Given an array that contains both positive and negative integers, find the product of the maximum product subarray. Expected Time complexity is O(n) and only O(1) extra space can be used.
Examples:
Input: arr[] = {6, -3, -10, 0, 2} Output: 180 // The subarray is {6, -3, -10} Input: arr[] = {-1, -3, -10, 0, 60} Output: 60 // The subarray is {60} Input: arr[] = {-2, -40, 0, -2, -3} Output: 80 // The subarray is {-2, -40}
Naive Solution:
The idea is to traverse over every contiguous subarrays, find the product of each of these subarrays and return the maximum product from these results.
Below is the implementation of the above approach.
<script> // 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;
document.write( "Maximum Sub array product is "
+ maxSubarrayProduct(arr, n));
// This code is contributed by Mayank Tyagi </script> |
Output:
Maximum Sub array product is 112
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Solution:
The following solution assumes that the given input array always has a positive output. The solution works for all cases mentioned above. It doesn’t work for arrays like {0, 0, -20, 0}, {0, 0, 0}.. etc. The solution can be easily modified to handle this case.
It is similar to Largest Sum Contiguous Subarray problem. The only thing to note here is, maximum product can also be obtained by minimum (negative) product ending with the previous element multiplied by this element. For example, in array {12, 2, -3, -5, -6, -2}, when we are at element -2, the maximum product is multiplication of, minimum product ending with -6 and -2.
<script> // 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;
document.write( "Maximum Sub array product is " +
maxSubarrayProduct(arr,n));
</script> |
Maximum Sub array product is 112
Time Complexity: O(n)
Auxiliary Space: O(1)
Please refer complete article on Maximum Product Subarray for more details!