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}

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++

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// C++ program to find Maximum Product Subarray
#include <bits/stdc++.h>
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 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;
}
  
// 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

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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[0]); printf("Maximum Sub array product is %d", maxSubarrayProduct(arr, n)); return 0; } [tabby title="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*/ [tabby title="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 [tabby title="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*/ [tabby title="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($arr[0]); echo("Maximum Sub array product is "); echo (maxSubarrayProduct($arr, $n)); // This code is contributed by nitin mittal ?>

Output:

Maximum Sub array product is 112

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

This article is compiled by Dheeraj Jain and reviewed by GeeksforGeeks team. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above



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