# Maximum Product Subarray | Set 2 (Using Two Traversals)

• Difficulty Level : Medium
• Last Updated : 17 Nov, 2021

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[] = {-1, -2, -3, 4}
Output:   24  // The subarray is {-2, -3, 4}

Input: arr[] = {-10}
Output:   0  // An empty array is also subarray
// and product of empty subarray is
// considered as 0.```

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We have discussed a solution of this problem here
In this post an interesting solution is discussed. The idea is based on the fact that overall maximum product is maximum of following two:

1. Maximum product in left to right traversal.
2. Maximum product in right to left traversal

For example, consider the above third sample input {-1, -2, -3, 4}. If we traverse the array only in forward direction (considering -1 as part of output), maximum product will be 2. If we traverse the array in backward direction (considering 4 as part of output), maximum product will be 24 i.e; { -2, -3, 4}.
One important thing is to handle 0’s. We need to compute fresh forward (or backward) sum whenever we see 0.
Below is the implementation of above idea :

## C++

 `// C++ program to find maximum product subarray``#include``using` `namespace` `std;` `// Function for maximum product``int` `max_product(``int` `arr[], ``int` `n)``{``    ``// Initialize maximum products in forward and``    ``// backward directions``    ``int` `max_fwd = INT_MIN, max_bkd = INT_MIN;` `    ``// Initialize current product``    ``int` `max_till_now = 1;` `    ``//check if zero is present in an array or not``    ``bool` `isZero=``false``;``    ` `    ``// max_fwd for maximum contiguous product in``    ``// forward direction``    ``// max_bkd for maximum contiguous product in``    ``// backward direction``    ``// iterating within forward direction in array``    ``for` `(``int` `i=0; i=0; i--)``    ``{``        ``max_till_now = max_till_now * arr[i];``        ``if` `(max_till_now == 0)``        ``{``            ``isZero=``true``;``            ``max_till_now = 1;``            ``continue``;``        ``}` `        ``// update max_bkd``        ``if` `(max_bkd < max_till_now)``            ``max_bkd = max_till_now;``    ``}` `    ``// return max of max_fwd and max_bkd``    ``int` `res =  max(max_fwd, max_bkd);` `    ``// Product should not be negative.``    ``// (Product of an empty subarray is``    ``// considered as 0)``    ``if``(isZero)``    ``return` `max(res, 0);` `    ``return` `res;``}` `// Driver Program to test above function``int` `main()``{``    ``int` `arr[] = {-1, -2, -3, 4};``    ``int` `n = ``sizeof``(arr)/``sizeof``(arr);``    ``cout << max_product(arr, n) << endl;``    ``return` `0;``}`

## Java

 `// Java program to find``// maximum product subarray``import` `java.io.*;` `class` `GFG``{` `// Function for maximum product``static` `int` `max_product(``int` `arr[], ``int` `n)``{``    ``// Initialize maximum products in``    ``// forward and backward directions``    ``int` `max_fwd = Integer.MIN_VALUE,``        ``max_bkd = Integer.MIN_VALUE;` `    ``//check if zero is present in an array or not``    ``boolean` `isZero=``false``;` `    ``// Initialize current product``    ``int` `max_till_now = ``1``;` `    ``// max_fwd for maximum contiguous``    ``// product in forward direction``    ``// max_bkd for maximum contiguous``    ``// product in backward direction``    ``// iterating within forward``    ``// direction in array``    ``for` `(``int` `i = ``0``; i < n; i++)``    ``{``        ``// if arr[i]==0, it is breaking``        ``// condition for contiguous subarray``        ``max_till_now = max_till_now * arr[i];``        ``if` `(max_till_now == ``0``)``        ``{``            ``isZero=``true``;``            ``max_till_now = ``1``;``            ``continue``;``        ``}``        ` `        ``// update max_fwd``        ``if` `(max_fwd < max_till_now)``            ``max_fwd = max_till_now;``    ``}` `    ``max_till_now = ``1``;` `    ``// iterating within backward``    ``// direction in array``    ``for` `(``int` `i = n - ``1``; i >= ``0``; i--)``    ``{``        ``max_till_now = max_till_now * arr[i];``        ``if` `(max_till_now == ``0``)``        ``{``            ``isZero=``true``;``            ``max_till_now = ``1``;``            ``continue``;``        ``}` `        ``// update max_bkd``        ``if` `(max_bkd < max_till_now)``            ``max_bkd = max_till_now;``    ``}` `    ``// return max of max_fwd and max_bkd``    ``int` `res = Math. max(max_fwd, max_bkd);` `    ``// Product should not be negative.``    ``// (Product of an empty subarray is``    ``// considered as 0)``    ``if``(isZero)``    ``return` `Math.max(res, ``0``);``    ` `    ``return` `res;``}` `// Driver Code``public` `static` `void` `main (String[] args)``{``    ``int` `arr[] = {-``1``, -``2``, -``3``, ``4``};``    ``int` `n = arr.length;``    ``System.out.println( max_product(arr, n) );``}``}` `// This code is contributed by anuj_67.`

