# 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, -3, 0, -2, -40}
Output:   80  // The subarray is {-2, -40}
```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the 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 = 1; ` `    ``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 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` `{ ` `            ``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 == 1) ` `        ``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); ` `    ``cout << ``"Maximum Sub array product is "` `         ``<< maxSubarrayProduct(arr, n); ` `    ``return` `0; ` `} ` ` `  `// This is code is contributed by rathbhupendra `

## C

 `// C program to find Maximum Product Subarray ` `#include ` ` `  `// Utility functions to get minimum of two integers ` `int` `min(``int` `x, ``int` `y) { ``return` `x < y ? x : y; } ` ` `  `// Utility functions to get maximum of two integers ` `int` `max(``int` `x, ``int` `y) { ``return` `x > y ? x : y; } ` ` `  `/* 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 = 1; ` `    ``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 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 min_ending_here will always be prev.  ` `        ``max_ending_here * arr[i] next max_ending_here  ` `        ``will be 1 if prev min_ending_here is 1, otherwise  ` `        ``next max_ending_here will be prev min_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 == 1) ` `        ``return` `0; ` `    ``return` `max_so_far; ` ` `  `    ``return` `max_so_far; ` `} ` ` `  `// Driver Program to test above function ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, -2, -3, 0, 7, -8, -2 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``printf``(``"Maximum Sub array product is %d"``, ` `           ``maxSubarrayProduct(arr, n)); ` `    ``return` `0; ` `} `

## Java

 `// Java program to find maximum product subarray ` `import` `java.io.*; ` ` `  `class` `ProductSubarray { ` ` `  `    ``// Utility functions to get minimum of two integers ` `    ``static` `int` `min(``int` `x, ``int` `y) { ``return` `x < y ? x : y; } ` ` `  `    ``// Utility functions to get maximum of two integers ` `    ``static` `int` `max(``int` `x, ``int` `y) { ``return` `x > y ? x : y; } ` ` `  `    ``/* Returns the product of max product subarray. ` `    ``Assumes that the given array always has a subarray ` `    ``with product more than 1 */` `    ``static` `int` `maxSubarrayProduct(``int` `arr[]) ` `    ``{ ` `        ``int` `n = arr.length; ` `        ``// 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 = ``1``; ` `        ``int` `flag = ``0``; ` ` `  `        ``/* Traverse through the array. Following ` `        ``values are maintained after the ith 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 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 min_ending_here will always be prev. ` `            ``max_ending_here * arr[i] ` `            ``next max_ending_here will be 1 if prev ` `            ``min_ending_here is 1, otherwise ` `            ``next max_ending_here will be  ` `                        ``prev min_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 == ``1``) ` `            ``return` `0``; ` `        ``return` `max_so_far; ` `    ``} ` ` `  `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` ` `  `        ``int` `arr[] = { ``1``, -``2``, -``3``, ``0``, ``7``, -``8``, -``2` `}; ` `        ``System.out.println(``"Maximum Sub array product is "` `                           ``+ maxSubarrayProduct(arr)); ` `    ``} ` `} ``/*This code is contributed by Devesh Agrawal*/`

