Longest subarray having maximum sum

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}

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 

Output:

Length of longest subarray having maximum sum = 4

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

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

Recommended Posts: