Largest Sum Contiguous Subarray

Write an efficient program to find the sum of the contiguous subarray within a one-dimensional array of numbers that has the largest sum. 

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

Explanation:
The simple idea of Kadane’s algorithm is to look for all positive contiguous segments of the array (max_ending_here is used for this). And keep track of the maximum sum contiguous segment among all positive segments (max_so_far is used for this). Each time we get a positive-sum compare it with max_so_far and update max_so_far if it is greater than max_so_far 

    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

Note: The above algorithm only works if and only if at least one positive number should be present otherwise it does not work i.e if an Array contains all negative numbers it doesn’t work.

Program:

C++

// C++ program to print largest contiguous array sum
#include<iostream>
#include<climits>
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 program to test maxSubArraySum*/
int main()
{
    int a[] = {-2, -3, 4, -1, -2, 1, 5, -3};
    int n = sizeof(a)/sizeof(a[0]);
    int max_sum = maxSubArraySum(a, n);
    cout << "Maximum contiguous sum is " << max_sum;
    return 0;
}

Java

import java.io.*;
// Java program to print largest contiguous array sum
import java.util.*;

class Kadane
{
    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));
    }

    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)

Another approach:

C++

int maxSubarraySum(int arr[], int size)
{
    int max_ending_here = 0, max_so_far = INT_MIN;
    for (int i = 0; i < size; i++) {

        // include current element to previous subarray only
        // when it can add to a bigger number than itself.
        if (arr[i] <= max_ending_here + arr[i]) {
            max_ending_here += arr[i];
        }

        // Else start the max subarray from current element
        else {
            max_ending_here = arr[i];
        }
        if (max_ending_here > max_so_far)
            max_so_far = max_ending_here;
    }
    return max_so_far;
} // contributed by Vipul Raj

Java

static int maxSubArraySum(int a[],int size)
{

    int max_so_far = a[0], max_ending_here = 0;

    for (int i = 0; i < size; i++)
    {
        max_ending_here = max_ending_here + a[i];
        if (max_ending_here < 0)
            max_ending_here = 0;

        /* Do not compare for all
           elements. Compare only
           when max_ending_here > 0 */
        else if (max_so_far < max_ending_here)
            max_so_far = max_ending_here;

    }
    return max_so_far;
}

// This code is contributed by ANKITRAI1

Python

def maxSubArraySum(a,size):

    max_so_far = a[0]
    max_ending_here = 0

    for i in range(0, size):
        max_ending_here = max_ending_here + a[i]
        if max_ending_here < 0:
            max_ending_here = 0

        # Do not compare for all elements. Compare only
        # when  max_ending_here > 0
        elif (max_so_far < max_ending_here):
            max_so_far = max_ending_here

    return max_so_far

C#

static int maxSubArraySum(int[] a, int size)
{
    int max_so_far = a[0], max_ending_here = 0;

    for (int i = 0; i < size; i++) {
        max_ending_here = max_ending_here + a[i];
        if (max_ending_here < 0)
            max_ending_here = 0;

        /* Do not compare for all
        elements. Compare only
        when max_ending_here > 0 */
        else if (max_so_far < max_ending_here)
            max_so_far = max_ending_here;
    }
    return max_so_far;
}

// This code is contributed
// by ChitraNayal

PHP

<?php
function maxSubArraySum(&$a, $size)
{
$max_so_far = $a[0];
$max_ending_here = 0;
for ($i = 0; $i < $size; $i++)
{
    $max_ending_here = $max_ending_here + $a[$i];
    if ($max_ending_here < 0)
        $max_ending_here = 0;

    /* Do not compare for all elements.
       Compare only when max_ending_here > 0 */
    else if ($max_so_far < $max_ending_here)
        $max_so_far = $max_ending_here;
}
return $max_so_far;

// This code is contributed
// by ChitraNayal
?>

Javascript

<script>
        // JavaScript Program to implement
        // the above approach

        function maxSubarraySum(arr, size)
        {
            let max_ending_here = 0, max_so_far = Number.MIN_VALUE;
            for (let i = 0; i < size; i++) {

                // include current element to previous subarray only
                // when it can add to a bigger number than itself.
                if (arr[i] <= max_ending_here + arr[i]) {
                    max_ending_here += arr[i];
                }

                // Else start the max subarray from current element
                else {
                    max_ending_here = arr[i];
                }
                if (max_ending_here > max_so_far) {
                    max_so_far = max_ending_here;
                }
            }
            return max_so_far;
        }

// This code is contributed by Potta Lokesh
    </script>

Algorithmic Paradigm: Dynamic Programming
Following is another simple implementation suggested by Mohit Kumar. The implementation handles the case when all numbers in the array are negative. 

