# Largest Sum Contiguous Subarray

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

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

```Initialize:
max_so_far = 0
max_ending_here = 0

Loop for each element of the array
(a) max_ending_here = max_ending_here + a[i]
(b) if(max_ending_here < 0)
max_ending_here = 0
(c) if(max_so_far < max_ending_here)
max_so_far = max_ending_here
return max_so_far
```

Explanation:
Simple idea of the 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 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 = 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

for i=1,  a[1] =  -3
max_ending_here = max_ending_here + (-3)
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 0 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
```

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.*;

{
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 = [-13, -3, -25, -20, -3, -16, -23, -12, -5, -22, -15, -4, -7]
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;
}

// Drive 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_
```

Output:
`Maximum contiguous sum is 7`

Above program can be optimized further, if we compare max_so_far with max_ending_here only if max_ending_here is greater than 0.

## C++

```int maxSubArraySum(int a[], int size)
{
int max_so_far = 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;
}
```

## Python

```def maxSubArraySum(a,size):

max_so_far = 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
```

Time Complexity: O(n)

Following is another simple implementation suggested by Mohit Kumar. The implementation handles the case when all numbers in 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 contributd 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;
}

// Drive 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_
```

Output:
`Maximum contiguous sum is 7`

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

## CPP

```// 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))
```

Output:
```Maximum contiguous sum is 7
Starting index 2
Ending index 6
```

Now try below question
Given an array of integers (possibly some of the 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 maximum product subarray.