最大乘积子数组

给定一个包含正整数和负整数的数组，找到最大乘积子数组的乘积。预期时间复杂度为 O(n)，只能使用 O(1) 额外空间。

例子：

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}

天真的解决方案：

这个想法是遍历每个连续的子数组，找到每个子数组的乘积并从这些结果中返回最大乘积。

下面是上述方法的实现。

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

Java

// Java program to find maximum product subarray
import java.io.*;
 
class GFG {
    /* Returns the product of max product subarray.*/
    static int maxSubarrayProduct(int arr[])
    {
        // Initializing result
        int result = arr[0];
        int n = arr.length;
 
        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 = Math.max(result, mul);
                mul *= arr[j];
            }
            // updating the result for (n-1)th index.
            result = Math.max(result, mul);
        }
        return result;
    }
 
    // Driver Code
    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 yashbeersingh42

Python3

# Python3 program to find Maximum Product Subarray
 
# Returns the product of max product subarray.
def maxSubarrayProduct(arr, n):
 
    # Initializing result
    result = arr[0]
 
    for i in range(n):
     
        mul = arr[i]
       
        # traversing in current subarray
        for j in range(i + 1, n):
         
            # 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
arr = [ 1, -2, -3, 0, 7, -8, -2 ]
n = len(arr)
print("Maximum Sub array product is" , maxSubarrayProduct(arr, n))
 
# This code is contributed by divyeshrabadiya07

C#

// C# program to find maximum product subarray
using System;
 
class GFG{
     
// Returns the product of max product subarray
static int maxSubarrayProduct(int []arr)
{
     
    // Initializing result
    int result = arr[0];
    int n = arr.Length;
 
    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 = Math.Max(result, mul);
            mul *= arr[j];
        }
         
        // Updating the result for (n-1)th index
        result = Math.Max(result, mul);
    }
    return result;
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 1, -2, -3, 0, 7, -8, -2 };
     
    Console.Write("Maximum Sub array product is " +
                  maxSubarrayProduct(arr));
}
}
 
// This code is contributed by shivanisinghss2110

Javascript

 
// Javascript program to find Maximum Product Subarray
 
/* Returns the product of max product subarray.*/
function maxSubarrayProduct(arr, n)
{
    // Initializing result
    let result = arr[0];
 
    for (let i = 0; i < n; i++)
    {
        let mul = arr[i];
        // traversing in current subarray
        for (let j = i + 1; j < n; j++)
        {
            // updating result every time
            // to keep an eye over the maximum product
            result = Math.max(result, mul);
            mul *= arr[j];
        }
        // updating the result for (n-1)th index.
        result = Math.max(result, mul);
    }
    return result;
}
 
// Driver code
 
    let arr = [ 1, -2, -3, 0, 7, -8, -2 ];
    let n = arr.length;
    document.write("Maximum Sub array product is "
        + maxSubarrayProduct(arr, n));
     
 
// This code is contributed by Mayank Tyagi
 

输出：

Maximum Sub array product is 112

时间复杂度： O(N ² )
辅助空间： O(1)

高效解决方案：

以下解决方案假定给定的输入数组始终具有正输出。该解决方案适用于上述所有情况。它不适用于 {0, 0, -20, 0}, {0, 0, 0}.. 等数组。可以轻松修改解决方案以处理这种情况。
它类似于最大和连续子数组问题。这里唯一需要注意的是，最大乘积也可以通过以前一个元素结尾的最小（负）乘积乘以这个元素来获得。例如，在数组{12, 2, -3, -5, -6, -2}中，当我们在元素-2处时，最大乘积是乘积，最小乘积以-6和-2结尾。

注意：如果数组的所有元素都是负数，则上述算法的最大乘积为 1。因此，如果最大乘积为 1，则我们必须返回数组的最大元素。

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)
{
    // 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 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 == 0)
        return 0;
    /* if all the array elements are negative */
    if (max_so_far == 1)
    {
       max_so_far = arr[0];
       for(int i = 1; i < n; i++)
         max_so_far = max(max_so_far, arr[i]);
    }
    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

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 = 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 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 == 0)
        return 0;
    return max_so_far;
 
    return max_so_far;
}
 
// Driver code
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 = 0;
        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 == 0)
            return 0;
        return max_so_far;
    }
 
    // Driver Code
    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*/

Python3

# 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 = 0
    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 == 0:
        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 = 0;
        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 == 0)
            return 0;
 
        return max_so_far;
    }
 
    // Driver Code
    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 = 0;
    $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==0) 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
?>

