Longest Subarray having sum of elements atmost ‘k’

• Difficulty Level : Medium
• Last Updated : 23 Apr, 2021

Given an array of integers, our goal is to find the length of the largest subarray having the sum of its elements at most ‘k’ where k>0.

Examples:

Input : arr[] = {1, 2, 1, 0, 1, 1, 0}, k = 4
Output : 5
Explanation:
{1, 2, 1} => sum = 4, length = 3
{1, 2, 1, 0}, {2, 1, 0, 1} => sum = 4, length = 4
{1, 0, 1, 1, 0} =>5 sum = 3, length = 5

Method 1 (Brute Force)
Find all the subarrays whose sum is less than or equal to k and return the one with the largest length.
Time Complexity : O(n^2)
Method 2 (Efficient):
An efficient approach is to use the sliding window technique

1. Traverse the array and check if on adding the current element its sum is less than or equal to k.
2. If it’s less than k then add it to the sum and increase the count.
3. Keep track of Maximum count.

C++

 // A C++ program to find longest subarray with// sum of elements at-least k.#include using namespace std; // function to find the length of largest subarray// having sum atmost k.int atMostSum(int arr[], int n, int k){    int sum = 0;    int cnt = 0, maxcnt = 0;     for (int i = 0; i < n; i++) {                 // If adding current element doesn't        // cross limit add it to current window        if ((sum + arr[i]) <= k) {            sum += arr[i];            cnt++;        }         // Else, remove first element of current        // window and add the current element        else if(sum!=0)        {            sum = sum - arr[i - cnt] + arr[i];        }         // keep track of max length.        maxcnt = max(cnt, maxcnt);    }    return maxcnt;} // Driver functionint main(){    int arr[] = {1, 2, 1, 0, 1, 1, 0};    int n = sizeof(arr) / sizeof(arr);    int k = 4;     cout << atMostSum(arr, n, k);    return 0;}

Java

 // Java program to find longest subarray with// sum of elements at-least k.import java.util.*; class GFG {         // function to find the length of largest    // subarray having sum atmost k.    public static int atMostSum(int arr[], int n,                                        int k)    {        int sum = 0;        int cnt = 0, maxcnt = 0;             for (int i = 0; i < n; i++) {                         // If adding current element doesn't            // cross limit add it to current window            if ((sum + arr[i]) <= k) {                sum += arr[i];                cnt++;            }                 // Else, remove first element of current            // window.            else if(sum!=0)           {            sum = sum - arr[i - cnt] + arr[i];           }                 // keep track of max length.            maxcnt = Math.max(cnt, maxcnt);        }        return maxcnt;    }         /* Driver program to test above function */    public static void main(String[] args)    {        int arr[] = { 1, 2, 1, 0, 1, 1, 0 };        int n = arr.length;        int k = 4;             System.out.print(atMostSum(arr, n, k));                 }}// This code is contributed by Arnav Kr. Mandal.

Python3

 # Python3 program to find longest subarray# with sum of elements at-least k. # function to find the length of largest# subarray having sum atmost k.def atMostSum(arr, n, k):    _sum = 0    cnt = 0    maxcnt = 0         for i in range(n):         # If adding current element doesn't        # Cross limit add it to current window        if ((_sum + arr[i]) <= k):            _sum += arr[i]            cnt += 1                 # Else, remove first element of current        # window and add the current element        elif(sum != 0):            _sum = _sum - arr[i - cnt] + arr[i]                 # keep track of max length.        maxcnt = max(cnt, maxcnt)     return maxcnt     # Driver functionarr = [1, 2, 1, 0, 1, 1, 0]n = len(arr)k = 4print(atMostSum(arr, n, k)) # This code is contributed by "Abhishek Sharma 44"

C#

 // C# program to find longest subarray// with sum of elements at-least k.using System; class GFG {         // function to find the length of largest    // subarray having sum atmost k.    public static int atMostSum(int []arr, int n,                                           int k)    {        int sum = 0;        int cnt = 0, maxcnt = 0;             for (int i = 0; i < n; i++) {                         // If adding current element doesn't            // cross limit add it to current window            if ((sum + arr[i]) <= k) {                sum += arr[i];                cnt++;            }                 // Else, remove first element            // of current window.            else if(sum!=0)            {                sum = sum - arr[i - cnt] + arr[i];            }                 // keep track of max length.            maxcnt = Math.Max(cnt, maxcnt);        }        return maxcnt;    }         // Driver Code    public static void Main()    {        int []arr = {1, 2, 1, 0, 1, 1, 0};        int n = arr.Length;        int k = 4;             Console.Write(atMostSum(arr, n, k));                 }} // This code is contributed by Nitin Mittal



Javascript



Output:

5

Time Complexity : O(n)
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