Solution2 Day2

Author: sahsudhanshu

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

@@ -0,0 +1,74 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+long long MOD = 1e9 + 7;
+long long fact[1005], invfact[1005];
+
+long long modpow(long long a, long long b)
+{
+    long long res = 1;
+    while (b)
+    {
+        if (b & 1)
+            res = res * a % MOD;
+        a = a * a % MOD;
+        b >>= 1;
+    }
+    return res;
+}
+
+long long nCr(int n, int r)
+{
+    if (r < 0 || r > n)
+        return 0;
+    return fact[n] * invfact[r] % MOD * invfact[n - r] % MOD;
+}
+
+int main()
+{
+    ios::sync_with_stdio(false);
+    cin.tie(NULL);
+
+    fact[0] = 1;
+    for (int i = 1; i <= 1000; i++)
+        fact[i] = fact[i - 1] * i % MOD;
+
+    invfact[1000] = modpow(fact[1000], MOD - 2);
+    for (int i = 999; i >= 0; i--)
+        invfact[i] = invfact[i + 1] * (i + 1) % MOD;
+
+    int t;
+    cin >> t;
+    while (t--)
+    {
+        int n, k;
+        cin >> n >> k;
+        vector<int> a(n);
+        for (int i = 0; i < n; i++)
+            cin >> a[i];
+
+        sort(a.begin(), a.end(), greater<int>());
+
+        int x = a[k - 1];
+
+        int total = 0, need = 0;
+        for (int i = 0; i < n; i++)
+            if (a[i] == x)
+                total++;
+        for (int i = 0; i < k; i++)
+            if (a[i] == x)
+                need++;
+
+        cout << nCr(total, need) << "\n";
+    }
+    return 0;
+}
+
+/*
+    Approach
+    - precompution factorials and inverse factorials
+    - in the question it can be seen that max value can be only be attain my specific no. only after decreasing order
+    - now it is possible there is a number which mai bi repeat multiple times so calculate the no. of ways using PnC using the precomputated factorials
+    
+    - TC : O(nlogn) SC: O(n)
+*/