Largest Sum Contiguous Subarray (Kadane’s Algorithm)

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. 

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:

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:

    Lets take the example: {-2, -3, 4, -1, -2, 1, 5, -3}
    max_so_far = INT_MIN
    max_ending_here = 0

    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 = 7
    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_

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.
?>

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>

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.

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
?>

Javascript

<script>
// javascript program to print largest
// contiguous array sum
function maxSubArraySum(a , size) {
        var max_so_far = Number.MIN_VALUE, 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>

Output

Maximum contiguous sum is 7
Starting index 2
Ending index 6

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

Kadane’s Algorithm can be viewed both as greedy and DP. As we can see that we are keeping a running sum of integers and when it becomes less than 0, we reset it to 0 (Greedy Part). This is because continuing with a negative sum is way worse than restarting with a new range. Now it can also be viewed as a DP, at each stage we have 2 choices: Either take the current element and continue with the previous sum OR restart a new range. Both choices are being taken care of in the implementation.

Recursive implementation of Kadane’s algorithm to find the maximum contiguous sum of an integer array.

To determine the maximum subarray sum of an integer array, Kadane’s Algorithm uses a Divide and Conquer strategy. This algorithm’s fundamental concept is to break the given array into smaller subarrays, recursively solve the issue for each subarray, and then combine the solutions to find the overall solution.

Recursive algorithm to find the maximum contiguous sum of an integer array:

1. The input array “arr” and its length “n” are the two parameters for the function “maxSubArraySum”.
2. When there is only one element in the array, the base case of the recursion occurs. In this situation, the function merely returns that element.
3. Finding the midpoint of the array “m” in the recursive case splits the problem into two smaller subproblems. (i.e., the index of the middle element).
4. To find the largest subarray sum in those subarrays, the function recursively calls itself on the left half of the array and the right half of the array (starting from the midpoint). “left_max” and “right_max” contain these values.
5. The function loops through the right half of the array, keeping track of the maximum subarray sum ending at each index, to determine the maximum subarray sum that crosses the middle element of the array. This value is stored in “right_sum”.
6. The function then performs a loop through the array’s left half, recording the highest subarray sum beginning at each index. “left_sum” is where this value is kept.
7. The sum of the maximum subarray sum beginning at the left half and ending at the right half (i.e., “left_sum + right_sum”) crosses the middle element. “cross_max” is where this value is kept.
8. The “left_max”, “right_max”, and “cross_max” subarray sums are returned by the function as their respective maximum values.

Below is the Implementation of above algorithm:

C++

#include <climits>
#include <iostream>

using namespace std;

int maxSubArraySum(int arr[], int n)
{
    // Base case: when there is only one element in the
    // array
    if (n == 1) {
        return arr[0];
    }
    // Recursive case: divide the problem into smaller
    // sub-problems
    int m = n / 2;
    // Find the maximum subarray sum in the left half
    int left_max = maxSubArraySum(arr, m);
    // Find the maximum subarray sum in the right half
    int right_max = maxSubArraySum(arr + m, n - m);
    // Find the maximum subarray sum that crosses the middle
    // element
    int left_sum = INT_MIN, right_sum = INT_MIN, sum = 0;
    for (int i = m; i < n; i++) {
        sum += arr[i];
        right_sum = max(right_sum, sum);
    }
    sum = 0;
    for (int i = m - 1; i >= 0; i--) {
        sum += arr[i];
        left_sum = max(left_sum, sum);
    }
    int cross_max = left_sum + right_sum;
    // Return the maximum of the three subarray sums
    return max(cross_max, max(left_max, right_max));
}

int main()
{
    int arr[] = { -2, -3, 4, -1, -2, 1, 5, -3 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int max_sum = maxSubArraySum(arr, n);
    cout << "Maximum contiguous sum is " << max_sum << endl;
    return 0;
}

Java

import java.util.Arrays;
public class Program {
    // Define a function to find the maximum subarray sum
    public static int maxSubArraySum(int[] arr)
    {
        // Base case: when there is only one element in the
        // array
        if (arr.length == 1) {
            return arr[0];
        }

        // Recursive case: divide the problem into smaller
        // sub-problems
        int m = arr.length / 2;

        // Find the maximum subarray sum in the left half
        int leftMax
            = maxSubArraySum(Arrays.copyOfRange(arr, 0, m));

        // Find the maximum subarray sum in the right half
        int rightMax = maxSubArraySum(
            Arrays.copyOfRange(arr, m, arr.length));

        // Find the maximum subarray sum that crosses the
        // middle element
        int leftSum = Integer.MIN_VALUE;
        int rightSum = Integer.MIN_VALUE;
        int sum = 0;

        // Traverse the array from the middle to the right
        for (int i = m; i < arr.length; i++) {
            sum += arr[i];
            rightSum = Math.max(rightSum, sum);
        }

        sum = 0;

