Merge pull request #247 from icyfire8/patch-2

Add Solution1.cpp for interesting pairs problem

Author: Dheeraj-06

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

@@ -0,0 +1,61 @@
+/*https://codeforces.com/contest/1771/submission/355306957
+PROBLEM STATEMENT:
+Hossam woke up bored, so he decided to create an interesting array with his friend Hazem.
+
+They are given an array 'a' of 'n' positive integers.
+Hossam chooses an element a[i] and Hazem chooses an element a[j].
+
+An ordered pair (a[i], a[j]) is considered interesting if it satisfies:
+1) 1 ≤ i, j ≤ n
+2) i ≠ j
+3) The absolute difference |a[i] − a[j]| is equal to the maximum absolute
+   difference among all possible pairs in the array, i.e.,
+   |a[i] − a[j]| = max₁≤p,q≤n |a[p] − a[q]|
+
+Input:
+- The first line contains an integer t (1 ≤ t ≤ 100), representing the number of test cases.
+- For each test case:
+  - The first line contains an integer n (2 ≤ n ≤ 10⁵).
+  - The second line contains n integers a₁, a₂, …, aₙ (1 ≤ aᵢ ≤ 10⁵).
+
+Constraints:
+- The total sum of n across all test cases does not exceed 10⁵.
+
+Output:
+- For each test case, output the number of interesting pairs (a[i], a[j]).
+*/
+
+//T.C.:O(t.nlogn)
+//S.C.:O(n)
+
+#include <bits/stdc++.h>
+#include <vector>
+using namespace std;
+
+int main() {
+    long long t, n, i, j, f, e, diff;
+	cin >> t;  
+
+    while(t--){
+        cin >> n;
+        vector <long long> v(n);
+        
+        for(i=0; i<n; i++)  cin >> v[i];
+        
+        f=0, e=0;
+        
+        sort(v.begin(),v.end()); //sorting the array from ascending to descending
+        diff = v[n-1] - v[0];  //maximum absolute difference
+        
+        if(diff == 0)   cout << n*(n-1) << endl; //if same element is present throughout the array  
+        
+        else    {
+            for(i=0; i<n; i++){
+                if(v[i] == v[0])    f++; //counting no. of smallest duplicate elements
+                if(v[i] == v[n-1])    e++; //counting no. of largest duplicate elements
+            }
+            
+            cout << 2*f*e << endl; //2 times the no. of pairs
+        }
+    }
+}