Largest Sum Contiguous Subarray



Largest Sum Contiguous Subarray

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

kadane-algorithm



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 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++ 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;
}

Output:

Maximum contiguous sum is 7

Another approach:

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

Time Complexity: O(n)

Algorithmic Paradigm: Dynamic Programming

The implementation handles the case when all numbers in the array are negative.

#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;
}

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

Output: 

Maximum contiguous sum is 7
Starting index 2
Ending index 6

Kadane’s Algorithm can be viewed both as a 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 more 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 previous sum OR restart a new range. These 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.

Last Updated on October 16, 2021 by admin

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