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Javascript Program for Largest Sum Contiguous Subarray

Last Updated : 30 Aug, 2024
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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: 

JavaScript
// JavaScript program to find maximum 
// contiguous subarray

// Function to find the maximum 
// contiguous subarray
function maxSubArraySum(a, size) {
    let maxint = Math.pow(2, 53)
    let max_so_far = -maxint - 1
    let max_ending_here = 0

    for (let 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
let a = [-2, -3, 4, -1, -2, 1, 5, -3]
console.log("Maximum contiguous sum is",
    maxSubArraySum(a, a.length))

// This code is contributed by AnkThon

Output
Maximum contiguous sum is 7

Another approach:

JavaScript
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;
}

// Example usage:
const arr = [-2, -3, 4, -1, -2, 1, 5, -3];
const size = arr.length;
const result = maxSubarraySum(arr, size);

console.log("Maximum subarray sum is:", result);

Output
Maximum subarray sum is: 7

 Complexity Analysis:

Time Complexity: O(n) 

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. 

JavaScript
// 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;
console.log("Maximum contiguous sum is ", maxSubArraySum(a, n));

Output
Maximum contiguous sum is  7

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

JavaScript
// javascript program to print largest 
// contiguous array sum    
function maxSubArraySum(a, size) {
    let max_so_far = Number.MIN_VALUE;
    let 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;
        }
    }
    console.log("Maximum contiguous sum is " + max_so_far);
    console.log("Starting index " + start);
    console.log("Ending index " + end);
}

// Driver code

let a = [-2, -3, 4, -1, -2, 1, 5, -3];
let n = a.length;
maxSubArraySum(a, n);

// This code is contributed by Rajput-Ji 

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. 

Complexity Analysis:

  • 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 refer complete article on Largest Sum Contiguous Subarray for more details!


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