to the left for (int j = 0; j < n - 1; j++) { arr[j] = arr[j + 1]; } // Place the first element at the end arr[n - 1] = temp; } // Print the rotated array for (i = 0; i < n; i++) { printf("%d ", arr[i]); } printf("\n"); } // Driver program int main() { int arr[] = { 1, 3, 5, 7, 9 }; int n = sizeof(arr) / sizeof(arr[0]); int k = 2; // Function Call leftRotate(arr, k, n); return 0; } // Java implementation for print left // rotation of any array K times import java.io.*; import java.util.*; class GFG{ // Function for the k times left rotation static void leftRotate(Integer arr[], int k, int n) { // In Collection class rotate function // takes two parameters - the name of // array and the position by which it // should be rotated // The below function will be rotating // the array left in linear time // Collections.rotate()rotate the // array from right hence n-k Collections.rotate(Arrays.asList(arr), n - k); // Print the rotated array from start position for(int i = 0; i < n; i++) System.out.print(arr[i] + " "); } // Driver code public static void main(String[] args) { Integer arr[] = { 1, 3, 5, 7, 9 }; int n = arr.length; int k = 2; // Function call leftRotate(arr, k, n); } } // This code is contributed by chahattekwani71 # Python3 implementation to print left # rotation of any array K times from collections import deque # Function For The k Times Left Rotation def leftRotate(arr, k, n): # The collections module has deque class # which provides the rotate(), which is # inbuilt function to allow rotation arr = deque(arr) # using rotate() to left rotate by k arr.rotate(-k) arr = list(arr) # Print the rotated array from # start position for i in range(n): print(arr[i], end = " ") # Driver Code if __name__ == '__main__': arr = [ 1, 3, 5, 7, 9 ] n = len(arr) k = 2 # Function Call leftRotate(arr, k, n) # This code is contributed by math_lover
// C# program for the above approach using System; class GFG { static void leftRotate(int[] arr, int d, int n) { for (int i = 0; i < d; i++) leftRotatebyOne(arr, n); } static void leftRotatebyOne(int[] arr, int n) { int i, temp = arr[0]; for (i = 0; i < n - 1; i++) arr[i] = arr[i + 1]; arr[n - 1] = temp; } /* utility function to print an array */ static void printArray(int[] arr, int size) { for (int i = 0; i < size; i++) Console.Write(arr[i] + " "); } // Driver Code public static void Main() { int[] arr = { 1, 3, 5, 7, 9 }; int n = arr.Length; int k = 2; // Function call leftRotate(arr, k, n); printArray(arr, n); } } // This code is contributed by avijitmondal1998. let arr = [4, 3, 7, 6, 2, 1, 5,8]; arr.sort((a, b) => a - b); // Sort the array let k = 6; let n = arr.length for (let i = 0; i < k; i++) { for (let j = 0; j < n-1; j++) { [arr[j + 1] , arr[j]] = [arr[j] , arr[j + 1]]; } } console.log("Final array:", arr); Note: the array itself gets updated after the rotation.
Time Complexity: O(n)
Auxiliary Space: O(1), since no extra space has been taken.
Method III(Using Reversal):
To left rotate an array by "k" units we will perform 3 simple reversals-
- Reverse the first "k" elements
- Reverse the last "n-k" elements where n is the size of the array
- Reverse the whole array
Code-
C++ // C++ Implementation For Print Left Rotation Of Any Array K // Times #include <bits/stdc++.h> #include <iostream> using namespace std; // Function For The k Times Left Rotation void leftRotate(int arr[], int k, int n) { // if k>n , k%n will bring k back in range k = (k%n); reverse(arr,arr+k); reverse(arr+k,arr+n); reverse(arr,arr+n); // Print the rotated array from start position for (int i = 0; i < n; i++) cout << arr[i] << " "; cout << "\n"; } // Driver program int main() { int arr[] = { 1, 3, 5, 7, 9 }; int n = sizeof(arr) / sizeof(arr[0]); int k = 2; // Function Call leftRotate(arr, k, n); return 0; } #include <stdio.h> // Function For k Times Left Rotation void leftRotate(int arr[], int k, int n) { // if k > n, k % n will bring k back in range k = (k % n); // Reverse the first part of the array (0 to k-1) for (int i = 0, j = k - 1; i < j; i++, j--) { int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; } // Reverse the second part of the array (k to n-1) for (int i = k, j = n - 1; i < j; i++, j--) { int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; } // Reverse the entire array for (int i = 0, j = n - 1; i < j; i++, j--) { int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; } // Print the rotated array for (int i = 0; i < n; i++) { printf("%d ", arr[i]); } printf("\n"); } // Driver program int main() { int arr[] = { 1, 3, 5, 7, 9 }; int n = sizeof(arr) / sizeof(arr[0]); int k = 2; // Function Call leftRotate(arr, k, n); return 0; } import java.util.*; public class Main { // Function for k times left rotation public static void leftRotate(int[] arr, int k) { // if k>arr.length,k%arr.length will bring k back to range k%=arr.length; // Reverse the first k elements reverseArray(arr, 0, k - 1); // Reverse the remaining n-k elements reverseArray(arr, k, arr.length - 1); // Reverse the entire array reverseArray(arr, 0, arr.length - 1); // Print the rotated array from start position String result = Arrays.toString(arr).replaceAll("\\[|\\]|,|\\s", " "); System.out.println(result); } // Helper function to reverse a section of an array from start to end (inclusive) public static void reverseArray(int[] arr, int start, int end) { while (start < end) { int temp = arr[start]; arr[start] = arr[end]; arr[end] = temp; start++; end--; } } // Driver code public static void main(String[] args) { int[] arr = {1, 3, 5, 7, 9}; int k = 2; // Function Call leftRotate(arr, k); } } # Function for k times left rotation def leftRotate(arr, k): # if k>len(arr) , k%=len(arr) bring k back to range k %= len(arr) # Reverse the first k elements arr = reverseArray(arr, 0, k - 1) # Reverse the remaining n-k elements arr = reverseArray(arr, k, len(arr) - 