Javascript Program for Maximum Product Subarray

// Javascript program to find Maximum Product Subarray
/* Returns the product of max product subarray.*/
function maxSubarrayProduct(arr, n) {
    // Initializing result
    let result = arr[0];

    for (let i = 0; i < n; i++) {
        let mul = arr[i];
        // traversing in current subarray
        for (let j = i + 1; j < n; j++) {
            // updating result every time
            // to keep an eye over the maximum product
            result = Math.max(result, mul);
            mul *= arr[j];
        }
        // updating the result for (n-1)th index.
        result = Math.max(result, mul);
    }
    return result;
}

// Driver code

let arr = [1, -2, -3, 0, 7, -8, -2];
let n = arr.length;
console.log("Maximum Sub array product is " + maxSubarrayProduct(arr, n));

// JavaScript program to find 
// Maximum Product Subarray

/* Returns the product 
  of max product subarray.
Assumes that the given 
array always has a subarray
with product more than 1 */
function maxSubarrayProduct(arr, n) {
    // max positive product 
    // ending at the current position
    let max_ending_here = 1;

    // min negative product ending 
    // at the current position
    let min_ending_here = 1;

    // Initialize overall max product
    let max_so_far = 0;
    let flag = 0;
    /* Traverse through the array. 
    Following values are
    maintained after the i'th iteration:
    max_ending_here is always 1 or 
    some positive product ending with arr[i]
    min_ending_here is always 1 or 
    some negative product ending with arr[i] */
    for (let i = 0; i < n; i++) {
        /* If this element is positive, update
        max_ending_here. Update min_ending_here only if
        min_ending_here is negative */
        if (arr[i] > 0) {
            max_ending_here = max_ending_here * arr[i];
            min_ending_here
                = Math.min(min_ending_here * arr[i], 1);
            flag = 1;
        }

        /* If this element is 0, then the maximum product
        cannot end here, make both max_ending_here and
        min_ending_here 0
        Assumption: Output is alway greater than or equal
                    to 1. */
        else if (arr[i] == 0) {
            max_ending_here = 1;
            min_ending_here = 1;
        }

        /* If element is negative. This is tricky
         max_ending_here can either be 1 or positive.
         min_ending_here can either be 1 or negative.
         next max_ending_here will always be prev.
         min_ending_here * arr[i] ,next min_ending_here
         will be 1 if prev max_ending_here is 1, otherwise
         next min_ending_here will be prev max_ending_here *
         arr[i] */

        else {
            let temp = max_ending_here;
            max_ending_here
                = Math.max(min_ending_here * arr[i], 1);
            min_ending_here = temp * arr[i];
        }

        // update max_so_far, if needed
        if (max_so_far < max_ending_here)
            max_so_far = max_ending_here;
    }
    if (flag == 0 && max_so_far == 0)
        return 0;
    return max_so_far;
}
// Driver program 

let arr = [1, -2, -3, 0, 7, -8, -2];
let n = arr.length;
console.log("Maximum Sub array product is " + maxSubarrayProduct(arr, n));