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Longest subarray having maximum sum

  • Difficulty Level : Hard
  • Last Updated : 21 May, 2021

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}

 

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Approach: Following are the steps:
 



  1. Find the maximum sum contiguous subarray. Let this sum be maxSum.
  2. 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).
 




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