day 2 p2

Author: Aaryan-Degama

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

@@ -1,68 +0,0 @@
-// link : https://codeforces.com/problemset/submission/1771/355643992
-
-#include <bits/stdc++.h>
-using namespace std;
-
-#define fastio ios::sync_with_stdio(false); cin.tie(NULL);
-
-using ll = long long;
-
-void solve() {
-    ll n;
-    cin >> n;
-
-    vector<ll> a(n);
-    for (ll i = 0; i < n; i++) {
-        cin >> a[i];
-    }
-
-    sort(a.begin(), a.end());
-
-    ll i = 1;
-    while (i < n && a[i] == a[i - 1]) i++;
-
-    if (i == n) {
-        cout << n * (n - 1) << "\n";
-        return;
-    }
-
-    ll j = 1;
-    while (j < n && a[n - j] == a[n - j - 1]) j++;
-
-    cout << 2LL * i * j << "\n";
-}
-
-int main() {
-    fastio
-    int t;
-    cin >> t;
-    while (t--) solve();
-}
-
-
-/*
-The maximum possible absolute difference in an array is obtained only by
-choosing the minimum and the maximum elements.
-
-So, all valid pairs must consist of:
-one minimum element
-one maximum element
-
-Approach:
-
-1. Sort the array.
-   Minimum elements appear at the beginning.
-   Maximum elements appear at the end.
-
-2. Count how many times the minimum element appears.
-   Let this count be 'i'.
-
-3. Count how many times the maximum element appears.
-   Let this count be 'j'.
-
-
-
-Time Complexity: O(n log n)
-Space Complexity: O(n)
-
-*/