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Maximum Product Subarray - Using Two Traversals in C++
In this problem, we are given an array arr[] of integers. Our task is to create a program to find the Maximum Product Subarray - Using Two Traversals in C++.
Problem description − Here in the array, we will find the maximum product subarray using two traversals one from index 0 and another index (n-1).
Let’s take an example to understand the problem,
Input
arr[] = {4, -2, 5, -6, 0, 8}Output
240
Example
Subarray = {4, -2, 5, -6}
Maximum product = 4 * (-2) * 5 * (-6) = 240Solution Approach
To solve this problem using two traversals. Here, we will find the maximum product using two local maximum values for traversal from left to right i.e. from index 0 to n-1. And one for traversal from right to left i.e. from index n-1 to 0. The rest algorithm is the same as finding the maximum product subarray.
Program to illustrate the working of our solution,
Example
#include<iostream>
using namespace std;
int CalcMaxProductSubArray(int arr[], int n) {
int frntMax = 1, rearMax = 1, maxVal = 1;
for (int i=0; i<n; i++) {
frntMax = frntMax*arr[i];
if (frntMax == 0)
frntMax = 1;
}
for (int i=n-1; i>=0; i--) {
rearMax = rearMax * arr[i];
if (rearMax == 0)
rearMax = 1;
}
maxVal = max(frntMax, rearMax);
return maxVal;
}
int main() {
int arr[] = {4, -2, 5, -6, 0, 8};
int n = sizeof(arr)/sizeof(arr[0]);
cout<<"Maximum product subarray is "<<CalcMaxProductSubArray(arr, n);
return 0;
}Output
Maximum product subarray is 240
Updated on: 15-Sep-2020
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