## Python3

 `# Python3 program to find``# maximum product subarray``import` `sys` `# Function for maximum product``def` `max_product(arr, n):` `    ``# Initialize maximum products``    ``# in forward and backward directions``    ``max_fwd ``=` `-``sys.maxsize ``-` `1``    ``max_bkd ``=` `-``sys.maxsize ``-` `1``    ` `    ``#check if zero is present in an array or not``    ``isZero``=``False``;` `    ``# Initialize current product``    ``max_till_now ``=` `1` `    ``# max_fwd for maximum contiguous``    ``# product in forward direction``    ``# max_bkd for maximum contiguous``    ``# product in backward direction``    ``# iterating within forward``    ``# direction in array``    ``for` `i ``in` `range``(n):``    ` `        ``# if arr[i]==0, it is breaking``        ``# condition for contiguous subarray``        ``max_till_now ``=` `max_till_now ``*` `arr[i]``        ``if` `(max_till_now ``=``=` `0``):``            ``isZero``=``True``            ``max_till_now ``=` `1``;``            ``continue``        ` `        ``if` `(max_fwd < max_till_now): ``#update max_fwd``            ``max_fwd ``=` `max_till_now``    ` `    ``max_till_now ``=` `1` `    ``# iterating within backward``    ``# direction in array``    ``for` `i ``in` `range``(n ``-` `1``, ``-``1``, ``-``1``):``        ``max_till_now ``=` `max_till_now ``*` `arr[i]``        ` `        ``if` `(max_till_now ``=``=` `0``):``            ``isZero``=``True``            ``max_till_now ``=` `1``            ``continue` `        ``# update max_bkd``        ``if` `(max_bkd < max_till_now) :``            ``max_bkd ``=` `max_till_now` `    ``# return max of max_fwd and max_bkd``    ``res ``=` `max``(max_fwd, max_bkd)` `    ``# Product should not be negative.``    ``# (Product of an empty subarray is``    ``# considered as 0)``    ``if` `isZero``=``=``True` `:``        ``return` `max``(res, ``0``)` `    ``return` `res` `# Driver Code``arr ``=` `[``-``1``, ``-``2``, ``-``3``, ``4``]``n ``=` `len``(arr)``print``(max_product(arr, n))` `# This code is contributed``# by Yatin Gupta`

## C#

 `// C# program to find maximum product``// subarray``using` `System;` `class` `GFG {` `    ``// Function for maximum product``    ``static` `int` `max_product(``int` `[]arr, ``int` `n)``    ``{``        ` `        ``// Initialize maximum products in``        ``// forward and backward directions``        ``int` `max_fwd = ``int``.MinValue,``            ``max_bkd = ``int``.MinValue;``    ` `        ``// Initialize current product``        ``int` `max_till_now = 1;``    ` `        ``// max_fwd for maximum contiguous``        ``// product in forward direction``        ``// max_bkd for maximum contiguous``        ``// product in backward direction``        ``// iterating within forward``        ``// direction in array``        ``for` `(``int` `i = 0; i < n; i++)``        ``{``            ` `            ``// if arr[i]==0, it is breaking``            ``// condition for contiguous subarray``            ``max_till_now = max_till_now * arr[i];``            ` `            ``if` `(max_till_now == 0)``            ``{``                ``max_till_now = 1;``                ``continue``;``            ``}``            ` `            ``// update max_fwd``            ``if` `(max_fwd < max_till_now)``                ``max_fwd = max_till_now;``        ``}``    ` `        ``max_till_now = 1;``    ` `        ``// iterating within backward``        ``// direction in array``        ``for` `(``int` `i = n - 1; i >= 0; i--)``        ``{``            ``max_till_now = max_till_now * arr[i];``            ``if` `(max_till_now == 0)``            ``{``                ``max_till_now = 1;``                ``continue``;``            ``}``    ` `            ``// update max_bkd``            ``if` `(max_bkd < max_till_now)``                ``max_bkd = max_till_now;``        ``}``    ` `        ``// return max of max_fwd and max_bkd``        ``int` `res = Math. Max(max_fwd, max_bkd);``    ` `        ``// Product should not be negative.``        ``// (Product of an empty subarray is``        ``// considered as 0)``        ``return` `Math.Max(res, 0);``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `Main ()``    ``{``        ``int` `[]arr = {-1, -2, -3, 4};``        ``int` `n = arr.Length;``        ` `        ``Console.Write( max_product(arr, n) );``    ``}``}` `// This code is contributed by nitin mittal.`

## PHP

 `= 0; ``\$i``--)``    ``{``        ``\$max_till_now` `= ``\$max_till_now` `* ``\$arr``[``\$i``];``        ``if` `(``\$max_till_now` `== 0)``        ``{``            ``\$max_till_now` `= 1;``            ``continue``;``        ``}` `        ``// update max_bkd``        ``if` `(``\$max_bkd` `< ``\$max_till_now``)``            ``\$max_bkd` `= ``\$max_till_now``;``    ``}` `    ``// return max of max_fwd``    ``// and max_bkd``    ``\$res` `= max(``\$max_fwd``, ``\$max_bkd``);` `    ``// Product should not be negative.``    ``// (Product of an empty subarray is``    ``// considered as 0)``    ``return` `max(``\$res``, 0);``}` `    ``// Driver Code``    ``\$arr` `= ``array``(-1, -2, -3, 4);``    ``\$n` `= ``count``(``\$arr``);``    ``echo` `max_product(``\$arr``, ``\$n``);` `// This code is contributed by anuj_67.``?>`

## Javascript

 ``

Output:

`24`

Time Complexity : O(n)
Auxiliary Space : O(1)
Note that the above solution requires two traversals of an array while the previous solution requires only one traversal.
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