#include <bits/stdc++.h>
using namespace std;

int main() {
    int t;
    cin >> t;
    while (t--) {
        int n, m, k, x, y;
        cin >> n >> m >> k >> x >> y;
        bool can_escape = true;
        for (int i = 0; i < k; i++) {
            int xi, yi;
            cin >> xi >> yi;
            int dx = abs(x - xi);
            int dy = abs(y - yi);
            if ((dx + dy) % 2 == 0) { // if Vika and friend are in same type of room
                can_escape = false;
            }
        }
        if (can_escape) {
            cout << "YES\n";
        } else {
            cout << "NO\n";
        }
    }
    return 0;
}
Codeforces A✅

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Python 3✅
(Rently)

Company Name: Newton School

Role: DSA Problem Setter

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Alien (Uber)
Python 3✅

vector<long long>solution(int t,vector<vector<long long>>carPrices){
    vector<long long>ans;
     for(int i=0;i<t;i++){
         vector<long long>arr=carPrices[i];
         long long val = 0;
         long long mx = INT_MIN;
         int N=arr.size();
        for (int i = 0; i < N; i++) {
    
            long long curr = arr[i];
            mx = max(mx, curr);
            val = max(val, mx - curr);
        }
        long long res = 0;
        while ((1LL << res) - 1 < val) {
            ++res;
        }
        cout<<res<<endl;
        ans.push_back(res);
     }
     return ans;
}

Uber ✅(Car)

Microsoft OA Material with Codes✅

String Replacement
C++✅

#include<bits/stdc++.h>
typedef double C;
typedef complex<C> P;
#define X real()
#define Y imag()
P a[205];
int n, x, y, k;
bool check(int r){
int c=0;
for(int i = 0; i < n; i++){
for(int j = i+1; j < n; j++){
P g = a[j] - a[i];
double d = abs(g);
if(d > 2*r) continue;
P mid = (a[i] + a[j]) * (1.0/2);
double h = sqrt(r*1.0*r - d*d/4);
P per = P(-g.Y,g.X) * (h/d);
int c1 = 2, c2 = 2;
for(int l = 0; l < n; l++){
if(l == i or l == j) continue;
if(abs(a[l] - (mid-per)) <= r)
c1++;
if(abs(a[l] - (mid+per)) <= r)
c2++;
}
c=max({c,c1,c2});
}
}
return c>=k;
}
int Solution::solve(vector<vector<int> > &A, int B ) {
n = A.size();
for(int i = 0; i < n;i++)
a[i] = {1.0*A[i][0], 1.0*A[i][1]};
k = B;
int l = 1, h = 2000000000, ans;
while(h >= l){
int m = (l+h)/2;
if(check(m)) {
ans = m;
h = m-1;
}
else l = m+1;
}
return ans;
}

Red Zone
Media. Net ✅

int closestToTarget(vector<int> oxygenLevels, int target) {
    int n = oxygenLevels.size();
    int minDiff = abs(oxygenLevels[0] - target); // Initialize the minimum difference with the first oxygen level

    // Iterate through the oxygen levels to find the minimum difference
    for (int i = 1; i < n; i++) {
        int diff = abs(oxygenLevels[i] - target);
        minDiff = min(minDiff, diff);
    }

    return minDiff;
}

Optimum oxygen
Salesforce ✅

Anyone Give Sprinklr/Salesforce OA Today?

Rabbit Code
Uber✅