Longest subarray having maximum sum
Last Updated :
11 Jul, 2025
Given an array arr[] containing n integers. The problem is to find the length of the subarray having maximum sum. If there exists two or more subarrays with maximum sum then print the length of the longest subarray.
Examples:
Input : arr[] = {5, -2, -1, 3, -4}
Output : 4
There are two subarrays with maximum sum:
First is {5}
Second is {5, -2, -1, 3}
Therefore longest one is of length 4.
Input : arr[] = {-2, -3, 4, -1, -2, 1, 5, -3}
Output : 5
The subarray is {4, -1, -2, 1, 5}
Approach: Following are the steps
- Find the maximum sum contiguous subarray. Let this sum be maxSum.
- Find the length of the longest subarray having sum equal to maxSum. Refer this post.
C++
// C++ implementation to find the length of the longest
// subarray having maximum sum
#include <bits/stdc++.h>
using namespace std;
// function to find the maximum sum that
// exists in a subarray
int maxSubArraySum(int arr[], int size)
{
int max_so_far = arr[0];
int curr_max = arr[0];
for (int i = 1; i < size; i++) {
curr_max = max(arr[i], curr_max + arr[i]);
max_so_far = max(max_so_far, curr_max);
}
return max_so_far;
}
// function to find the length of longest
// subarray having sum k
int lenOfLongSubarrWithGivenSum(int arr[], int n, int k)
{
// unordered_map 'um' implemented
// as hash table
unordered_map<int, int> um;
int sum = 0, maxLen = 0;
// traverse the given array
for (int i = 0; i < n; i++) {
// accumulate sum
sum += arr[i];
// when subarray starts from index '0'
if (sum == k)
maxLen = i + 1;
// make an entry for 'sum' if it is
// not present in 'um'
if (um.find(sum) == um.end())
um[sum] = i;
// check if 'sum-k' is present in 'um'
// or not
if (um.find(sum - k) != um.end()) {
// update maxLength
if (maxLen < (i - um[sum - k]))
maxLen = i - um[sum - k];
}
}
// required maximum length
return maxLen;
}
// function to find the length of the longest
// subarray having maximum sum
int lenLongSubarrWithMaxSum(int arr[], int n)
{
int maxSum = maxSubArraySum(arr, n);
return lenOfLongSubarrWithGivenSum(arr, n, maxSum);
}
// Driver program to test above
int main()
{
int arr[] = { 5, -2, -1, 3, -4 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << "Length of longest subarray having maximum sum = "
<< lenLongSubarrWithMaxSum(arr, n);
return 0;
}
// Java implementation to find
// the length of the longest
// subarray having maximum sum
import java.util.*;
class GFG
{
// function to find the
// maximum sum that
// exists in a subarray
static int maxSubArraySum(int arr[],
int size)
{
int max_so_far = arr[0];
int curr_max = arr[0];
for (int i = 1; i < size; i++)
{
curr_max = Math.max(arr[i],
curr_max + arr[i]);
max_so_far = Math.max(max_so_far,
curr_max);
}
return max_so_far;
}
// function to find the
// length of longest
// subarray having sum k
static int lenOfLongSubarrWithGivenSum(int arr[],
int n, int k)
{
// unordered_map 'um' implemented
// as hash table
HashMap<Integer,
Integer> um = new HashMap<Integer,
Integer>();
int sum = 0, maxLen = 0;
// traverse the given array
for (int i = 0; i < n; i++)
{
// accumulate sum
sum += arr[i];
// when subarray starts
// from index '0'
if (sum == k)
maxLen = i + 1;
// make an entry for 'sum' if
// it is not present in 'um'
if (um.containsKey(sum))
um.put(sum, i);
// check if 'sum-k' is present
// in 'um' or not
if (um.containsKey(sum - k))
{
// update maxLength
if (maxLen < (i - um.get(sum - k)))
maxLen = i - um.get(sum - k);
}
}
// required maximum length
return maxLen;
}
// function to find the length
// of the longest subarray
// having maximum sum
static int lenLongSubarrWithMaxSum(int arr[], int n)
{
int maxSum = maxSubArraySum(arr, n);
return lenOfLongSubarrWithGivenSum(arr, n, maxSum);
}
// Driver Code
public static void main(String args[])
{
int arr[] = { 5, -2, -1, 3, -4 };
int n = arr.length;
System.out.println("Length of longest subarray " +
"having maximum sum = " +
lenLongSubarrWithMaxSum(arr, n));
}
}
// This code is contributed by Arnab Kundu
# Python3 implementation to find the length
# of the longest subarray having maximum sum
# function to find the maximum sum that
# exists in a subarray
def maxSubArraySum(arr, size):
max_so_far = arr[0]
curr_max = arr[0]
for i in range(1,size):
curr_max = max(arr[i], curr_max + arr[i])
max_so_far = max(max_so_far, curr_max)
