Largest Sum Contiguous Subarray (Kadane’s Algorithm)

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Given an array arr[] of size N. The task is to find the sum of the contiguous subarray within a arr[] with the largest sum.

kadane-algorithm

The idea of Kadane’s algorithm is to maintain a variable max_ending_here that stores the maximum sum contiguous subarray ending at current index and a variable max_so_far stores the maximum sum of contiguous subarray found so far, Everytime there is a positive-sum value in max_ending_here compare it with max_so_far and update max_so_far if it is greater than max_so_far.

So the main Intuition behind Kadane’s Algorithm is,

The subarray with negative sum is discarded (by assigning max_ending_here = 0 in code).
We carry subarray till it gives positive sum.

Pseudocode of Kadane’s algorithm:

Initialize:
max_so_far = INT_MIN
max_ending_here = 0

Loop for each element of the array

(a) max_ending_here = max_ending_here + a[i]
(b) if(max_so_far < max_ending_here)
max_so_far = max_ending_here
(c) if(max_ending_here < 0)
max_ending_here = 0
return max_so_far

Illustration of Kadane’s Algorithm:

Lets take the example: {-2, -3, 4, -1, -2, 1, 5, -3}

Note: in the image max_so_far is represented by Max_Sum and max_ending_here by Curr_Sum

For i=0, a[0] = -2

max_ending_here = max_ending_here + (-2)

Set max_ending_here = 0 because max_ending_here < 0

and set max_so_far = -2

For i=1, a[1] = -3

max_ending_here = max_ending_here + (-3)

Since max_ending_here = -3 and max_so_far = -2, max_so_far will remain -2

Set max_ending_here = 0 because max_ending_here < 0

For i=2, a[2] = 4

max_ending_here = max_ending_here + (4)

max_ending_here = 4

max_so_far is updated to 4 because max_ending_here greater than max_so_far which was -2 till now

For i=3, a[3] = -1

max_ending_here = max_ending_here + (-1)

max_ending_here = 3

For i=4, a[4] = -2

max_ending_here = max_ending_here + (-2)

max_ending_here = 1

For i=5, a[5] = 1

max_ending_here = max_ending_here + (1)

max_ending_here = 2

For i=6, a[6] = 5

max_ending_here = max_ending_here + (5)

max_ending_here =

max_so_far is updated to 7 because max_ending_here is greater than max_so_far

For i=7, a[7] = -3

max_ending_here = max_ending_here + (-3)

max_ending_here = 4

Follow the below steps to Implement the idea:

Initialize the variables max_so_far = INT_MIN and max_ending_here = 0
Run a for loop from 0 to N-1 and for each index i:
- Add the arr[i] to max_ending_here.
- If max_so_far is less than max_ending_here then update max_so_far to max_ending_here.
- If max_ending_here < 0 then update max_ending_here = 0
Return max_so_far

Below is the Implementation of the above approach.

C++

// C++ program to print largest contiguous array sum
#include <bits/stdc++.h>
using namespace std;
 
int maxSubArraySum(int a[], int size)
{
    int max_so_far = INT_MIN, max_ending_here = 0;
 
    for (int i = 0; i < size; i++) {
        max_ending_here = max_ending_here + a[i];
        if (max_so_far < max_ending_here)
            max_so_far = max_ending_here;
 
        if (max_ending_here < 0)
            max_ending_here = 0;
    }
    return max_so_far;
}
 
// Driver Code
int main()
{
    int a[] = { -2, -3, 4, -1, -2, 1, 5, -3 };
    int n = sizeof(a) / sizeof(a[0]);
 
    // Function Call
    int max_sum = maxSubArraySum(a, n);
    cout << "Maximum contiguous sum is " << max_sum;
    return 0;
}

Java

// Java program to print largest contiguous array sum
import java.io.*;
import java.util.*;
 
class Kadane {
    // Driver Code
    public static void main(String[] args)
    {
        int[] a = { -2, -3, 4, -1, -2, 1, 5, -3 };
        System.out.println("Maximum contiguous sum is "
                           + maxSubArraySum(a));
    }
 
