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, -40, 0, -2, -3}
Output: 80 // The subarray is {-2, -40}
Naive Solution:
The idea is to traverse over every contiguous subarrays, find the product of each of these subarrays and return the maximum product from these results.
Below is the implementation of the above approach.
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 codearr = [ 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 |
Output:
Maximum Sub array product is 112
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient 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.
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 1def 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 always 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 functionarr = [1, -2, -3, 0, 7, -8, -2]print "Maximum product subarray is", maxsubarrayproduct(arr)# This code is contributed by Devesh Agrawal |
Maximum Sub array product is 112
Time Complexity: O(n)
Auxiliary Space: O(1)
Please refer complete article on Maximum Product Subarray for more details!
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