return max_so_far
# function to find the length of longest
# subarray having sum k
def lenOfLongSubarrWithGivenSum(arr, n, k):
# unordered_map 'um' implemented
# as hash table
um = dict()
Sum, maxLen = 0, 0
# traverse the given array
for i in range(n):
# accumulate Sum
Sum += arr[i]
# when subarray starts from index '0'
if (Sum == k):
maxLen = i + 1
# make an entry for 'Sum' if it is
# not present in 'um'
if (Sum not in um.keys()):
um[Sum] = i
# check if 'Sum-k' is present in 'um'
# or not
if (Sum in um.keys()):
# update maxLength
if ((Sum - k) in um.keys() and
maxLen < (i - um[Sum - k])):
maxLen = i - um[Sum - k]
# required maximum length
return maxLen
# function to find the length of the longest
# subarray having maximum Sum
def lenLongSubarrWithMaxSum(arr, n):
maxSum = maxSubArraySum(arr, n)
return lenOfLongSubarrWithGivenSum(arr, n, maxSum)
# Driver Code
arr = [5, -2, -1, 3, -4]
n = len(arr)
print("Length of longest subarray having maximum sum = ",
lenLongSubarrWithMaxSum(arr, n))
# This code is contributed by mohit kumar
// C# implementation to find
// the length of the longest
// subarray having maximum sum
using System;
using System.Collections.Generic;
public class GFG{
// function to find the
// maximum sum that
// exists in a subarray
static int maxSubArraySum(int []arr,
int size)
{
int max_so_far = arr[0];
int curr_max = arr[0];
for (int i = 1; i < size; i++)
{
curr_max = Math.Max(arr[i],
curr_max + arr[i]);
max_so_far = Math.Max(max_so_far,
curr_max);
}
return max_so_far;
}
// function to find the
// length of longest
// subarray having sum k
static int lenOfLongSubarrWithGivenSum(int []arr,
int n, int k)
{
// unordered_map 'um' implemented
// as hash table
Dictionary<int,
int> um = new Dictionary<int,
int>();
int sum = 0, maxLen = 0;
// traverse the given array
for (int i = 0; i < n; i++)
{
// accumulate sum
sum += arr[i];
// when subarray starts
// from index '0'
if (sum == k)
maxLen = i + 1;
// make an entry for 'sum' if
// it is not present in 'um'
if (um.ContainsKey(sum))
um.Add(sum, i);
// check if 'sum-k' is present
// in 'um' or not
if (um.ContainsKey(sum - k))
{
// update maxLength
if (maxLen < (i - um[sum - k]))
maxLen = i - um[sum - k];
}
}
// required maximum length
return maxLen;
}
// function to find the length
// of the longest subarray
// having maximum sum
static int lenLongSubarrWithMaxSum(int []arr, int n)
{
int maxSum = maxSubArraySum(arr, n);
return lenOfLongSubarrWithGivenSum(arr, n, maxSum);
}
// Driver Code
public static void Main()
{
int []arr = { 5, -2, -1, 3, -4 };
int n = arr.Length;
Console.WriteLine("Length of longest subarray " +
"having maximum sum = " +
lenLongSubarrWithMaxSum(arr, n));
}
}
// This code is contributed by Rajput-Ji
<script>
// Javascript implementation to find the length of the longest
// subarray having maximum sum
// function to find the maximum sum that
// exists in a subarray
function maxSubArraySum(arr, size)
{
var max_so_far = arr[0];
var curr_max = arr[0];
for (var i = 1; i < size; i++) {
curr_max = Math.max(arr[i], curr_max + arr[i]);
max_so_far = Math.max(max_so_far, curr_max);
}
return max_so_far;
}
// function to find the length of longest
// subarray having sum k
function lenOfLongSubarrWithGivenSum( arr, n, k)
{
// unordered_map 'um' implemented
// as hash table
var um = new Map();
var sum = 0, maxLen = 0;
// traverse the given array
for (var i = 0; i < n; i++) {
// accumulate sum
sum += arr[i];
// when subarray starts from index '0'
if (sum == k)
maxLen = i + 1;
// make an entry for 'sum' if it is
// not present in 'um'
if (!um.has(sum))
um.set(sum, i);
// check if 'sum-k' is present in 'um'
// or not
if (um.has(sum - k)) {
// update maxLength
if (maxLen < (i - um.get(sum-k)))
maxLen = i - um.get(sum-k)
}
}
// required maximum length
return maxLen;
}
// function to find the length of the longest
// subarray having maximum sum
function lenLongSubarrWithMaxSum(arr, n)
{
var maxSum = maxSubArraySum(arr, n);
return lenOfLongSubarrWithGivenSum(arr, n, maxSum);
}
// Driver program to test above
var arr = [5, -2, -1, 3, -4];
var n = arr.length;
document.write( "Length of longest subarray having maximum sum = "
+ lenLongSubarrWithMaxSum(arr, n));
// This code is contributed by rrrtnx.
</script>
OutputLength of longest subarray having maximum sum = 4
Time Complexity: O(n).
Auxiliary Space: O(n).
