Merge pull request #438 from Naman2251/main

Create day005sol01.cpp

Author: zonedoutdg

Problems/Binary-Search/Day-06/sol/Naman2251/day006sol01.cpp

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+// Submission Link: https://codeforces.com/contest/2014/submission/356013624
+/*
+It is only not possible to call robinhood if n is 2 or 1.
+avg=(sum+k)/n.
+Since we need strictly more than half of the total population strictly less than half of the average wealth
+therefore we would just make our avg equal to a[n/2]+1 as then more than hal would be unhappy.
+(sum+k)/(n*2)=a[n/2]+1
+On solving, k=a[n/2]*n*2+1-sum
+If k<0 then no need of any coin so k=0, therefore final ans is max(k, 0).
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+ 
+int main() {
+    long long t;
+    cin>>t;
+    while(t--) {
+        long long n, sum=0;
+        cin>>n;
+        vector <long long> a(n);
+        for(int i=0; i<n; i++) {
+            cin>>a[i];
+            sum+=a[i];
+        }
+        if(n<=2) {
+            cout<<-1<<endl;
+            continue;
+        }
+        sort(a.begin(), a.end());
+        long long k= a[n/2]*n*2+1-sum;
+        cout<<max(k, 0LL)<<endl;
+    }
+}

Problems/Data-structures/Day-05/sol/Naman2251/day005sol01.cpp

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+//Submission Link: https://cses.fi/problemset/result/15792090/
+
+/*
+we are going through the array ans checking if the number right left to it is smaller to it then answer for that is (i-1) else we will check position ans[i-1]
+if we encounter ans[i]=-1 we know there is no number smaller than that so the ans is -1
+*/
+
+#include<bits/stdc++.h>
+using namespace std;
+void solve(){
+    int n;
+    cin>>n;
+    vector<int> vec(n);
+    for(int i=0;i<n;i++){
+        cin>>vec[i];
+    }
+    vector<int> ans(n);
+    ans[0]=-1;
+    for(int i=1;i<n;i++){
+        if(vec[i]>vec[i-1]){
+            ans[i]=i-1;
+        }else{
+            int a=ans[i-1];
+            bool ok=true;
+            while(a>-1){
+                if(vec[a]<vec[i]){
+                    ans[i]=a;
+                    ok=false;
+                    break;
+                }
+                a=ans[a];
+            }
+            if(ok) {
+                ans[i]=-1;
+            }
+        }
+    }
+    for(int i=0;i<n;i++){
+        cout<<ans[i]+1<<" ";
+    }
+}
+int main(){
+    solve();
+}

Problems/Data-structures/Day-05/sol/Naman2251/day005sol02.cpp

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+Submission Link: https://cses.fi/problemset/result/15799460/
+
+/*
+-> I made a vector which stores pair(pair(arrival, departue), person no.).
+-> I sorted so that person coming first gets room first.
+-> I made a priority queue min-heap(element with lower value have higher priority) and stored element as 
+   pair(departure, room no.)
+-> Now I compared each incoming element with the top one only as it has been emptied first. If a[i].first.first then remove 
+   first element and pushed the incoming element and in ans vector put the room else I would add no. of room and push the element 
+   in pq and in ans vector put the room. Now I printed rm(no. of room) and vector ans(in which room each person stays). 
+*/
+
+
+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+    int n;
+    cin>>n;
+    vector<pair<pair<int,int>,int>>a(n);
+    for(int i=0;i<n;i++) {
+        cin>>a[i].first.first;
+        cin>>a[i].first.second;
+        a[i].second=i;
+    }
+    sort(a.begin(),a.end());
+    priority_queue<pair<int,int>,vector<pair<int,int>>,greater<pair<int,int>>> pq;
+    int rm=1;
+    pq.push({a[0].first.second,rm});
+    vector <int> ans(n);
+    ans[a[0].second]=1;
+    for(int i=1;i<n;i++) {
+        int d=pq.top().first;
+        int r=pq.top().second;
+        if(a[i].first.first>d) {
+            pq.pop();
+            pq.push({a[i].first.second,r});
+            ans[a[i].second]=r;
+        }  
+        else {
+            rm++;
+            pq.push({a[i].first.second,rm});
+            ans[a[i].second]=rm;
+        }
+    }
+    cout<<rm<<endl;
+    for(int i=0;i<n;i++) cout<<ans[i]<<" ";
+    cout<<endl;
+}