Hi Codeforces Community,

This solution tackles a problem where we are given two arrays, a and b, both of size n. Our goal is to check for each element in b whether it meets or exceeds the corresponding element in a. If b[i] < a[i], we count it as a mismatch, which we'll store in a variable c.

Problem Summary:: We have two arrays of integers, a and b, each with n elements. We need to compare each element in b with the corresponding element in a: If b[i] >= a[i], it’s considered a valid match, and we move to the next element in both arrays. If b[i] < a[i], we count it as a mismatch and move only i forward to check the next b element without incrementing j. //(https://mirror.codeforces.com/contest/1972/problem/A question link)

Code Walkthrough:: Here’s a structured solution in C++ with explanations:

//Absar Ali

include <bits/stdc++.h>

using namespace std;

void ans() { int n; cin >> n; int a[n]; int b[n]; for(int i=0;i<n;i++) { cin >> a[i]; }for(int i=0;i<n;i++) { cin >> b[i]; } int c=0,j=0; for (int i = 0; i < n;) { if(b[i]>=a[j]) { j++; i++; } else { c++; i++; } } cout << c << endl; }

int32_t main() { int t; cin >> t; while (t--) { ans(); } return 0; }

Explanation of the Solution:: _ Inputs and Initialization:_ We read n, the size of the arrays. Arrays a and b store n integers each.

_ Looping Logic:_ We use two pointers: i for array b and j for array a. The pointer j advances only when we find a match (b[i] >= a[j]). If b[i] < a[j], it’s counted as a mismatch (incrementing c), and we move only i forward.

Output: At the end of each test case, c is printed, showing the total mismatches where b[i] < a[i].

Complexity::

This solution runs in O(n), as we process each element of b and a once, making it efficient for larger values of n.