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Author: capricemoto

Problems/Mathematics/Day-08/sol/Samarth/sol1.cpp

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+#include <bits/stdc++.h>
+#define endl '\n'
+#define in(a) for(int i = 0; i<a.size(); i++) cin>>a[i];
+typedef long long ll;
+
+using namespace std;
+
+/*
+PROBLEM STATEMENT:
+Sasha wants to buy an apartment on a street where the houses are numbered from 1 to 109
+
+from left to right.
+
+There are n
+bars on this street, located in houses with numbers a1,a2,…,an
+
+. Note that there might be multiple bars in the same house, and in this case, these bars are considered distinct.
+
+Sasha is afraid that by the time he buys the apartment, some bars may close, but no more than k
+
+bars can close.
+
+For any house with number x
+, define f(x) as the sum of |x−y| over all open bars y
+
+(that is, after closing some bars).
+
+Sasha can potentially buy an apartment in a house with number x
+(where 1≤x≤109) if and only if it is possible to close at most k bars so that after that f(x)
+
+becomes minimal among all houses.
+
+Determine how many different houses Sasha can potentially buy an apartment in.
+
+
+APPROACH:
+the median is the answer(if n is even then elemnts btwn & including n/2 &n/2+1) if no bars are closed;
+and we observe a pattern that if even bars are closed in even n the range increases to n/2-k/2 to n/2+1 +k/2;
+and if odd bars are closed in odd n then range becomes (n+1)/2-(k+1)/2 to (n+1)/2 +(k+1)/2;
+so we just implement the same ;
+
+
+TIME COMPLEXITY:O(nlogn) (to sort);
+
+Submission link:
+https://codeforces.com/contest/2098/submission/356268443
+
+
+*/
+
+
+
+
+void solve(){
+	int n,k;cin>>n>>k;vector<int>a(n);in(a);
+	sort(a.begin(),a.end());
+	if(n&1){
+		cout<<a[(n+1)/2+(k+1)/2-1]-a[(n+1)/2-(k+1)/2-1]+1<<endl;
+	}
+	else{
+		cout<<a[n/2+k/2]-a[n/2-k/2-1]+1<<endl;
+	}
+}
+
+int main(){
+    ios::sync_with_stdio(0); cin.tie(0);
+	int t;cin>>t;
+	while(t--)
+		solve();
+	
+}