# Largest Sum contiguous subarray

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74 # 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. ```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 =  -2
max_ending_here = max_ending_here + (-2)
Set max_ending_here = 0 because max_ending_here < 0

for i=1,  a =  -3
max_ending_here = max_ending_here + (-3)
Set max_ending_here = 0 because max_ending_here < 0

for i=2,  a =  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 =  -1
max_ending_here = max_ending_here + (-1)
max_ending_here = 3

for i=4,  a =  -2
max_ending_here = max_ending_here + (-2)
max_ending_here = 1

for i=5,  a =  1
max_ending_here = max_ending_here + (1)
max_ending_here = 2

for i=6,  a =  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 =  -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);`
`    ``int` `max_sum = maxSubArraySum(a, n);`
`    ``cout << ``"Maximum contiguous sum is "` `<< max_sum;`
`    ``return` `0;`
`}`

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;`
`}`

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;`
`   ``int` `curr_max = a;`
`   ``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);`
`   ``int` `max_sum = maxSubArraySum(a, n);`
`   ``cout << ``"Maximum contiguous sum is "` `<< max_sum;`
`   ``return` `0;`
`}`

Output:

`Maximum contiguous sum is 7`

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

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

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 adding ‘n’ consecutive integers in the array, n <= ARRAY_SIZE. Also give where in the array this sequence of n integers starts.