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 subarrayint 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 kint 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 sumint lenLongSubarrWithMaxSum(int arr[], int n){ int maxSum = maxSubArraySum(arr, n); return lenOfLongSubarrWithGivenSum(arr, n, maxSum);}// Driver program to test aboveint 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 Codepublic 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 subarraydef 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 kdef 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 Sumdef lenLongSubarrWithMaxSum(arr, n): maxSum = maxSubArraySum(arr, n) return lenOfLongSubarrWithGivenSum(arr, n, maxSum)# Driver Codearr = [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 subarrayfunction 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 kfunction 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 sumfunction lenLongSubarrWithMaxSum(arr, n){ var maxSum = maxSubArraySum(arr, n); return lenOfLongSubarrWithGivenSum(arr, n, maxSum);}// Driver program to test abovevar 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 codeint 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 approachpublic 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 codeif __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 codeconst 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)
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