    // Function Call
    static int maxSubArraySum(int a[])
    {
        int size = a.length;
        int max_so_far = Integer.MIN_VALUE, max_ending_here
                                            = 0;
 
        for (int i = 0; i < size; i++) {
            max_ending_here = max_ending_here + a[i];
            if (max_so_far < max_ending_here)
                max_so_far = max_ending_here;
            if (max_ending_here < 0)
                max_ending_here = 0;
        }
        return max_so_far;
    }
}

Python

# Python program to find maximum contiguous subarray
 
# Function to find the maximum contiguous subarray
from sys import maxint
 
 
def maxSubArraySum(a, size):
 
    max_so_far = -maxint - 1
    max_ending_here = 0
 
    for i in range(0, size):
        max_ending_here = max_ending_here + a[i]
        if (max_so_far < max_ending_here):
            max_so_far = max_ending_here
 
        if max_ending_here < 0:
            max_ending_here = 0
    return max_so_far
 
# Driver function to check the above function
 
 
a = [-2, -3, 4, -1, -2, 1, 5, -3]
 
print "Maximum contiguous sum is", maxSubArraySum(a, len(a))
 
# This code is contributed by _Devesh Agrawal_

C#

// C# program to print largest
// contiguous array sum
using System;
 
class GFG {
    static int maxSubArraySum(int[] a)
    {
        int size = a.Length;
        int max_so_far = int.MinValue, max_ending_here = 0;
 
        for (int i = 0; i < size; i++) {
            max_ending_here = max_ending_here + a[i];
 
            if (max_so_far < max_ending_here)
                max_so_far = max_ending_here;
 
            if (max_ending_here < 0)
                max_ending_here = 0;
        }
 
        return max_so_far;
    }
 
    // Driver code
    public static void Main()
    {
        int[] a = { -2, -3, 4, -1, -2, 1, 5, -3 };
        Console.Write("Maximum contiguous sum is "
                      + maxSubArraySum(a));
    }
}
 
// This code is contributed by Sam007_

Javascript

<script>
 
// JavaScript program to find maximum 
// contiguous subarray
  
// Function to find the maximum 
// contiguous subarray
function maxSubArraySum(a, size)
{
    var maxint = Math.pow(2, 53)
    var max_so_far = -maxint - 1
    var max_ending_here = 0
      
    for (var i = 0; i < size; i++)
    {
        max_ending_here = max_ending_here + a[i]
        if (max_so_far < max_ending_here)
            max_so_far = max_ending_here
 
        if (max_ending_here < 0)
            max_ending_here = 0 
    }
    return max_so_far
}
  
// Driver code
var a = [ -2, -3, 4, -1, -2, 1, 5, -3 ]
document.write("Maximum contiguous sum is", 
               maxSubArraySum(a, a.length))
  
// This code is contributed by AnkThon
 
</script>

PHP

<?php
// PHP program to print largest
// contiguous array sum
 
function maxSubArraySum($a, $size)
{
    $max_so_far = PHP_INT_MIN; 
    $max_ending_here = 0;
 
    for ($i = 0; $i < $size; $i++)
    {
        $max_ending_here = $max_ending_here + $a[$i];
        if ($max_so_far < $max_ending_here)
            $max_so_far = $max_ending_here;
 
        if ($max_ending_here < 0)
            $max_ending_here = 0;
    }
    return $max_so_far;
}
 
// Driver code
$a = array(-2, -3, 4, -1,
           -2, 1, 5, -3);
$n = count($a);
$max_sum = maxSubArraySum($a, $n);
echo "Maximum contiguous sum is " , 
                          $max_sum;
 
// This code is contributed by anuj_67.
?>

Output

Maximum contiguous sum is 7

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

Print the Largest Sum Contiguous Subarray:

To print the subarray with the maximum sum the idea is to maintain start index of maximum_sum_ending_here at current index so that whenever maximum_sum_so_far is updated with maximum_sum_ending_here then start index and end index of subarray can be updated with start and current index.

