Merge pull request #505 from Abh-igyan/patch-3

Add Solution2.cpp for binary array median problem

Author: Dheeraj-06

Problems/Mathematics/Day-08/sol/Abhigyan Tiwari/Solution2.cpp

@@ -0,0 +1,83 @@
+//submission: https://codeforces.com/problemset/submission/1999/356309382
+// P.S.: Given a binary array arr of length n and an  integer k. Need to 
+// find sum of median of all subsequences of arr. Ans. should be modulo 1e9+7
+
+#include <bits/stdc++.h>
+#define ll long long 
+
+using namespace std;
+
+const int MOD = 1e9+7;
+// The precomputation is O(N) and each test case is O(K)
+ll power(ll base, ll exp) {
+    ll res = 1;
+    base %= MOD;
+    while (exp > 0) {
+        if (exp % 2 == 1) res = (res * base) % MOD;
+        base = (base * base) % MOD;
+        exp /= 2;
+    }
+    return res;
+}
+
+ll modInverse(ll n) {
+    return power(n, MOD - 2);
+}
+
+const int MAX = 200005;
+ll fact[MAX], invFact[MAX];
+
+void precompute() {
+    fact[0] = 1;
+    for (int i = 1; i < MAX; i++) fact[i] = (fact[i - 1] * i) % MOD;
+    invFact[MAX - 1] = modInverse(fact[MAX - 1]);
+    for (int i = MAX - 2; i >= 0; i--) invFact[i] = (invFact[i + 1] * (i + 1)) % MOD;
+}
+
+ll nCr(int n, int r) {
+    if (r < 0 || r > n) return 0;
+    ll num = fact[n];
+    ll den = (invFact[r] * invFact[n - r]) % MOD;
+    return (num * den) % MOD;
+}
+
+void solve() {
+    int n, k;
+    cin>>n>>k;
+    int c1 = 0;
+    for (int i = 0; i < n; i++) {
+        int x;
+        cin >> x;
+        if (x == 1) c1++;
+    }
+    int c0 = n - c1;
+
+    ll ans = 0;
+    int min_ones = (k + 1) / 2;
+
+    // Median is 1 if we pick i ones where i >= (k+1)/2
+    for (int i = min_ones; i <= k; i++) {
+        if (i <= c1 && (k - i) <= c0) {
+            ll ways = (nCr(c1, i) * nCr(c0, k - i)) % MOD;
+            ans = (ans + ways) % MOD;
+        }
+    }
+    cout << ans << "\n";
+}
+// T.C.: O(MAX+T*K)
+// S.C.: O(MAX)
+
+int main() {
+    // Fast I/O
+    ios_base::sync_with_stdio(false);
+    cin.tie(NULL);
+    
+    precompute();
+    
+    int t;
+    if (!(cin >> t)) return 0;
+    while (t--) {
+        solve();
+    }
+    return 0;
+}