Upper Bound

Given a sorted array arr[] and a number target, find the upper bound of the target in this given array. The upper bound of a number is defined as the smallest index in the sorted array where the element is greater than the given number.

Note: If all the elements in the given array are smaller than or equal to the target, the upper bound will be the length of the array.

Examples:

Input: arr[] = [2, 3, 7, 10, 11, 11, 25], target = 9
Output: 3
Explanation: 3 is the smallest index in arr[] at which element (arr[3] = 10) is larger than 9.

Input: arr[] = [2, 3, 7, 10, 11, 11, 25], target = 11
Output: 6
Explanation: 6 is the smallest index in arr[] at which element (arr[6] = 25) is larger than 11.

Input: arr[] = [2, 3, 7, 10, 11, 11, 25], target = 100
Output: 7
Explanation: As no element in arr[] is greater than 100, return the length of array.

[Naive Approach] Using Linear Search - O(n) Time and O(1) Space

The idea is to use linear search. We compare each element of the given array with the target and find the first index where the element is greater than target.

#include <iostream> #include <vector> using namespace std; int upperBound(vector<int> &arr, int target) {  int n = arr.size();   // Compare target with each element in array  for (int i = 0; i < n; ++i) {    // If larger value found, return its index  if(arr[i] > target) {  return i;  }  }    // If all elements are smaller, return length  return n; } int main() {  vector<int> arr = {2, 3, 7, 10, 11, 11, 25};  int target = 11;    cout << upperBound(arr, target);  return 0; } 

#include <stdio.h> int upperBound(int arr[], int n, int target) {  // Compare target with each element in array  for (int i = 0; i < n; ++i) {  // If larger value found, return its index  if (arr[i] > target) {  return i;  }  }  // If all elements are smaller, return length  return n; } int main() {  int arr[] = {2, 3, 7, 10, 11, 11, 25};  int target = 11;  int n = sizeof(arr) / sizeof(arr[0]);  printf("%d", upperBound(arr, n, target));  return 0; } 

class GfG {  // Function to find the upper bound of a number  static int upperBound(int[] arr, int target) {  int n = arr.length;  // Compare target with each element in array  for (int i = 0; i < n; ++i) {  // If larger value found, return its index  if (arr[i] > target) {  return i;  }  }  // If all elements are smaller, return length  return n;  }  public static void main(String[] args) {  int[] arr = {2, 3, 7, 10, 11, 11, 25};  int target = 11;  System.out.println(upperBound(arr, target));  } } 

def upperBound(arr, target): n = len(arr) # Compare target with each element in array for i in range(n): # If larger value found, return its index if arr[i] > target: return i # If all elements are smaller, return length return n if __name__ == "__main__": arr = [2, 3, 7, 10, 11, 11, 25] target = 11 print(upperBound(arr, target)) 

using System; class GfG {    static int upperBound(int[] arr, int target) {  int n = arr.Length;  // Compare target with each element in array  for (int i = 0; i < n; ++i) {  // If larger value found, return its index  if (arr[i] > target) {  return i;  }  }  // If all elements are smaller, return length  return n;  }  static void Main() {  int[] arr = {2, 3, 7, 10, 11, 11, 25};  int target = 11;  Console.WriteLine(upperBound(arr, target));  } } 

function upperBound(arr, target) {  let n = arr.length;  // Compare target with each element in array  for (let i = 0; i < n; ++i) {  // If larger value found, return its index  if (arr[i] > target) {  return i;  }  }  // If all elements are smaller, return length  return n; } // Driver Code let arr = [2, 3, 7, 10, 11, 11, 25]; let target = 11; console.log(upperBound(arr, target)); 

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[Expected Approach - 1] Using Binary Search - O(log n) Time and O(1) Space

The idea is to use the fact that the given array is sorted. We can apply binary search to find the index of the element just larger than the target.

Step-by-step implementation:

1. Set variables lo and hi to the starting and ending of array.
2. Find mid = (lo + hi) / 2 and compare arr[mid] with target
   => if arr[mid] <= target, then all elements in the range arr[lo...mid] will also be <= target, so update lo = mid+1.
   => if arr[mid] > target, then upper bound will lie in the range arr[lo...mid], so update result to mid and update hi = mid - 1.
3. Continue step 2 till lo <= hi.
4. Return result as the upper bound.

