day 2 solution 2

Author: Aaryan-Degama

Problems/Mathematics/Day-02/sol/Aaryan_Degama/Solution2.cpp

@@ -0,0 +1,87 @@
+// Submission link: https://codeforces.com/contest/1475/submission/355643230
+
+#include <bits/stdc++.h>
+using namespace std;
+
+#define fastio ios::sync_with_stdio(false); cin.tie(NULL);
+
+using ll = long long;
+const ll MOD = 1e9 + 7;
+
+ll binexp(ll a, ll b) {
+    ll res = 1;
+    a %= MOD;
+    while (b) {
+        if (b & 1) res = res * a % MOD;
+        a = a * a % MOD;
+        b >>= 1;
+    }
+    return res;
+}
+
+vector<ll> fac(1001);
+
+ll nCr(ll n, ll r) {
+    if (r < 0 || r > n) return 0;
+    return fac[n] * binexp(fac[r], MOD - 2) % MOD * binexp(fac[n - r], MOD - 2) % MOD;
+}
+
+void solve() {
+    ll n, k;
+    cin >> n >> k;
+    vector<ll> v(n);
+    for (ll &x : v) cin >> x;
+
+    sort(v.begin(), v.end());
+
+    ll ele = v[n - k];
+    ll total = 0;
+
+    for (ll x : v) {
+        if (x == ele) total++;
+    }
+
+    ll taken = 0;
+    for (ll i = n - k; i < n; i++) {
+        if (v[i] == ele) taken++;
+    }
+
+    cout << nCr(total, taken) << "\n";
+}
+
+int main() {
+    fastio
+
+    fac[0] = 1;
+    for (int i = 1; i <= 1000; i++) {
+        fac[i] = fac[i - 1] * i % MOD;
+    }
+
+    int t;
+    cin >> t;
+    while (t--) solve();
+    return 0;
+}
+
+
+/*
+Explaination : 
+
+To maximize the sum, we must always select the k largest elements of the array.
+Any selection that excludes one of the k largest elements will produce a smaller sum.
+If all elements were distinct, the answer would be 1, since the k largest elements
+are uniquely determined. However, when duplicate values exist, multiple selections
+can result in the same maximum sum.
+
+Approach:
+1. Sort the array in non-decreasing order.
+2. The k largest elements are the last k elements of the sorted array.
+3. Let 'ele' be the smallest element among these k elements (i.e., ele = v[n - k]).
+4. Some elements larger than 'ele' are forced to be chosen, but occurrences of 'ele'
+   introduce flexibility.
+
+
+Time Complexity :  O(n log(n))
+Space Complexity : O(n)
+
+*/