[14 Feb 2020] Maximum Product Subarray

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++ 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 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[0]);     cout << "Maximum Sub array product is "          << maxSubarrayProduct(arr, n);     return 0; }    // This is code is contributed by rathbhupendra

Output:

Maximum Sub array product is 112

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

// 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*/

Output:

Maximum Sub array product is 112

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

# 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

Output:

Maximum Sub array product is 112

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