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}

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.
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// 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;
}
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// 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
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# 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

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// 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
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Output:
Length of longest subarray having maximum sum = 4

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




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