solved first question of day2

Author: ishantripathi64

Problems/Mathematics/Day-02/sol/Ishantripathi/solution1.cpp

@@ -0,0 +1,110 @@
+#include <bits/stdc++.h>
+
+#include <ext/pb_ds/assoc_container.hpp>
+#include <ext/pb_ds/tree_policy.hpp>
+
+using namespace std;
+using namespace __gnu_pbds;
+
+typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds; // find_by_order, order_of_key
+// less makes it sorted greater makes it reverse sorted less equal makes it like multiset
+// we can also change the data type of the pbds
+
+// Fast I/O
+#define FAST_IO                  \
+    ios::sync_with_stdio(false); \
+    cin.tie(NULL);
+#define pb push_back
+#define in insert
+#define int long long
+#define all(x) (x).begin(), (x).end()
+
+// Typedefs for convenience
+typedef vector<int> vi;
+typedef pair<int, int> pii;
+
+// Constants
+const int INF = 1e18;
+const int MOD = 1e9 + 7;
+
+// Debug (can be disabled in contests)
+#ifdef DEBUG
+#define dbg(x) cerr << #x << " = " << x << '\n';
+#else
+#define dbg(x)
+#endif
+
+// gcd
+int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
+
+// lcm
+int lcm(int a, int b)
+{
+    if (a == 0 || b == 0)
+        return 0;
+    return abs(a * b) / gcd(a, b);
+}
+
+// Binary exponentiation (modular)
+long long binexp(long long a, long long b, long long m)
+{
+    a %= m;
+    long long res = 1;
+    while (b > 0)
+    {
+        if (b & 1)
+            res = res * a % m;
+        a = a * a % m;
+        b >>= 1;
+    }
+    return res;
+}
+
+void solve()
+{
+    int n;
+    cin >> n;
+    vi v(n);
+    for (int i = 0; i < n; i++)
+    {
+        cin >> v[i];
+    }
+    int mx = *max_element(all(v)), mi = *min_element(all(v));
+    int cnt = 0, cnt2 = 0;
+    for (int i = 0; i < n; i++)
+    {
+        if (v[i] == mx)
+            cnt++;
+        if (v[i] == mi)
+            cnt2++;
+    }
+    if (mi == mx)
+    {
+        cout << (cnt * (cnt - 1)) << endl;
+    }
+    else
+    {
+        cout << 2 * (cnt * cnt2) << endl;
+    }
+}
+
+signed main()
+{
+    FAST_IO
+    int t = 1;
+    cin >> t;
+    while (t--)
+        solve();
+    return 0;
+}
+/*
+    so basically we're asked in this question to compute the pairs whose diff is max(abs(ap-aq)) where p and q varies over all n
+    now a simple thought would tell us that we need to count number of pairs such that one is max and other is min since max diff
+    is only occured when substracting max - min
+
+    time complexity is o(n);
+    space complexity is o(1);
+
+    submission id https://codeforces.com/contest/1771/submission/355258380
+
+*/