Added my solution for q001 day 2

Author: lavaygarg

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

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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`
+
+---*/
+#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)