Merge branch 'main' into main

Author: Dheeraj-06

Problems/Mathematics/Day-01/sol/Abhay/Q1sol.cpp

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+#include <bits/stdc++.h>
+using namespace std;
+/*
+Logic:
+The main key in solving the problem is that a,b will be enought to know that the size of the area.
+Another observation would be that 1 is always facing the element  more than the element which faces 
+at the last element or the size of the circle.
+Once, you get the size of the array, we just have to check if the any of the elements are larger than 
+the size of the array. 
+On deeper checks like- a=3, b=2,c=1;
+we can observe how this isn't true but how does the conditon fulfill the case?
+--> the max check covers all small checks, like the circle don being possible
+in cases like a=1,b=6, c=100
+--> the circle condition is also checked by the max case. 
+
+So, after the checking, we need to see the answer, that can be easily given as the
+difference is same for all pairs of elements  modular nature of the circle( cool words that make sense)
+ 
+Time complexity-->
+
+The time complexity is O(1)
+The space complexity is O(1)
+
+Submission link- https://codeforces.com/contest/1560/submission/355188467
+*/
+//Soln->
+int main() {
+    ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    int t;
+    cin >> t;
+
+    while (t--) {
+        int a,b,c;
+        cin>>a;
+        cin>>b;
+        cin>>c;
+        int x=abs(b-a);
+    if(max({a,b,c})>(2*x))
+    cout<<-1<<endl;
+    else
+    {
+        if(c>x)
+        cout<<c-x<<endl;
+        else
+        cout<<c+x<<endl;
+    }
+    
+    }
+    return 0;
+}
+    

Problems/Mathematics/Day-01/sol/Harshit Sethi/Solution1.cpp

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+// https://codeforces.com/problemset/submission/1560/355220642
+
+#include <iostream>
+#include <cmath>
+using namespace std;
+int main()
+{
+    int t;
+    cin >> t;
+    while (t--)
+    {
+        long long a, b, c;
+        cin >> a >> b >> c;
+        long long order = fabs(a - b);
+        if ((2 * order < a) || (2 * order < b) || (2 * order < c))
+            cout << -1 << endl;
+        else if (order < c)
+            cout << c - order << endl;
+        else
+            cout << c + order << endl;
+    }
+}

Problems/Mathematics/Day-01/sol/Kushagra/Solution1.cpp

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+/*PROBLEM STATEMENT :
+There are n people standing in a circle. They are numbered from 1 to n in a clockwise order. The circle is even (i.e. n is even).
+
+Each person is looking at the person standing exactly opposite him in the circle.
+
+You are given three distinct integers a, b, and c. It is known that person a is looking at person b.
+
+Determine the number of the person that person c is looking at.
+If there are multiple answers, print any of them.
+If there is no answer, print -1.
+
+Approach : the length of the circle will be |a-b|*2 so if any of a,b,c is greater than this we print -1 if this is not true we print c+|a-b| if it exists in the circle otherwise we print c-|a-b| 
+Time Complexity : O(1)
+Space Complexity : O(1)
+Input
+7
+6 2 4
+2 3 1
+2 4 10
+5 3 4
+1 3 2
+2 5 4
+4 3 2
+Output
+8
+-1
+-1
+-1
+4
+1
+-1
+https://codeforces.com/problemset/submission/1560/355227237
+*/
+#include <iostream>
+using namespace std;
+int main() {
+    int t;
+    cin >>t;
+    while(t--)
+    {
+        int a,b,c;
+        cin >>a >>b>>c;
+        int len=abs(a-b)*2;
+        if(c>len||a>len||b>len)
+        cout <<-1<<"\n";
+        else
+        {
+            if(c+len/2<=len)
+            cout <<c+len/2<<"\n";
+            else
+            cout << c-len/2<<"\n";
+        }        
+        
+    }
+}

Problems/Mathematics/Day-01/sol/Priyanshi_Giri/question1.cpp

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+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+    ios_base::sync_with_stdio(false);
+    cin.tie(NULL);
+
+    int t;
+    cin >> t;
+    while (t--) {
+        int a, b, c;
+        cin >> a >> b >> c;
+        int n = 2*(abs(a-b));
+
+        if(a == b){
+            cout << -1 << endl;
+            continue;
+        }
+
+        if(a > n || b > n || c > n){
+            cout  << -1 << endl;
+            continue;
+        }
+
+        if(c + (n/2) <= n){
+            cout << (c+(n/2)) << endl;
+        }else{
+            cout << (c-(n/2)) << endl;
+        }
+
+    }
+    return 0;
+}
+
+
+
+
+/* Explaination : The opposite numbers must be having exactly n/2 numbers in between, hence the total number, ie. n = 2*abs(a-b).
+
+Time Complexity : O(t), t is the number of test cases. 
+
+Submission Link : https://codeforces.com/contest/1560/submission/355247479
+
+*/