Follow the below steps to implement the idea:

Initialize the variables s, start, and end with 0 and max_so_far = INT_MIN and max_ending_here = 0
Run a for loop from 0 to N-1 and for each index i:
- Add the arr[i] to max_ending_here.
- If max_so_far is less than max_ending_here then update max_so_far to max_ending_here and update start to s and end to i .
- If max_ending_here < 0 then update max_ending_here = 0 and s with i+1.
Print values from index start to end.

Below is the Implementation of above approach:

C++

// C++ program to print largest contiguous array sum
 
#include <climits>
#include <iostream>
using namespace std;
 
void maxSubArraySum(int a[], int size)
{
    int max_so_far = INT_MIN, max_ending_here = 0,
        start = 0, end = 0, s = 0;
 
    for (int i = 0; i < size; i++) {
        max_ending_here += a[i];
 
        if (max_so_far < max_ending_here) {
            max_so_far = max_ending_here;
            start = s;
            end = i;
        }
 
        if (max_ending_here < 0) {
            max_ending_here = 0;
            s = i + 1;
        }
    }
    cout << "Maximum contiguous sum is " << max_so_far
         << endl;
    cout << "Starting index " << start << endl
         << "Ending index " << end << endl;
}
 
/*Driver program to test maxSubArraySum*/
int main()
{
    int a[] = { -2, -3, 4, -1, -2, 1, 5, -3 };
    int n = sizeof(a) / sizeof(a[0]);
    maxSubArraySum(a, n);
    return 0;
}

Java

// Java program to print largest
// contiguous array sum
import java.io.*;
import java.util.*;
class GFG {
 
    static void maxSubArraySum(int a[], int size)
    {
        int max_so_far = Integer.MIN_VALUE,
            max_ending_here = 0, start = 0, end = 0, s = 0;
 
        for (int i = 0; i < size; i++) {
            max_ending_here += a[i];
 
            if (max_so_far < max_ending_here) {
                max_so_far = max_ending_here;
                start = s;
                end = i;
            }
 
            if (max_ending_here < 0) {
                max_ending_here = 0;
                s = i + 1;
            }
        }
        System.out.println("Maximum contiguous sum is "
                           + max_so_far);
        System.out.println("Starting index " + start);
        System.out.println("Ending index " + end);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int a[] = { -2, -3, 4, -1, -2, 1, 5, -3 };
        int n = a.length;
        maxSubArraySum(a, n);
    }
}
 
// This code is contributed by  prerna saini

Python3

# Python program to print largest contiguous array sum
 
from sys import maxsize
 
# Function to find the maximum contiguous subarray
# and print its starting and end index
 
 
def maxSubArraySum(a, size):
 
    max_so_far = -maxsize - 1
    max_ending_here = 0
    start = 0
    end = 0
    s = 0
 
    for i in range(0, size):
 
        max_ending_here += a[i]
 
        if max_so_far < max_ending_here:
            max_so_far = max_ending_here
            start = s
            end = i
 
        if max_ending_here < 0:
            max_ending_here = 0
            s = i+1
 
    print("Maximum contiguous sum is %d" % (max_so_far))
    print("Starting Index %d" % (start))
    print("Ending Index %d" % (end))
 
 
# Driver program to test maxSubArraySum
a = [-2, -3, 4, -1, -2, 1, 5, -3]
maxSubArraySum(a, len(a))

C#

// C# program to print largest
// contiguous array sum
using System;
 
class GFG {
    static void maxSubArraySum(int[] a, int size)
    {
        int max_so_far = int.MinValue, max_ending_here = 0,
            start = 0, end = 0, s = 0;
 
        for (int i = 0; i < size; i++) {
            max_ending_here += a[i];
 
            if (max_so_far < max_ending_here) {
                max_so_far = max_ending_here;
                start = s;
                end = i;
            }
 
            if (max_ending_here < 0) {
                max_ending_here = 0;
                s = i + 1;
            }
        }
        Console.WriteLine("Maximum contiguous "
                          + "sum is " + max_so_far);
        Console.WriteLine("Starting index " + start);
        Console.WriteLine("Ending index " + end);
    }
 
    // Driver code
    public static void Main()
    {
        int[] a = { -2, -3, 4, -1, -2, 1, 5, -3 };
        int n = a.Length;
        maxSubArraySum(a, n);
    }
}
 
// This code is contributed
// by anuj_67.

