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,
arr[] = {4, -2, 5, -6, 0, 8}
240
Subarray = {4, -2, 5, -6} Maximum product = 4 * (-2) * 5 * (-6) = 240
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,
#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; }
Maximum product subarray is 240