Fix: Solved Mathematics Day-02 q001 (Issue #185)

Author: MK-codes365

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

@@ -0,0 +1,52 @@
+/*
+Problem: counting pairs where absolute difference equals max possible difference in array
+
+My approach:
+basically max diff is always gonna be max_element - min_element
+so we just need pairs where one is max and other is min
+first find both values then count how many times each appears
+if everything is same number then answer is just n*(n-1) (all pairs work)
+otherwise its 2*count_max*count_min (since both (i,j) and (j,i) count as different)
+
+TimeComplexity: O(n) 
+SpaceC: O(n)
+
+CodeForces submission: https://codeforces.com/contest/1771/submission/355388765
+*/
+
+#include<bits/stdc++.h>
+using namespace std;
+
+int main(){
+    int t;
+    cin >> t;
+    
+    while(t--){
+        int n;
+        cin >> n;
+        vector<int> arr(n);
+        
+        for(int i=0; i<n; i++){
+            cin >> arr[i];
+        }
+        int maxi = arr[0], mini = arr[0];
+        for(int i=1; i<n; i++){
+            if(arr[i] > maxi) maxi = arr[i];
+            if(arr[i] < mini) mini = arr[i];
+        }
+        if(maxi == mini){
+            long long result = (long long)n * (n-1);
+            cout << result << endl;
+            continue;
+        }
+        long long max_count = 0, min_count = 0;
+        for(int i=0; i<n; i++){
+            if(arr[i] == maxi) max_count++;
+            if(arr[i] == mini) min_count++;
+        }
+        long long answer = 2 * max_count * min_count;
+        cout << answer << endl;
+    }
+    
+    return 0;
+}