# 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++

```// C program to find Maximum Product Subarray
#include <stdio.h>

// 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;

/* 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);
}

/* 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;
}

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[0]);
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;

/* 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);
}

/* 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;
}

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

# 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)

# 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
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
```

Output:

`Maximum Sub array product is 112`

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

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