Merge branch 'opencodeiiita:main' into issue2

Author: Humanshere

Problems/Binary-Search/Day-06/Samarth/sol1.cpp

@@ -0,0 +1,75 @@
+#include <bits/stdc++.h>
+#define endl '\n'
+typedef long long ll;
+
+using namespace std;
+
+/*
+PROBLEM STATEMENT:
+There are n people living in the town. Just now, the wealth of the i-th person was ai
+
+gold. But guess what? The richest person has found an extra pot of gold!
+
+More formally, find an aj=max(a1,a2,…,an)
+, change aj to aj+x, where x
+
+is a non-negative integer number of gold found in the pot. If there are multiple maxima, it can be any one of them.
+
+A person is unhappy if their wealth is strictly less than half of the average wealth∗
+
+.
+
+If strictly more than half of the total population n
+
+are unhappy, Robin Hood will appear by popular demand.
+
+Determine the minimum value of x
+for Robin Hood to appear, or output −1 if it is impossible.
+
+
+APPROACH:
+just following the conditions of the ques after sorting the array we get
+that x>2*n*a[mid]- sum where mid=n/2 and sum is sum of elements of array;
+the reason to sort them was that if the largest element satisfied the condition then all the subsequent
+smaller will do so automatically;
+and if n==1||n==2 ans would -1 reason simple most of them will be richer
+and if x is less than zero ans zero.
+
+
+TIME COMPLEXITY:O(n +nlogn(for sorting))so O(nlogn);
+SPACE COMPLEXITY:O(n);
+
+
+
+SUBMISSION LINK:https://codeforces.com/contest/2014/submission/356057548
+
+
+*/
+
+
+
+
+
+ll solve(){
+	int n;cin>>n;int a[n];ll sum=0;
+	for(int i=0;i<n;i++){
+		cin>>a[i];sum+=a[i];
+	}
+	if(n==1||n==2)return -1;
+	else{
+		sort(a,a+n);
+		int mid=n/2;
+		ll ans=(ll)2*n*a[mid]-sum+1;
+		if(ans>0)return ans;
+		else return 0;
+	}
+}
+
+int main(){
+    int t;cin>>t;
+    while(t--){
+    	cout<<solve()<<endl;
+    }
+
+	
+}

Problems/Binary-Search/Day-06/sol/Aiyaan_Mahajan/Q1sol.cpp

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+/*
+DAY 6
+Q1 Robin hood in the town
+.*/
+/*
+We need more than half the people to have wealth
+< half of the new average.
+
+Adding gold only increases the average, so poor people become “unhappy” first.
+
+So we find the smallest 
+𝑥
+x such that the poorest majority are still below:
+half-average= (S + x)/2*n
+
+Solve for 
+x, and if it’s negative → answer is 0.
+If 𝑛≤2
+ it’s impossible → -1.
+
+That’s it.
+
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+  
+    int t;
+    cin >> t;
+    while (t--) {
+        int n;
+        cin >> n;
+        vector<long long> a(n);
+        long long S = 0;
+        for (auto &x : a) {
+            cin >> x;
+            S += x;
+        }
+
+        if (n <= 2) {
+            cout << -1 << '\n';
+            continue;
+        }
+
+        sort(a.begin(), a.end());
+        int m = n / 2 + 1;
+        long long v = a[m - 1];
+
+        long long x = 2LL * n * v - S + 1;
+        if (x < 0) x = 0;
+
+        cout << x << '\n';
+    }
+    return 0;
+}
+
+
+
+
+
+//TC =O(Nlog(N)) and SC = O(N)
+/*
+My submission : https://codeforces.com/contest/2014/submission/356052582
+*/

Problems/Binary-Search/Day-06/sol/Himansh_Arora/Solution1.cpp

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+//problem link- https://codeforces.com/contest/2014/problem/C
+//submission link- https://codeforces.com/contest/2014/submission/356102395
+//approach- (sum+x)/2*n needs to be greater than median for valid x
+//          so I wrote binary search to find minimum x such that sum+x>median*2*n 
+//          as x will follow predicate function(FFFFTTTTTT)
+//Time complexity- O(nlogn)
+
+
+
+
+
+
+
+
+
+#include <bits/stdc++.h>
+using namespace std;
+#define endl '\n'
+#define sp << " " <<
+#define int long long
+#define fora(a) for(auto i:a)
+#define gcd(a,b) __gcd(a,b)
+#define lcm(a,b) (a*(b/gcd(a,b)))
+#define forn(i,n) for(int i=0; i<n; i++)
+#define printv(a) {for(auto u:a) cout<<u<<" "; cout<<endl;}
+#define inputv(a) {for(int i=0;i<n;i++) cin>>a[i];}
+#define ll long long
+#define vi vector<int>
+#define vll vector<long long>
+#define pii pair<int, int>
+#define pll pair<long long, long long>
+#define all(x) x.begin(),x.end()
+
+void fast_io() {
+    ios_base::sync_with_stdio(false);
+    cin.tie(NULL);
+    cout.tie(NULL);
+}
+
+
+const int MOD = 1e9 + 7;
+ll addMod(ll a, ll b) { return (a + b) % MOD; }
+ll subMod(ll a, ll b) { return (a - b + MOD) % MOD; }
+ll mulMod(ll a, ll b) { return (a * b) % MOD; }
+
+ll powerMod(ll a, ll b) {
+    ll res = 1;
+    while (b > 0) {
+        if (b & 1) res = mulMod(res, a);
+        a = mulMod(a, a);
+        b >>= 1;
+    }
+    return res;
+}
+
+
+
+
+
+void solve() {
+    int n;
+    cin>>n;
+    vi a(n);
+    inputv(a);
+    sort(all(a));
+    int sum=accumulate(all(a),0LL);
+    if(n==1||n==2){cout<<-1<<endl;return;}
+    int low=0;
+    int high=2*n*a[n/2];
+    while(high-low>=0){
+        int mid=low+(high-low)/2;
+        if(sum+mid>a[n/2]*2*n){
+            high=mid-1;
+        }else{
+            low=mid+1;
+        }
+    }
+
+
+     cout<<low<<endl;
+}
+
+
+
+
+
+signed main() {
+    fast_io();
+    
+    int t;
+    cin >> t;
+    
+    while (t--) {
+        solve();
+    }
+    
+    return 0;
+}