## Python

 `# Python 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 ` `def` `maxsubarrayproduct(arr): ` ` `  `    ``n ``=` `len``(arr) ` ` `  `    ``# max positive product ending at the current position ` `    ``max_ending_here ``=` `1` ` `  `    ``# min positive product ending at the current position ` `    ``min_ending_here ``=` `1` ` `  `    ``# Initialize maximum so far ` `    ``max_so_far ``=` `1` `    ``flag ``=` `0` ` `  `    ``# Traverse throughout the array. Following values ` `    ``# are maintained after the ith 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` `i ``in` `range``(``0``, n): ` ` `  `        ``# 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 alway greater than or equal to 1. ` `        ``elif` `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 min_ending_here will always be prev. ` `        ``# max_ending_here * arr[i] ` `        ``# next max_ending_here will be 1 if prev ` `        ``# min_ending_here is 1, otherwise ` `        ``# next max_ending_here will be prev min_ending_here * arr[i] ` `        ``else``: ` `            ``temp ``=` `max_ending_here ` `            ``max_ending_here ``=` `max` `(min_ending_here ``*` `arr[i], ``1``) ` `            ``min_ending_here ``=` `temp ``*` `arr[i] ` `        ``if` `(max_so_far < max_ending_here): ` `            ``max_so_far ``=` `max_ending_here ` `             `  `    ``if` `flag ``=``=` `0` `and` `max_so_far ``=``=` `1``: ` `        ``return` `0` `    ``return` `max_so_far ` ` `  `# Driver function to test above function ` `arr ``=` `[``1``, ``-``2``, ``-``3``, ``0``, ``7``, ``-``8``, ``-``2``] ` `print` `"Maximum product subarray is"``, maxsubarrayproduct(arr) ` ` `  `# This code is contributed by Devesh Agrawal `

## C#

 `// C# program to find maximum product subarray ` `using` `System; ` ` `  `class` `GFG { ` ` `  `    ``// Utility functions to get minimum of two integers ` `    ``static` `int` `min(``int` `x, ``int` `y) { ``return` `x < y ? x : y; } ` ` `  `    ``// Utility functions to get maximum of two integers ` `    ``static` `int` `max(``int` `x, ``int` `y) { ``return` `x > y ? x : y; } ` ` `  `    ``/* Returns the product of max product subarray. ` `    ``Assumes that the given array always has a subarray ` `    ``with product more than 1 */` `    ``static` `int` `maxSubarrayProduct(``int``[] arr) ` `    ``{ ` `        ``int` `n = arr.Length; ` `        ``// 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 = 1; ` `        ``int` `flag = 0; ` ` `  `        ``/* Traverse through the array. Following ` `        ``values are maintained after the ith 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 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 min_ending_here will always be prev. ` `            ``max_ending_here * arr[i] ` `            ``next max_ending_here will be 1 if prev ` `            ``min_ending_here is 1, otherwise ` `            ``next max_ending_here will be  ` `            ``prev min_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 == 1) ` `            ``return` `0; ` ` `  `        ``return` `max_so_far; ` `    ``} ` ` `  `    ``public` `static` `void` `Main() ` `    ``{ ` ` `  `        ``int``[] arr = { 1, -2, -3, 0, 7, -8, -2 }; ` ` `  `        ``Console.WriteLine(``"Maximum Sub array product is "` `                          ``+ maxSubarrayProduct(arr)); ` `    ``} ` `} ` ` `  `/*This code is contributed by vt_m*/`

## PHP

 ` ``\$y``? ``\$x` `: ``\$y``;  ` `} ` ` `  `/* 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 ` `    ``\$max_ending_here` `= 1; ` ` `  `    ``// min negative product ending at ` `    ``// the current position ` `    ``\$min_ending_here` `= 1; ` ` `  `    ``// Initialize overall max product ` `    ``\$max_so_far` `= 1; ` `    ``\$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` `(``\$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 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 min_ending_here will ` `        ``always be prev. max_ending_here *  ` `        ``arr[i] next max_ending_here will be ` `        ``1 if prev min_ending_here is 1, ` `        ``otherwise next max_ending_here will ` `        ``be prev min_ending_here * arr[i] */` `        ``else` `        ``{ ` `            ``\$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``==1) ``return` `0;  ` `    ``return` `\$max_so_far``; ` `} ` ` `  `// Driver Program to test above function ` `    ``\$arr` `= ``array``(1, -2, -3, 0, 7, -8, -2); ` `    ``\$n` `= sizeof(``\$arr``) / sizeof(``\$arr``); ` `    ``echo``(``"Maximum Sub array product is "``); ` `    ``echo` `(maxSubarrayProduct(``\$arr``, ``\$n``)); ` ` `  `// This code is contributed by nitin mittal  ` `?> `

Output:

```Maximum Sub array product is 112
```

Time Complexity: O(n)
Auxiliary Space: O(1)