        // Traverse the array from the middle to the left
        for (int i = m - 1; i >= 0; i--) {
            sum += arr[i];
            leftSum = Math.max(leftSum, sum);
        }

        int crossMax = leftSum + rightSum;

        // Return the maximum of the three subarray sums
        return Math.max(crossMax,
                        Math.max(leftMax, rightMax));
    }

    // Example usage
    public static void main(String[] args)
    {
        int[] arr = { -2, -3, 4, -1, -2, 1, 5, -3 };
        int maxSum = maxSubArraySum(arr);
        System.out.println("Maximum contiguous sum is "
                           + maxSum);
    }
}

Python

import sys

# Define a function to find the maximum subarray sum


def maxSubArraySum(arr):
    # Base case: when there is only one element in the array
    if len(arr) == 1:
        return arr[0]

    # Recursive case: divide the problem into smaller sub-problems
    m = len(arr) // 2

    # Find the maximum subarray sum in the left half
    left_max = maxSubArraySum(arr[:m])

    # Find the maximum subarray sum in the right half
    right_max = maxSubArraySum(arr[m:])

    # Find the maximum subarray sum that crosses the middle element
    left_sum = -sys.maxsize - 1
    right_sum = -sys.maxsize - 1
    sum = 0

    # Traverse the array from the middle to the right
    for i in range(m, len(arr)):
        sum += arr[i]
        right_sum = max(right_sum, sum)

    sum = 0

    # Traverse the array from the middle to the left
    for i in range(m - 1, -1, -1):
        sum += arr[i]
        left_sum = max(left_sum, sum)

    cross_max = left_sum + right_sum

    # Return the maximum of the three subarray sums
    return max(cross_max, max(left_max, right_max))


# Example usage
arr = [-2, -3, 4, -1, -2, 1, 5, -3]
max_sum = maxSubArraySum(arr)
print("Maximum contiguous sum is", max_sum)

C#

using System;
using System.Collections.Generic;
using System.Linq;

class Program {
    // Define a function to find the Maximum subarray sum
    public static int MaxSubArraySum(int[] arr)
    {
        // Base case: when there is only one element in the
        // array
        if (arr.Length == 1) {
            return arr[0];
        }

        // Recursive case: divide the problem into smaller
        // sub-problems
        int m = arr.Length / 2;

        // Find the Maximum subarray sum in the left half
        int leftMax = MaxSubArraySum(arr.Take(m).ToArray());

        // Find the Maximum subarray sum in the right half
        int rightMax
            = MaxSubArraySum(arr.Skip(m).ToArray());

        // Find the Maximum subarray sum that crosses the
        // middle element
        int leftSum = int.MinValue;
        int rightSum = int.MinValue;
        int sum = 0;

        // Traverse the array from the middle to the right
        for (int i = m; i < arr.Length; i++) {
            sum += arr[i];
            rightSum = Math.Max(rightSum, sum);
        }

        sum = 0;

        // Traverse the array from the middle to the left
        for (int i = m - 1; i >= 0; i--) {
            sum += arr[i];
            leftSum = Math.Max(leftSum, sum);
        }

        int crossMax = leftSum + rightSum;

        // Return the Maximum of the three subarray sums
        return Math.Max(crossMax,
                        Math.Max(leftMax, rightMax));
    }

    // Example usage
    public static void Main(string[] args)
    {
        int[] arr = { -2, -3, 4, -1, -2, 1, 5, -3 };
        int MaxSum = MaxSubArraySum(arr);
        Console.WriteLine("Maximum contiguous sum is "
                          + MaxSum);
    }
}

// This code is contributed by Tapesh(tapeshdua420)

Javascript

function maxSubArraySum(arr, n) {
  // Base case: when there is only one element in the array
  if (n === 1) {
    return arr[0];
  }
  // Recursive case: divide the problem into smaller sub-problems
  const m = Math.floor(n / 2);
  // Find the maximum subarray sum in the left half
  const leftMax = maxSubArraySum(arr.slice(0, m), m);
  // Find the maximum subarray sum in the right half
  const rightMax = maxSubArraySum(arr.slice(m), n - m);
  // Find the maximum subarray sum that crosses the middle element
  let leftSum = -Infinity,
      rightSum = -Infinity,
      sum = 0;
  for (let i = m; i < n; i++) {
    sum += arr[i];
    rightSum = Math.max(rightSum, sum);
  }
  sum = 0;
  for (let i = m - 1; i >= 0; i--) {
    sum += arr[i];
    leftSum = Math.max(leftSum, sum);
  }
  const crossMax = leftSum + rightSum;
  // Return the maximum of the three subarray sums
  return Math.max(crossMax, leftMax, rightMax);
}

const arr = [-2, -3, 4, -1, -2, 1, 5, -3];
const n = arr.length;
const maxSum = maxSubArraySum(arr, n);
console.log(`Maximum contiguous sum is ${maxSum}`);

Output

Maximum contiguous sum is 7

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.

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