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Sum of count of persons still performing the job whenever a person finishes job

  • Difficulty Level : Easy
  • Last Updated : 18 Nov, 2021

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) = 5

Input: 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 job
int 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 Code
int main()
{
    int P = 22, K = 9, X = 1;
 
    cout << countPersons(P, K, X);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
public class GFG {
 
// Function to find the sum of count
// of persons that are still
// performing the job
static 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 Code
public 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 job
def 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 Code
if __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 approach
using System;
class GFG {
 
// Function to find the sum of count
// of persons that are still
// performing the job
static 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 Code
public 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 job
function 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 Code
let P = 22, K = 9, X = 1;
document.write(countPersons(P, K, X));
 
// This code is contributed by Samim Hossain Mondal.
</script>
Output
153

Time Complexity: O(1) 
Auxiliary Space: O(1)


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