C++ Program for Maximum Product Subarray
Last Updated :
17 Aug, 2023
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.
C++
#include <bits/stdc++.h>
using namespace std;
int maxSubarrayProduct( int arr[], int n)
{
int result = arr[0];
for ( int i = 0; i < n; i++)
{
int mul = arr[i];
for ( int j = i + 1; j < n; j++)
{
result = max(result, mul);
mul *= arr[j];
}
result = max(result, mul);
}
return result;
}
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;
}
|
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.
C++
#include <bits/stdc++.h>
using namespace std;
int maxSubarrayProduct( int arr[], int n)
{
int max_ending_here = 1;
int min_ending_here = 1;
int max_so_far = 0;
int flag = 0;
for ( int i = 0; i < n; i++)
{
if (arr[i] > 0)
{
max_ending_here = max_ending_here * arr[i];
min_ending_here
= min(min_ending_here * arr[i], 1);
flag = 1;
}
else if (arr[i] == 0) {
max_ending_here = 1;
min_ending_here = 1;
}
else {
int 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 && max_so_far == 0)
return 0;
return max_so_far;
}
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;
}
|
Output
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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