Problems/Mathematics/Day-01/sol/sol1.cpp

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+#include <bits/stdc++.h>
+using namespace std;
+
+#define umpp(i, j) unordered_map<i, j, custom_hash>
+#define mpp(i, j) map<i, j>
+#define int long long
+#define ll long long
+#define vi vector<int>
+#define pii pair<int, int>
+#define vpii vector<pair<int,int>>
+#define pb push_back
+#define pob pop_back
+#define sz(x) ((int)(x).size())
+#define all(v) v.begin(), v.end()
+#define F first
+#define S second
+#define sortall(x) sort(all(x))
+#define tr(it, a) for(auto it = a.begin(); it != a.end(); it++)
+#define forx(x, n) for (int x = 0; x < (n); x++)
+#define PI 3.1415926535897932384626
+const int mod = 1e9 + 7;
+
+void solve() {
+    int a,b,c; cin >> a>>b>>c;
+    int diff = abs(a - b);
+    int t = 2 * diff;
+    if(a>t || b>t||c>t){
+        cout << -1 << endl;
+    }
+    else{
+        int a1 = c - diff;
+        int a2 = c + diff;
+        if(a1>=1 && a1<=t){
+            cout << a1 << endl;
+            return;
+        }
+        cout << a2 << endl;
+    }
+}
+
+signed main() {
+    ios_base::sync_with_stdio(0);
+    cin.tie(0);
+    cout.tie(0);
+    srand(chrono::high_resolution_clock::now().time_since_epoch().count());
+    int t = 1;
+    cin >> t;
+    while (t--) solve();
+    return 0;
+}

Problems/Mathematics/Day-01/sol/solution1.cpp

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+/*
+Problem: 
+Given an even number of people standing in a circle, numbered clockwise from 1, 
+each person is looking at the person directly opposite. 
+Given numbers a,b,c,find the number being looked at by c, 
+if a is looking at b.Output -1 if no such circle exists.
+
+Approach :
+1. Let n be the total number of people in the circle.Since opposite person is at n/2 distance
+    n=2*|a-b|
+2. If a,b,c are greater than n,output -1.
+3. The person opposite c is at position c + n/2.If this exceeds n,subtract n. 
+
+Time Complexity: O(1) per test case
+Space Complexity: O(1)
+
+https://github.com/Tabassumasra05/CP-Chronicles/blob/main/Problems/Mathematics/Day-01/sol/solution1.cpp
+
+*/
+
+
+#include<bits/stdc++.h>
+using namespace std;
+
+void solve(){
+    long long a,b,c,d;
+    cin>>a>>b>>c;
+    long long n=2*abs(a-b);
+    if(a>n||b>n||c>n){
+        cout<<-1<<endl;
+        return;
+    }
+    d=c+(n/2);
+    if(d>n){
+        d-=n;
+    }
+    cout<<d<<endl;
+    return;
+    
+
+}
+
+int main(){
+    int t;
+    cin>>t;
+    while(t--){
+        solve();
+    }
+}

Problems/Mathematics/Day-02/q001.md

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+## Problem Description
+
+Hossam woke up bored, so he decided to create an interesting array with his friend Hazem.
+
+You are given an array of `n` positive integers  
+`a1, a2, ..., an`.
+
+Hossam will choose a number `ai` and Hazem will choose a number `aj`.
+
+Your task is to count the number of index pairs `(i, j)` such that:
+
+- `1 ≤ i, j ≤ n`
+- `i ≠ j`
+- The absolute difference `|ai − aj|` is equal to the **maximum absolute difference** over all pairs in the array
+
+---
+
+## Input Format
+
+- The first line contains a single integer `t` — the number of test cases.
+- For each test case:
+  - The first line contains an integer `n`
+  - The second line contains `n` integers `a1, a2, ..., an`
+
+---
+
+## Output Format
+
+- For each test case, print a single integer — the number of valid pairs `(i, j)`.
+
+---
+
+## Constraints
+
+- `1 ≤ t ≤ 100`
+- `2 ≤ n ≤ 100000`
+- `1 ≤ ai ≤ 100000`
+- The sum of `n` over all test cases does not exceed `100000`
+
+---
+
+
+## Sample Input
+
+```
+2
+5
+6 2 3 8 1
+6
+7 2 8 3 2 10
+```
+---
+
+## Output
+
+```
+2
+4
+```
+
+[PROBLEM LINK](https://codeforces.com/problemset/problem/1771/A)
+