#include <iostream> #include <vector> using namespace std; int upperBound(vector<int> &arr, int target) {  int lo = 0, hi = arr.size() - 1;   int res = arr.size();    while(lo <= hi) {  int mid = lo + (hi - lo)/2;    // If arr[mid] > target, then arr[mid] can be  // the upper bound so store mid in result and  // search in left half, i.e. arr[lo...mid-1]  if(arr[mid] > target) {  res = mid;  hi = mid - 1;  }    // If arr[mid] <= target, then upper bound   // cannot lie in the range [lo...mid], so   // search in right half, i.e. arr[mid+1...hi]  else  lo = mid + 1;  }  return res; } int main() {  vector<int> arr = {2, 3, 7, 10, 11, 11, 25};  int target = 11;    cout << upperBound(arr, target);  return 0; } 

#include <stdio.h> int upperBound(int arr[], int n, int target) {  int lo = 0, hi = n - 1;  int res = n;  while (lo <= hi) {  int mid = lo + (hi - lo) / 2;  // If arr[mid] > target, then arr[mid] can be  // the upper bound so store mid in result and  // search in left half, i.e. arr[lo...mid-1]  if (arr[mid] > target) {  res = mid;  hi = mid - 1;  }  // If arr[mid] <= target, then upper bound   // cannot lie in the range [lo...mid], so   // search in right half, i.e. arr[mid+1...hi]  else {  lo = mid + 1;  }  }  return res; } int main() {  int arr[] = {2, 3, 7, 10, 11, 11, 25};  int target = 11;  int n = sizeof(arr) / sizeof(arr[0]);  printf("%d", upperBound(arr, n, target));  return 0; } 

class GfG {  static int upperBound(int[] arr, int target) {  int lo = 0, hi = arr.length - 1;  int res = arr.length;  while (lo <= hi) {  int mid = lo + (hi - lo) / 2;  // If arr[mid] > target, then arr[mid] can be  // the upper bound so store mid in result and  // search in left half, i.e. arr[lo...mid-1]  if (arr[mid] > target) {  res = mid;  hi = mid - 1;  }  // If arr[mid] <= target, then upper bound   // cannot lie in the range [lo...mid], so   // search in right half, i.e. arr[mid+1...hi]  else {  lo = mid + 1;  }  }  return res;  }  public static void main(String[] args) {  int[] arr = {2, 3, 7, 10, 11, 11, 25};  int target = 11;  System.out.println(upperBound(arr, target));  } } 

def upperBound(arr, target): lo, hi = 0, len(arr) - 1 res = len(arr) while lo <= hi: mid = lo + (hi - lo) // 2 # If arr[mid] > target, then arr[mid] can be # the upper bound so store mid in result and # search in left half, i.e. arr[lo...mid-1] if arr[mid] > target: res = mid hi = mid - 1 # If arr[mid] <= target, then upper bound  # cannot lie in the range [lo...mid], so  # search in right half, i.e. arr[mid+1...hi] else: lo = mid + 1 return res if __name__ == "__main__": arr = [2, 3, 7, 10, 11, 11, 25] target = 11 print(upperBound(arr, target)) 

using System; class GfG {  static int UpperBound(int[] arr, int target) {  int lo = 0, hi = arr.Length - 1;  int res = arr.Length;    while (lo <= hi) {  int mid = lo + (hi - lo) / 2;    // If arr[mid] > target, then arr[mid] can be  // the upper bound so store mid in result and  // search in left half, i.e. arr[lo...mid-1]  if (arr[mid] > target) {  res = mid;  hi = mid - 1;  }    // If arr[mid] <= target, then upper bound   // cannot lie in the range [lo...mid], so   // search in right half, i.e. arr[mid+1...hi]  else {  lo = mid + 1;  }  }    return res;  }  static void Main(string[] args) {  int[] arr = { 2, 3, 7, 10, 11, 11, 25 };  int target = 11;    Console.WriteLine(UpperBound(arr, target));  } } 

function upperBound(arr, target) {  let lo = 0, hi = arr.length - 1;  let res = arr.length;  while (lo <= hi) {  let mid = lo + Math.floor((hi - lo) / 2);  // If arr[mid] > target, then arr[mid] can be  // the upper bound so store mid in result and  // search in left half, i.e. arr[lo...mid-1]  if (arr[mid] > target) {  res = mid;  hi = mid - 1;  }    // If arr[mid] < target, then upper bound   // cannot lie in the range [lo...mid], so   // search in right half, i.e. arr[mid+1...hi]  else {  lo = mid + 1;  }  }  return res; } // Driver Code let arr = [2, 3, 7, 10, 11, 11, 25]; let target = 11; console.log(upperBound(arr, target)); 

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[Expected Approach - 2] Using Built-In Methods - O(log n) Time and O(1) Space

We can use built-in functions to find the Upper bound of an element in a sorted array efficiently.

• C++: std::upper_bound(v.begin(), v.end(), x) (from <algorithm>)
• Python: bisect.bisect_right(arr, x) (from bisect module)

For more details see upper_bound in c++, bisect in python

#include <iostream> #include <vector> #include <algorithm> using namespace std; int upperBound(vector<int> &arr, int target) {    // Using inbuilt method  int index = upper_bound(arr.begin(), arr.end(),   target) - arr.begin();    return index; } int main() {  vector<int> arr = {2, 3, 7, 10, 11, 11, 25};  int target = 11;    cout << upperBound(arr, target);  return 0; } 

import bisect def upperBound(arr, target): # Using bisect_right to get the upper bound index index = bisect.bisect_right(arr, target) return index if __name__ == "__main__": arr = [2, 3, 7, 10, 11, 11, 25] target = 11 print(upperBound(arr, target)) 

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• Java Equivalent of C++’s upper_bound() Method