🚀 22-Sep-2020

Author: sureshmangs

Diff for: competitive programming/hackerearth/Can you Guess.cpp

@@ -0,0 +1,71 @@
+No problem statement.
+
+Find the logic from the given sample input/output.
+
+And answer Q queries.
+
+Constraints :
+
+1 <= Value <= 100000
+1<=nunber of query<=10000
+
+SAMPLE INPUT
+8
+10
+30
+45
+9
+69
+77
+127
+150
+SAMPLE OUTPUT
+8
+42
+33
+4
+27
+19
+1
+222
+
+
+
+
+
+
+
+
+
+
+#include<bits/stdc++.h>
+using namespace std;
+
+int solve(int n){
+    int sum=0;
+    for(int i=1;i*i<=n;i++){
+        if(n%i==0){
+            sum+=i;
+            if(i!=n/i)
+            sum+=n/i;
+        }
+    }
+    sum-=n;
+    return sum;
+}
+
+
+int main(){
+    ios_base::sync_with_stdio(false);
+    cin.tie(NULL);
+
+    int t;
+    cin>>t;
+    while(t--){
+        int n;
+        cin>>n;
+        cout<<solve(n)<<endl;
+    }
+
+    return 0;
+}

Diff for: competitive programming/hackerearth/basic_number_theory_Can you Guess.cpp

@@ -0,0 +1,71 @@
+No problem statement.
+
+Find the logic from the given sample input/output.
+
+And answer Q queries.
+
+Constraints :
+
+1 <= Value <= 100000
+1<=nunber of query<=10000
+
+SAMPLE INPUT
+8
+10
+30
+45
+9
+69
+77
+127
+150
+SAMPLE OUTPUT
+8
+42
+33
+4
+27
+19
+1
+222
+
+
+
+
+
+
+
+
+
+
+#include<bits/stdc++.h>
+using namespace std;
+
+int solve(int n){
+    int sum=0;
+    for(int i=1;i*i<=n;i++){
+        if(n%i==0){
+            sum+=i;
+            if(i!=n/i)
+            sum+=n/i;
+        }
+    }
+    sum-=n;
+    return sum;
+}
+
+
+int main(){
+    ios_base::sync_with_stdio(false);
+    cin.tie(NULL);
+
+    int t;
+    cin>>t;
+    while(t--){
+        int n;
+        cin>>n;
+        cout<<solve(n)<<endl;
+    }
+
+    return 0;
+}

Diff for: competitive programming/hackerearth/basic_number_theory_Love Triangle.cpp

@@ -0,0 +1,52 @@
+Two boys, Venky and Sagar, are at war over a girl, Riu.
+Driven by their feelings, they decided to confess to Riu.
+Since both of them were equally dumb, Riu decided that she would go out with that boy
+ who would successfully find the pattern and thus prove that he is less dumb.
+
+Read input until end of file.
+
+Given : 0<=N<=10^18
+
+SAMPLE INPUT
+1
+2
+3
+10
+100
+SAMPLE OUTPUT
+1
+2
+3
+11
+121
+
+
+
+
+
+
+
+#include<bits/stdc++.h>
+using namespace std;
+
+long long solve(long long n){
+    if(n<9) return n;
+    return (n%9)+10*solve(n/9);
+}
+
+int main(){
+    ios_base::sync_with_stdio(false);
+    cin.tie(NULL);
+
+    long long n;
+    while(cin>>n){
+        long long sum=0;
+        cout<<solve(n)<<endl;
+    }
+
+    return 0;
+}
+
+
+
+

Diff for: competitive programming/hackerearth/basic_number_theory_Mystery.cpp

@@ -0,0 +1,71 @@
+Solve the mystery. Using sample input and output, figure out the logic to solve the question.
+
+Input
+First number is T, the number of test cases. Next T numbers are integers, N.
+
+Output
+Print T lines containing an answer corresponding to the T input numbers.
+
+Constraints
+1<=T<=10^4
+
+1<=N<=10^8
+
+SAMPLE INPUT
+10
+2
+6
+12
+60
+5
+169
+1
+8
+23
+100
+SAMPLE OUTPUT
+2
+4
+6
+12
+2
+3
+1
+4
+2
+9
+
+
+
+
+
+
+
+#include<bits/stdc++.h>
+using namespace std;
+
+int solve(int n){
+    int cnt=0;
+    for(int i=1;i*i<=n;i++){
+        if(n%i==0){
+            cnt++;
+            if(i!=n/i) cnt++;
+        }
+    }
+    return cnt;
+}
+
+int main(){
+    ios_base::sync_with_stdio(false);
+    cin.tie(NULL);
+
+    int t;
+    cin>>t;
+    while(t--){
+        int n;
+        cin>>n;
+        cout<<solve(n)<<endl;
+    }
+
+    return 0;
+}

Diff for: competitive programming/leetcode/1094. Car Pooling.cpp

@@ -0,0 +1,83 @@
+You are driving a vehicle that has capacity empty seats initially available for passengers.  The vehicle only drives east (ie. it cannot turn around and drive west.)
+
+Given a list of trips, trip[i] = [num_passengers, start_location, end_location] contains information about the i-th trip: the number of passengers that must be picked up, and the locations to pick them up and drop them off.  The locations are given as the number of kilometers due east from your vehicle's initial location.
+
+Return true if and only if it is possible to pick up and drop off all passengers for all the given trips.
+
+
+
+Example 1:
+
+Input: trips = [[2,1,5],[3,3,7]], capacity = 4
+Output: false
+Example 2:
+
+Input: trips = [[2,1,5],[3,3,7]], capacity = 5
+Output: true
+Example 3:
+
+Input: trips = [[2,1,5],[3,5,7]], capacity = 3
+Output: true
+Example 4:
+
+Input: trips = [[3,2,7],[3,7,9],[8,3,9]], capacity = 11
+Output: true
+
+
+
+Constraints:
+
+trips.length <= 1000
+trips[i].length == 3
+1 <= trips[i][0] <= 100
+0 <= trips[i][1] < trips[i][2] <= 1000
+1 <= capacity <= 100000
+   Hide Hint #1
+Sort the pickup and dropoff events by location, then process them in order.
+
+
+
+
+
+
+
+
+
+
+
+
+class Solution {
+public:
+    bool carPooling(vector<vector<int>>& trips, int capacity) {
+        vector<int> stops (10001, 0);
+
+        for(auto &t: trips){
+            stops[t[1]]+=t[0];
+            stops[t[2]]-=t[0];
+        }
+
+        int occupied=0;
+
+        for(int i=0;capacity>=0 && i<1001;i++)
+            capacity-=stops[i];
+
+        return capacity>=0;
+    }
+};
+
+
+
+
+
+
+/*
+We only have 1001 stops
+we can just figure out how many people get it and out in each location.
+
+Process all trips, adding passenger count to the start location, and removing it from the end location.
+After processing all trips, a positive value for the specific location tells that we are getting more passengers;
+negative - more empty seats.
+
+Source:   https://leetcode.com/problems/car-pooling/discuss/317611/C%2B%2BJava-O(n)-Thousand-and-One-Stops
+*/
+