Count ways to distribute exactly one coin to each worker

Given two arrays coins[] and salaries[] where coins[i] represents the value of ith coin and salaries[j] represents the minimum value of the coin that jth worker will accept. The task is to compute the number of ways to distribute exactly one coin to each worker. Since the answer can be large, print it modulo 109 + 7.
Examples:

Input: coins[] = {1, 2, 3}, salaries[] = {1, 2} 
Output:
Explanation: 
If the coin with value 1 is not used, then the remaining two coins are acceptable by both workers, contributing two possible ways to pay the workers. 
If the coin with value 1 is used, then it can only be used for the first worker. Then either of the remaining coins can be used to pay the second worker. This also contributes to two possible ways. 
Therefore, the four ways to pay the two workers are [2, 3], [3, 2], [1, 2], [1, 3].
Input: coins[] = {1, 2}, salaries[] = {2} 
Output: 1

Approach: The idea is to use the Sorting and Two Pointers technique to solve the problem. Follow the steps below to solve the problem:

  • Sort the salaries in descending order.
  • Let f(i) be the number of coins used to pay the ith worker in the absence of any other workers. Since the salary[] array is sorted so the first worker demands the highest salary and the last worker demands the lowest salary. Therefore, the result is:

\prod_{i=0}^{m-1} (f(i) - i)

  • For the function f(i), i is equal to the number of coins that are available to use to pay the current worker assuming that all previous workers have been paid.
  • As the workers are sorted in non-increasing order of salaries, so any coin used to pay a former worker is guaranteed to be usable to pay the current worker. So the number of coins that can be used to pay the current worker will always be equal to f(i) − i, independent of the previous choices made.
  • To compute f(i) efficiently, use two pointers to track the number of coins that are currently valid.
  • Print the total count of ways after all the above steps.

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
const int MOD = 1000000007;
 
// Function to find number of way to
// distribute coins giving exactly one
// coin to each person
int solve(vector<int>& values,
          vector<int>& salary)
{
    long long ret = 1;
    int amt = 0;
 
    // Sort the given arrays
    sort(values.begin(), values.end());
    sort(salary.begin(), salary.end());
 
    // Start from bigger salary
    while (salary.size()) {
        while (values.size()
               && values.back()
                      >= salary.back()) {
 
            // Increment the amount
            amt++;
            values.pop_back();
        }
        if (amt == 0)
            return 0;
 
        // Reduce amount of valid
        // coins by one each time
        ret *= amt--;
        ret %= MOD;
        salary.pop_back();
    }
 
    // Return the result
    return ret;
}
 
// Driver code
int main()
{
    // Given two arrays
    vector<int> values{ 1, 2 }, salary{ 2 };
 
    // Function Call
    cout << solve(values, salary);
 
    return 0;
}

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Java

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// Java program for
// the above approach
import java.util.*;
class GFG{
 
static int MOD = 1000000007;
 
// Function to find number of way to
// distribute coins giving exactly one
// coin to each person
static int solve(Vector<Integer> values,
                 Vector<Integer> salary)
{
  int ret = 1;
  int amt = 0;
 
  // Sort the given arrays
  Collections.sort(values);
  Collections.sort(salary);
 
  // Start from bigger salary
  while (salary.size() > 0)
  {
    while (values.size() > 0 &&
           values.get(values.size() - 1) >=
           salary.get(salary.size() - 1))
    {
      // Increment the amount
      amt++;
      values.remove(values.size() - 1);
    }
     
    if (amt == 0)
      return 0;
 
    // Reduce amount of valid
    // coins by one each time
    ret *= amt--;
    ret %= MOD;
    salary.remove(salary.size() - 1);
  }
 
  // Return the result
  return ret;
}
 
// Driver code
public static void main(String[] args)
{
  // Given two arrays
  Vector<Integer> values = new Vector<Integer>();
  values.add(1);
  values.add(2);
  Vector<Integer> salary = new Vector<Integer>();
  salary.add(2);
   
  // Function Call
  System.out.print(solve(values, salary));
}
}
 
// This code is contributed by Princi Singh

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Python3

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# Python3 program for the above approach
MOD = 1000000007
 
# Function to find number of way to
# distribute coins giving exactly one
# coin to each person
def solve(values, salary):
     
    ret = 1
    amt = 0
 
    # Sort the given arrays
    values = sorted(values)
    salary = sorted(salary)
 
    # Start from bigger salary
    while (len(salary) > 0):
        while ((len(values) and
                values[-1] >= salary[-1])):
 
            # Increment the amount
            amt += 1
            del values[-1]
 
        if (amt == 0):
            return 0
 
        # Reduce amount of valid
        # coins by one each time
        ret *= amt
        amt -= 1
        ret %= MOD
        del salary[-1]
 
    # Return the result
    return ret
 
# Driver code
if __name__ == '__main__':
     
    # Given two arrays
    values = [ 1, 2 ]
    salary = [2]
 
    # Function call
    print(solve(values, salary))
 
# This code is contributed by mohit kumar 29

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C#

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// C# program for
// the above approach
using System;
using System.Collections;
class GFG{
  
static int MOD = 1000000007;
  
// Function to find number of way to
// distribute coins giving exactly one
// coin to each person
static int solve(ArrayList values,
                 ArrayList salary)
{
  int ret = 1;
  int amt = 0;
  
  // Sort the given arrays
  values.Sort();
  salary.Sort();
  
  // Start from bigger salary
  while (salary.Count > 0)
  {
    while (values.Count > 0 &&
           (int)values[values.Count - 1] >=
           (int)salary[salary.Count - 1])
    {
      // Increment the amount
      amt++;
      values.RemoveAt(values.Count - 1);
    }
      
    if (amt == 0)
      return 0;
  
    // Reduce amount of valid
    // coins by one each time
    ret *= amt--;
    ret %= MOD;
    salary.RemoveAt(salary.Count - 1);
  }
  
  // Return the result
  return ret;
}
  
// Driver code
public static void Main(string[] args)
{
  // Given two arrays
  ArrayList values = new ArrayList();
  values.Add(1);
  values.Add(2);
  ArrayList salary = new ArrayList();
  salary.Add(2);
 
  // Function Call
  Console.Write(solve(values, salary));
}
}
 
// This code is contributed by Rutvik_56

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Output: 

1




Time Complexity: O(N*log N) 
Auxiliary Space: O(1)

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