C++

#include<iostream>
using namespace std;

int maxSubArraySum(int a[], int size)
{
   int max_so_far = a[0];
   int curr_max = a[0];

   for (int i = 1; i < size; i++)
   {
        curr_max = max(a[i], curr_max+a[i]);
        max_so_far = max(max_so_far, curr_max);
   }
   return max_so_far;
}

/* Driver program to test maxSubArraySum */
int main()
{
   int a[] =  {-2, -3, 4, -1, -2, 1, 5, -3};
   int n = sizeof(a)/sizeof(a[0]);
   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.*;

class GFG {

    static int maxSubArraySum(int a[], int size)
    {
    int max_so_far = a[0];
    int curr_max = a[0];

    for (int i = 1; i < size; i++)
    {
           curr_max = Math.max(a[i], curr_max+a[i]);
        max_so_far = Math.max(max_so_far, curr_max);
    }
    return max_so_far;
    }

    /* Driver program to test maxSubArraySum */
    public static void main(String[] args)
    {
    int a[] = {-2, -3, 4, -1, -2, 1, 5, -3};
    int n = a.length;
    int max_sum = maxSubArraySum(a, n);
    System.out.println("Maximum contiguous sum is "
                       + max_sum);
    }
}

// This code is contributed by Prerna Saini

Python

# Python program to find maximum contiguous subarray

def maxSubArraySum(a,size):

    max_so_far =a[0]
    curr_max = a[0]

    for i in range(1,size):
        curr_max = max(a[i], curr_max + a[i])
        max_so_far = max(max_so_far,curr_max)

    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)
    {
    int max_so_far = a[0];
    int curr_max = a[0];

    for (int i = 1; i < size; i++)
    {
        curr_max = Math.Max(a[i], curr_max+a[i]);
        max_so_far = Math.Max(max_so_far, curr_max);
    }

    return max_so_far;
    }

    // Driver code
    public static void Main ()
    {
        int []a = {-2, -3, 4, -1, -2, 1, 5, -3};
        int n = a.Length;
        Console.Write("Maximum contiguous sum is "
                           + maxSubArraySum(a, n));
    }

}

// This code is contributed by Sam007_

PHP

<?php
function maxSubArraySum($a, $size)
{
    $max_so_far = $a[0];
    $curr_max = $a[0];

    for ($i = 1; $i < $size; $i++)
    {
        $curr_max = max($a[$i],
                        $curr_max + $a[$i]);
        $max_so_far = max($max_so_far,
                          $curr_max);
    }
    return $max_so_far;
}

// Driver Code
$a = array(-2, -3, 4, -1,
           -2, 1, 5, -3);
$n = sizeof($a);
$max_sum = maxSubArraySum($a, $n);
echo "Maximum contiguous sum is " .
                          $max_sum;

// This code is contributed
// by Akanksha Rai(Abby_akku)
?>

Javascript

<script>
// C# program to print largest
// contiguous array sum

function maxSubArraySum(a,size)
{
  let max_so_far = a[0];
  let curr_max = a[0];

  for (let i = 1; i < size; i++)
  {
      curr_max = Math.max(a[i], curr_max+a[i]);
      max_so_far = Math.max(max_so_far, curr_max);
  }

return max_so_far;
}

// Driver code

let a = [-2, -3, 4, -1, -2, 1, 5, -3];
let n = a.length;
document.write("Maximum contiguous sum is ",maxSubArraySum(a, n));

</script>

Output

Maximum contiguous sum is 7

Time complexity: O(n), where n is the size of the array.
Auxiliary Space: O(1)

To print the subarray with the maximum sum, we maintain indices whenever we get the maximum sum.  

C++

// C++ program to print largest contiguous array sum
#include<iostream>
#include<climits>
using namespace std;

int 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]);
    int max_sum = maxSubArraySum(a, n);
    return 0;
}

Java

// Java program to print largest
// contiguous array sum
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);
$max_sum = maxSubArraySum($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

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

Now try the below question 
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.