Merge pull request #338 from sahsudhanshu/Sol1Day2

Sol1 day2

Author: Dheeraj-06

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

@@ -0,0 +1,65 @@
+#include <bits/stdc++.h>
+typedef long long ll;
+typedef long double ld;
+using namespace std;
+
+int main()
+{
+    ios::sync_with_stdio(false);
+    cin.tie(NULL);
+
+    ll t;
+    cin >> t;
+    while (t--)
+    {
+        ll n;
+        cin >> n;
+        vector<ll> v(n);
+
+        for (ll i = 0; i < n; i++)
+            cin >> v[i];
+
+        sort(v.begin(), v.end());
+
+        ll low = v[0];
+        ll high = v[n - 1];
+
+        ll cLow = 0;
+        ll cHigh = 0;
+
+        for (auto i : v)
+        {
+            if (i != low)
+                break;
+            cLow++;
+        }
+        reverse(v.begin(), v.end());
+
+        for (auto i : v)
+        {
+            if (i != high)
+                break;
+            cHigh++;
+        }
+
+        cout << ((low == high) ? (cLow * (cLow - 1)) : (cLow * cHigh * 2)) << endl;
+    }
+
+    return 0;
+}
+
+/*
+APPROACH:
+
+- Sort the array to easily identify minimum and maximum elements
+- Find the minimum (low) and maximum (high) values
+- Count occurrences of minimum value (cLow) and maximum value (cHigh)
+- If all elements are equal (low == high), we need to choose any 2 elements from n,
+    so answer = n * (n-1) (dividing by 2 would give combinations, but we want pairs)
+- Otherwise, the answer is cLow * cHigh * 2 (considering both orderings of pairs)
+
+TC: O(nlog n)
+SC: O(n)
+
+https://codeforces.com/contest/1771/submission/355619015
+*/