Merge branch 'main' into main

Author: Dheeraj-06

Problems/Mathematics/Day-01/sol/Apoorv012/q1.cpp

@@ -0,0 +1,38 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+using ll = long long;
+const ll MOD = 998244353;
+
+void solve() {
+    int a, b, c;
+    cin >> a >> b >> c;
+
+    int diff = abs(a-b);
+    if (a > 2*diff || b > 2*diff || c > 2*diff) {
+        cout << -1 << endl;
+        return;
+    }
+
+    int ans = c + diff;
+    if (ans > 2*diff) {
+        ans -= 2*diff;
+    }
+
+    cout << ans << endl;
+
+}
+
+int main() {
+    ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+    
+    int t = 1;
+    cin >> t;
+    
+    while (t--) {
+        solve();
+    }
+    
+    return 0;
+}

Problems/Mathematics/Day-01/sol/Nitin Kumar/ques1.cpp

@@ -0,0 +1,30 @@
+/* Submission link: https://codeforces.com/contest/1560/submission/355246381 
+
+Time Complexity : O(1);
+
+*/
+
+#include<bits/stdc++.h>
+using namespace std;
+int main(){
+    int t;
+    cin>>t;
+    while(t--){
+        long long int a,b,c;
+         cin>>a>>b>>c;
+         long long int maxval=2* abs(a-b);
+         if(maxval>=a&&maxval>=b&&maxval>=c){
+            long long int temp=2*abs(a-b);
+            long long int ans=abs(a-b)+c;
+            if(ans%temp==0){
+                cout<<temp<<endl;
+            }
+            else{
+                cout<<ans%temp<<endl;
+            }
+         }
+         else{
+            cout<<"-1\n";
+         }
+    }
+}

Problems/Mathematics/Day-02/sol/Lavay Garg/Solution1.cpp

@@ -0,0 +1,74 @@
+/*## 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`
+
+---*/
+#include<bits/stdc++.h>
+using namespace std;
+int main(){
+    int t;
+    cin>>t;
+    while(t--){
+        int n;
+        cin>>n;
+        long long a[n];
+        for(int i=0;i<n;i++){
+            cin>>a[i];
+        }
+        sort(a,a+n);
+        if (a[0] == a[n - 1]) {
+            cout << 1LL*n*(n-1)<< "\n";
+            continue;
+        }
+        long long maxdiff=a[n-1]-a[0];
+        long long min=0;
+        long long max=0;
+        for(int i=0;i<n;i++){
+            if(a[i]==a[0]){
+                min++;
+            }
+            if(a[i]==a[n-1]){
+                max++;
+            }
+           
+        }
+        cout<<2LL*max*min<<endl;
+    }
+}
+//Time Complexity=O(t*nlogn+t*n)
+//space complexity=O(n)

Problems/Mathematics/Day-02/sol/Lavay Garg/run

@@ -0,0 +1,100 @@
+/*
+Problem Statement:
+Hossam and Hazem want to count the number of interesting pairs (i, j) such that:
+1. 1 <= i, j <= n
+2. i != j
+3. |a[i] - a[j]| == max(|a[p] - a[q]|) for all 1 <= p, q <= n.
+
+---------------------------------------
+Approach:
+1.  Find the maximum and minimum values in the array. Let them be maxi and mini.
+2.  The maximum absolute difference is diff = maxi - mini.
+3.  We need to find pairs (a[i], a[j]) such that |a[i] - a[j]| = diff.
+4.  Case 1: If maxi == mini (all elements are same), then any pair (i, j) with i != j works.
+    The count is n * (n - 1).
+5.  Case 2: If maxi != mini, we need one element to be maxi and the other to be mini.
+    Count the occurrences of maxi (cntmax) and mini (cntmin).
+    The number of pairs is 2 * cntmin * cntmax (since order matters.
+
+----------------------------------------
+Time Complexity: O(n) - find min, max and their counts.
+Space Complexity: O(n) - store the input array.
+
+----------------------------------------
+Example:
+Input:
+5
+6 2 3 8 1
+
+Process:
+1. mini = 1, maxi = 8.
+2. diff = 7.
+3. cntmin (1) = 1.
+4. cntmax (8) = 1.
+5. Result = 2 * 1 * 1 = 2.
+
+Output:
+2
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+#define ll long long
+#define vll vector<ll>
+#define out(x) cout << (x) << endl
+#define outr(x)            \
+    {                      \
+        cout << x << endl; \
+        return;            \
+    }
+
+void solve()
+{
+    ll n;
+    cin >> n;
+    vll a(n);
+    int mini = INT_MAX;
+    int maxi = INT_MIN;
+
+    for (int i = 0; i < n; i++)
+    {
+        cin >> a[i];
+        if (a[i] < mini)
+            mini = a[i];
+        if (a[i] > maxi)
+            maxi = a[i];
+    }
+
+    if (mini == maxi)
+    {
+        cout <<  n * (n - 1) << endl;
+        return;
+    }
+
+    ll cntmin = 0;
+    ll cntmax = 0;
+    for (int x : a)
+    {
+        if (x == mini)
+            cntmin++;
+        if (x == maxi)
+            cntmax++;
+    }
+
+    cout << 2*cntmin*cntmax << endl;
+}
+
+int main()
+{
+    ios_base::sync_with_stdio(false);
+    cin.tie(NULL);
+    int t;
+    cin >> t;
+    while (t--)
+    {
+        solve();
+    }
+    return 0;
+}
+
+//submission link: https://codeforces.com/problemset/submission/1771/355269691

Problems/Mathematics/Day-02/sol/Satwik_Santosh/Solution1.cpp

@@ -53,6 +53,9 @@
 |Priyanshu Ranjan        | justpriyanshu            | BIT Mesra
 | Vrajkumar Shah        | vraj826                   | DDU, Nadiad
 | Atharva Mehta          | Atharva-insane           |  IIIT Allahabad                                                  |                                               
-| Kushagra Raghuvanshi   | Taskmaster-afk          | IIIT Allahabad                                                   |
+| Kushagra Raghuvanshi   | Taskmaster-afk          | IIIT Allahabad                                                   |                                                | 
+| Apoorv Mittal          | Apoorv012                |  JIIT Noida                                                  |                                               
+|Aman Kumar Mehta       |AmanMehta22                |Amity University Jharkhand                                         |                                                                                             |     
+| Naman Pal              | Naman2251                | IIIT Allahabad
 <!-- ADD ABOVE THIS -->
 <!-- example | Korvac | Betty | Reyansh College | -->