Given two integers P denoting the number of people and a positive integer K. Each person is assigned with the same job which takes exactly K hours to finish. A condition is given that all of them can start performing their jobs exactly in X hours of interval. Count the number of persons still performing the job whenever a prior person finishes his/her job. The task is to find the sum of such counts.
Examples:
Input: P = 4, K = 6, X = 3
Output: 5
Explanation: Let the four persons be P1, P2, P3, P4
- P1 starts at 0 and finishes at 6
- P2 starts at 3 and finishes at 9
- P3 starts at 6 and finishes at 12
- P4 starts at 9 and finishes at 15
So, when P1 finishes, P2 and P3 started performing their respective job, count for P1 = 2
when P2 finishes, P3 and P4 started performing their respective job, count for P2 = 2
when P3 finishes, only P4 started performing the job, count for P3 = 1
when P4 finishes, there is no person who starts at this point, so count for P4 = 0
Therefore, Total counts = (2 + 2 + 1 + 0) = 5Input: P = 9, K = 72, X = 8
Output: 36
Approach: The given problem can be solved by analyzing the problem with the concept of math. Follow the steps below to solve the problem:
- Find the minimum of total persons excluding the first one(because P1 always starts at 0) and K/X, store it in a variable say a.
- Check if a is equal to 0
- if yes, return 0.
- else, calculate the total sum of count by making mathematical formula.
- Return the final sum of count as the required answer.
Below is the implementation of the above approach:
C++
// C++ implementation for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the sum of count// of persons that are still// performing the jobint countPersons(int P, int K, int X){ // Find the minimum of total persons // excluding the first one and K/X int a = min(P - 1, K / X); // If a is equal to 0, return 0 if (a == 0) { return 0; } // Else calculate the total sum of // count by the making a formula int ans = max(0, a * (a - 1) / 2) + a * (P - a); // Return the sum of count return ans;}// Driver Codeint main(){ int P = 22, K = 9, X = 1; cout << countPersons(P, K, X); return 0;} |
Java
// Java program for the above approachimport java.util.*;public class GFG {// Function to find the sum of count// of persons that are still// performing the jobstatic int countPersons(int P, int K, int X){ // Find the minimum of total persons // excluding the first one and K/X int a = Math.min(P - 1, K / X); // If a is equal to 0, return 0 if (a == 0) { return 0; } // Else calculate the total sum of // count by the making a formula int ans = Math.max(0, a * (a - 1) / 2) + a * (P - a); // Return the sum of count return ans;}// Driver Codepublic static void main(String args[]){ int P = 22, K = 9, X = 1; System.out.println(countPersons(P, K, X));}}// This code is contributed by Samim Hossain Mondal. |
Python3
# Python program for the above approach# Function to find the sum of count# of persons that are still# performing the jobdef countPersons(P, K, X): # Find the minimum of total persons # excluding the first one and K/X a = min(P - 1, K / X) # If a is equal to 0, return 0 if a == 0: return 0 # Else calculate the total sum of # count by the making a formula ans = max(0, a * (a - 1) / 2) # Return the sum of count return ans + a * (P - a)# Driver Codeif __name__ == "__main__": P = 22 K = 9 X = 1 print(int(countPersons(P, K, X)))# This code is contributed by Potta Lokesh |
C#
// C# program for the above approachusing System;class GFG {// Function to find the sum of count// of persons that are still// performing the jobstatic int countPersons(int P, int K, int X){ // Find the minimum of total persons // excluding the first one and K/X int a = Math.Min(P - 1, K / X); // If a is equal to 0, return 0 if (a == 0) { return 0; } // Else calculate the total sum of // count by the making a formula int ans = Math.Max(0, a * (a - 1) / 2) + a * (P - a); // Return the sum of count return ans;}// Driver Codepublic static void Main(){ int P = 22, K = 9, X = 1; Console.Write(countPersons(P, K, X));}}// This code is contributed by Samim Hossain Mondal. |
Javascript
<script>// Javascript implementation for the above approach// Function to find the sum of count// of persons that are still// performing the jobfunction countPersons(P, K, X){ // Find the minimum of total persons // excluding the first one and K/X let a = Math.min(P - 1, K / X); // If a is equal to 0, return 0 if (a == 0) { return 0; } // Else calculate the total sum of // count by the making a formula let ans = Math.max(0, a * (a - 1) / 2) + a * (P - a); // Return the sum of count return ans;}// Driver Codelet P = 22, K = 9, X = 1;document.write(countPersons(P, K, X));// This code is contributed by Samim Hossain Mondal.</script> |
153
Time Complexity: O(1)
Auxiliary Space: O(1)
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