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# C++ Program for Maximum Product Subarray

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.

## C++

 `// C++ program to find Maximum Product Subarray``#include ``using` `namespace` `std;` `/* Returns the product of max product subarray.*/``int` `maxSubarrayProduct(``int` `arr[], ``int` `n)``{``    ``// Initializing result``    ``int` `result = arr[0];` `    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``int` `mul = arr[i];``        ``// traversing in current subarray``        ``for` `(``int` `j = i + 1; j < n; j++)``        ``{``            ``// updating result every time``            ``// to keep an eye over the maximum product``            ``result = max(result, mul);``            ``mul *= arr[j];``        ``}``        ``// updating the result for (n-1)th index.``        ``result = max(result, mul);``    ``}``    ``return` `result;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 1, -2, -3, 0, 7, -8, -2 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``cout << ``"Maximum Sub array product is "``         ``<< maxSubarrayProduct(arr, n);``    ``return` `0;``}` `// This code is contributed by yashbeersingh42`

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.

## C++

 `// C++ program to find Maximum Product Subarray``#include ``using` `namespace` `std;` `/* Returns the product``  ``of max product subarray.``Assumes that the given``array always has a subarray``with product more than 1 */``int` `maxSubarrayProduct(``int` `arr[], ``int` `n)``{``    ``// max positive product``    ``// ending at the current position``    ``int` `max_ending_here = 1;` `    ``// min negative product ending``    ``// at the current position``    ``int` `min_ending_here = 1;` `    ``// Initialize overall max product``    ``int` `max_so_far = 0;``    ``int` `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` `(``int` `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``                ``= 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 always 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` `{``            ``int` `temp = max_ending_here;``            ``max_ending_here``                ``= 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 code``int` `main()``{``    ``int` `arr[] = { 1, -2, -3, 0, 7, -8, -2 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``cout << ``"Maximum Sub array product is "``         ``<< maxSubarrayProduct(arr, n);``    ``return` `0;``}` `// This is code is contributed by rathbhupendra`

Output

`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!

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