Javascript

 
// JavaScript 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 */
function maxSubarrayProduct(arr, n)
{
    // max positive product
    // ending at the current position
    let max_ending_here = 1;
 
    // min negative product ending
    // at the current position
    let min_ending_here = 1;
 
    // Initialize overall max product
    let max_so_far = 0;
    let 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 (let 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
                = Math.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 {
            let temp = max_ending_here;
            max_ending_here
                = Math.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 program
     
    let arr = [ 1, -2, -3, 0, 7, -8, -2 ];
    let n = arr.length;
    document.write("Maximum Sub array product is " +
                    maxSubarrayProduct(arr,n));
     

输出

Maximum Sub array product is 112

时间复杂度： O(n)
辅助空间： O(1)

高效解决方案：

上面的解决方案假设给定数组总是有一个积极的结果，这不适用于数组只包含非正元素的情况，如 {0, 0, -20, 0}, {0, 0, 0}..等。修改后的解决方案也类似于使用 Kadane 算法的最大和连续子阵列问题。为了便于理解，我们没有像之前的解决方案那样使用任何标志。在这里，我们使用了 3 个名为max_so_far、max_ending_here 和 min_ending_here的变量。对于每个索引，以该索引结尾的最大数字将是maximum(arr[i], max_ending_here * arr[i], min_ending_here[i]*arr[i]) 。同样，此处结束的最小数字将是这 3 个中的最小值。因此我们得到最大乘积子数组的最终值。

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)
{
    // max positive product
    // ending at the current position
    int max_ending_here = arr[0];
 
    // min negative product ending
    // at the current position
    int min_ending_here = arr[0];
 
    // Initialize overall max product
    int max_so_far = arr[0];
    /* Traverse through the array.
    the maximum product subarray ending at an index
    will be the maximum of the element itself,
    the product of element and max product ending previously
    and the min product ending previously. */
    for (int i = 1; i < n; i++)
    {
        int temp = max({arr[i], arr[i] * max_ending_here, arr[i] * min_ending_here});
        min_ending_here = min({arr[i], arr[i] * max_ending_here, arr[i] * min_ending_here});
        max_ending_here = temp;
        max_so_far = max(max_so_far, max_ending_here);
    }
    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 kaustav

Python3

# Python3 program to find Maximum Product Subarray
 
#  Returns the product
# of max product subarray.
def maxSubarrayProduct(arr, n):
 
    # max positive product
    # ending at the current position
    max_ending_here = arr[0]
 
    # min negative product ending
    # at the current position
    min_ending_here = arr[0]
 
    # Initialize overall max product
    max_so_far = arr[0]
     
    # /* Traverse through the array.
    # the maximum product subarray ending at an index
    # will be the maximum of the element itself,
    # the product of element and max product ending previously
    # and the min product ending previously. */
    for i in range(1, n):
        temp = max(max(arr[i], arr[i] * max_ending_here), arr[i] * min_ending_here)
        min_ending_here = min(min(arr[i], arr[i] * max_ending_here), arr[i] * min_ending_here)
        max_ending_here = temp
        max_so_far = max(max_so_far, max_ending_here)
     
    return max_so_far
 
# Driver code
arr = [ 1, -2, -3, 0, 7, -8, -2 ]
n = len(arr)
print(f"Maximum Sub array product is {maxSubarrayProduct(arr, n)}")
 
# This code is contributed by shinjanpatra

Javascript

 
// JavaScript program to find Maximum Product Subarray
 
/* Returns the product
of max product subarray. */
function maxSubarrayProduct(arr, n)
{
 
    // max positive product
    // ending at the current position
    let max_ending_here = arr[0];
 
    // min negative product ending
    // at the current position
    let min_ending_here = arr[0];
 
    // Initialize overall max product
    let max_so_far = arr[0];
     
    /* Traverse through the array.
    the maximum product subarray ending at an index
    will be the maximum of the element itself,
    the product of element and max product ending previously
    and the min product ending previously. */
    for (let i = 1; i < n; i++)
    {
        let temp = Math.max(Math.max(arr[i], arr[i] * max_ending_here), arr[i] * min_ending_here);
        min_ending_here = Math.min(Math.min(arr[i], arr[i] * max_ending_here), arr[i] * min_ending_here);
        max_ending_here = temp;
        max_so_far = Math.max(max_so_far, max_ending_here);
    }
    return max_so_far;
}
 
// Driver code
let arr = [ 1, -2, -3, 0, 7, -8, -2 ]
let n = arr.length
document.write("Maximum Sub array product is "+maxSubarrayProduct(arr, n));
 
// This code is contributed by shinjanpatra