Approach: Kadane's algorithm
Following are the steps:
- traverse the array.
- Keep track of the maximum sum subarray ending at each index.
- Then, we return the length of the subarray with the maximum sum among all the subarrays.
Below is the implementation:
C++
// C++ program to find the length of the subarray having maximum sum
#include <iostream>
#include <algorithm>
using namespace std;
int maxSubArrayLen(int arr[], int n) {
// Initializing the variables
int max_so_far = arr[0];
int max_ending_here = arr[0];
int max_len = 1;
int curr_len = 1;
// Traversing through the array to find the subarray
for (int i = 1; i < n; i++) {
if (max_ending_here < 0) {
max_ending_here = arr[i];
curr_len = 1;
} else {
max_ending_here += arr[i];
curr_len++;
}
if (max_ending_here > max_so_far) {
max_so_far = max_ending_here;
max_len = curr_len;
} else if (max_ending_here == max_so_far) {
max_len = max(max_len, curr_len);
}
}
return max_len;
}
// Driver code
int main() {
// Input array
int arr[] = {5, -2, -1, 3, -4};
int n = sizeof(arr) / sizeof(arr[0]);
int max_len = maxSubArrayLen(arr, n);
cout << "Length of the subarray with maximum sum = " << max_len << endl;
return 0;
}
//Code in java for above approach
public class Main {
public static int maxSubArrayLen(int[] arr, int n) {
// Initializing the variables
int max_so_far = arr[0];
int max_ending_here = arr[0];
int max_len = 1;
int curr_len = 1;
// Traversing through the array to find the subarray
for (int i = 1; i < n; i++) {
if (max_ending_here < 0) {
max_ending_here = arr[i];
curr_len = 1;
} else {
max_ending_here += arr[i];
curr_len++;
}
if (max_ending_here > max_so_far) {
max_so_far = max_ending_here;
max_len = curr_len;
} else if (max_ending_here == max_so_far) {
max_len = Math.max(max_len, curr_len);
}
}
return max_len;
}
public static void main(String[] args) {
// Input array
int[] arr = {5, -2, -1, 3, -4};
int n = arr.length;
int max_len = maxSubArrayLen(arr, n);
System.out.println("Length of the subarray with maximum sum = " + max_len);
}
}
def maxSubArrayLen(arr):
# Initializing the variables
max_so_far = arr[0]
max_ending_here = arr[0]
max_len = 1
curr_len = 1
# Traversing through the array to find the subarray
for i in range(1, len(arr)):
if max_ending_here < 0:
max_ending_here = arr[i]
curr_len = 1
else:
max_ending_here += arr[i]
curr_len += 1
if max_ending_here > max_so_far:
max_so_far = max_ending_here
max_len = curr_len
elif max_ending_here == max_so_far:
max_len = max(max_len, curr_len)
return max_len
# Driver code
if __name__ == '__main__':
# Input array
arr = [5, -2, -1, 3, -4]
max_len = maxSubArrayLen(arr)
print(f"Length of the subarray with maximum sum = {max_len}")
using System;
public class GFG
{
public static int MaxSubArrayLen(int[] arr, int n)
{
// Initializing the variables
int maxSoFar = arr[0];
int maxEndingHere = arr[0];
int maxLen = 1;
int currLen = 1;
// Traversing through the array to find the subarray
for (int i = 1; i < n; i++)
{
if (maxEndingHere < 0)
{
maxEndingHere = arr[i];
currLen = 1;
}
else
{
maxEndingHere += arr[i];
currLen++;
}
if (maxEndingHere > maxSoFar)
{
maxSoFar = maxEndingHere;
maxLen = currLen;
}
else if (maxEndingHere == maxSoFar)
{
maxLen = Math.Max(maxLen, currLen);
}
}
return maxLen;
}
// Driver code
public static void Main(string[] args)
{
// Input array
int[] arr = { 5, -2, -1, 3, -4 };
int n = arr.Length;
int maxLen = MaxSubArrayLen(arr, n);
Console.WriteLine("Length of the subarray with maximum sum = " + maxLen);
}
}
function maxSubArrayLen(arr) {
// Initializing the variables
let max_so_far = arr[0];
let max_ending_here = arr[0];
let max_len = 1;
let curr_len = 1;
// Traversing through the array to find the subarray
for (let i = 1; i < arr.length; i++) {
if (max_ending_here < 0) {
max_ending_here = arr[i];
curr_len = 1;
} else {
max_ending_here += arr[i];
curr_len++;
}
if (max_ending_here > max_so_far) {
max_so_far = max_ending_here;
max_len = curr_len;
} else if (max_ending_here === max_so_far) {
max_len = Math.max(max_len, curr_len);
}
}
return max_len;
}
// Driver code
const arr = [5, -2, -1, 3, -4];
const max_len = maxSubArrayLen(arr);
console.log("Length of the subarray with maximum sum = " + max_len);
OutputLength of the subarray with maximum sum = 4
Time Complexity: O(n)
Auxiliary Space: O(1)
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