1) # Reverse the entire array arr = reverseArray(arr, 0, len(arr) - 1) # Print the rotated array from start position print(" ".join(map(str,arr))) # Helper function to reverse a section of an array from start to end (inclusive) def reverseArray(arr, start, end): while start < end: temp = arr[start] arr[start] = arr[end] arr[end] = temp start += 1 end -= 1 return arr # Driver code arr = [1, 3, 5, 7, 9] k = 2 # Function Call leftRotate(arr, k) // C# Implementation For Print Left Rotation Of Any Array K // Times using System; using System.Collections.Generic; class Program { // Driver program static void Main(string[] args) { int[] arr = { 1, 3, 5, 7, 9 }; int n = arr.Length; int k = 2; leftRotate(arr, k, n); Console.ReadKey(); } // Function For The k Times Left Rotation static void leftRotate(int[] arr, int k, int n) { k%=n; Array.Reverse(arr, 0, k); Array.Reverse(arr, k, n - k); Array.Reverse(arr, 0, n); // Print the rotated array from start position for (int i = 0; i < n; i++) Console.Write(arr[i] + " "); Console.WriteLine(); } } // This code is contributed by Tapesh(tapeshdua420) // Function for k times left rotation function leftRotate(arr, k) { k%=arr.length // Reverse the first k elements arr = reverseArray(arr, 0, k - 1); // Reverse the remaining n-k elements arr = reverseArray(arr, k, arr.length - 1); // Reverse the entire array arr = reverseArray(arr, 0, arr.length - 1); // Print the rotated array from start position console.log(arr.join(" ")); } // Helper function to reverse a section of an array from start to end (inclusive) function reverseArray(arr, start, end) { while (start < end) { let temp = arr[start]; arr[start] = arr[end]; arr[end] = temp; start++; end--; } return arr; } // Driver code let arr = [1, 3, 5, 7, 9 ]; let n = arr.length; let k = 2; // Function Call leftRotate(arr, k, n); fun leftRotate(arr: IntArray, k: Int, n: Int) { // If k > n, k % n will bring k back in range val rotation = k % n // Reverse the first part of the array (0 to rotation-1) for (i in 0 until rotation / 2) { val temp = arr[i] arr[i] = arr[rotation - i - 1] arr[rotation - i - 1] = temp } // Reverse the second part of the array (rotation to n-1) for (i in 0 until (n - rotation) / 2) { val temp = arr[rotation + i] arr[rotation + i] = arr[n - i - 1] arr[n - i - 1] = temp } // Reverse the entire array for (i in 0 until n / 2) { val temp = arr[i] arr[i] = arr[n - i - 1] arr[n - i - 1] = temp } // Print the rotated array for (i in arr) { print("$i ") } println() } fun main() { val arr = intArrayOf(1, 3, 5, 7, 9) val n = arr.size val k = 2 // Function Call leftRotate(arr, k, n) } fun leftRotate(arr: IntArray, k: Int, n: Int) { // If k > n, k % n will bring k back in range val rotation = k % n // Reverse the first part of the array (0 to rotation-1) for (i in 0 until rotation / 2) { val temp = arr[i] arr[i] = arr[rotation - i - 1] arr[rotation - i - 1] = temp } // Reverse the second part of the array (rotation to n-1) for (i in 0 until (n - rotation) / 2) { val temp = arr[rotation + i] arr[rotation + i] = arr[n - i - 1] arr[n - i - 1] = temp } // Reverse the entire array for (i in 0 until n / 2) { val temp = arr[i] arr[i] = arr[n - i - 1] arr[n - i - 1] = temp } // Print the rotated array for (i in arr) { print("$i ") } println() } fun main() { val arr = intArrayOf(1, 3, 5, 7, 9) val n = arr.size val k = 2 // Function Call leftRotate(arr, k, n) }
Note: the array itself gets updated after the rotation.
Time Complexity: O(n)
Auxiliary Space: O(1), since no extra space has been taken.
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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
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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
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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
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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
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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,
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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
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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
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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
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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
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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]
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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
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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.
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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
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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
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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
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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
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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
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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
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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
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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
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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
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