Longest subarray having maximum sum

Last Updated : 15 Oct, 2023

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

// 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

# 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#

// 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

Javascript

<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>

Output

Length 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;
}

Java

//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);
    }
}

Python3

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}")

C#

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);
    }
}

Javascript

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);

Output

Length of the subarray with maximum sum = 4

Time Complexity: O(n)
Auxiliary Space: O(1)

Reverse a doubly circular linked list

ayushjauhari14

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Longest subarray having maximum sum

C++

Java

Python3

C#

Javascript

Approach: Kadane’s algorithm

C++

Java

Python3

C#

Javascript

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