Javascript

<script>
// javascript program to print largest 
// contiguous array sum    
function maxSubArraySum(a , size) {
        var max_so_far = Number.MIN_SAFE_INTEGER, max_ending_here = 0, start = 0, end = 0, s = 0;
 
        for (i = 0; i < size; i++) {
            max_ending_here += a[i];
 
            if (max_so_far < max_ending_here) {
                max_so_far = max_ending_here;
                start = s;
                end = i;
            }
 
            if (max_ending_here < 0) {
                max_ending_here = 0;
                s = i + 1;
            }
        }
        document.write("Maximum contiguous sum is " + max_so_far);
        document.write("<br/>Starting index " + start);
        document.write("<br/>Ending index " + end);
    }
 
    // Driver code
     
        var a = [ -2, -3, 4, -1, -2, 1, 5, -3 ];
        var n = a.length;
        maxSubArraySum(a, n);
 
// This code is contributed by Rajput-Ji 
</script>

PHP

<?php 
// PHP program to print largest 
// contiguous array sum
 
function maxSubArraySum($a, $size)
{
    $max_so_far = PHP_INT_MIN;
    $max_ending_here = 0;
    $start = 0;
    $end = 0;
    $s = 0;
 
    for ($i = 0; $i < $size; $i++)
    {
        $max_ending_here += $a[$i];
 
        if ($max_so_far < $max_ending_here)
        {
            $max_so_far = $max_ending_here;
            $start = $s;
            $end = $i;
        }
 
        if ($max_ending_here < 0)
        {
            $max_ending_here = 0;
            $s = $i + 1;
        }
    }
    echo "Maximum contiguous sum is ".
                     $max_so_far."\n";
    echo "Starting index ". $start . "\n".
            "Ending index " . $end . "\n";
}
 
// Driver Code
$a = array(-2, -3, 4, -1, -2, 1, 5, -3);
$n = sizeof($a);
maxSubArraySum($a, $n);
 
// This code is contributed
// by ChitraNayal
?>

Output

Maximum contiguous sum is 7
Starting index 2
Ending index 6

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

Largest Sum Contiguous Subarray using Dynamic Programming:

For each index i, DP[i] stores the maximum possible Largest Sum Contiguous Subarray ending at index i, and therefore we can calculate DP[i] using the mentioned state transition:

DP[i] = max(DP[i-1] + arr[i] , arr[i] )

Below is the implementation:

C++

// C++ program to print largest contiguous array sum
#include <bits/stdc++.h>
using namespace std;
 
void maxSubArraySum(int a[], int size)
{
    vector<int> dp(size, 0);
    dp[0] = a[0];
    int ans = dp[0];
    for (int i = 1; i < size; i++) {
        dp[i] = max(a[i], a[i] + dp[i - 1]);
        ans = max(ans, dp[i]);
    }
    cout << ans;
}
 
/*Driver program to test maxSubArraySum*/
int main()
{
    int a[] = { -2, -3, 4, -1, -2, 1, 5, -3 };
    int n = sizeof(a) / sizeof(a[0]);
    maxSubArraySum(a, n);
    return 0;
}

Output

Practice Problem:

Given an array of integers (possibly some elements negative), write a C program to find out the *maximum product* possible by multiplying ‘n’ consecutive integers in the array where n ≤ ARRAY_SIZE. Also, print the starting point of the maximum product subarray.

Please write comments if you find anything incorrect, or if you want to share more information about the topic discussed above.

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Last Updated : 30 Oct, 2023