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Find number of Employees Under every Employee

  • Difficulty Level : Hard
  • Last Updated : 10 Jun, 2021

Given a dictionary that contains mapping of employee and his manager as a number of (employee, manager) pairs like below. 

{ "A", "C" },
{ "B", "C" },
{ "C", "F" },
{ "D", "E" },
{ "E", "F" },
{ "F", "F" } 

In this example C is manager of A, 
C is also manager of B, F is manager 
of C and so on.

 

Write a function to get no of employees under each manager in the hierarchy not just their direct reports. It may be assumed that an employee directly reports to only one manager. In the above dictionary the root node/ceo is listed as reporting to himself. 
Output should be a Dictionary that contains following. 

A - 0  
B - 0
C - 2
D - 0
E - 1
F - 5 

Source: Microsoft Interview
This question might be solved differently but i followed this and found interesting, so sharing:
 

 1. Create a reverse map with Manager->DirectReportingEmployee 
    combination. Off-course employee will be multiple so Value 
    in Map is List of Strings.
        "C" --> "A", "B",
        "E" --> "D" 
        "F" --> "C", "E", "F"

 
2. Now use the given employee-manager map to iterate  and at 
   the same time use newly reverse map to find the count of 
   employees under manager.
   
   Let the map created in step 2 be 'mngrEmpMap' 
   Do following for every employee 'emp'.
     a) If 'emp' is not present in 'mngrEmpMap' 
          Count under 'emp' is 0 [Nobody reports to 'emp']
     b) If 'emp' is present in 'mngrEmpMap' 
          Use the list of direct reports from map 'mngrEmpMap'
          and recursively calculate number of total employees
          under 'emp'. 

A trick in step 2.b is to use memorization(Dynamic programming) while finding number of employees under a manager so that we don’t need to find number of employees again for any of the employees. In the below code populateResultUtil() is the recursive function that uses memoization to avoid re-computation of same results.
Below is Java implementation of above ideas
 



Java




// Java program to find number of persons under every employee
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
 
public class NumberEmployeeUnderManager
{
    // A hashmap to store result. It stores count of employees
    // under every employee, the count may by 0 also
    static Map<String,Integer> result =
                             new HashMap<String, Integer>();
 
    // Driver function
    public static void main(String[] args)
    {
        Map<String, String> dataSet = new HashMap<String, String>();
        dataSet.put("A", "C");
        dataSet.put("B", "C");
        dataSet.put("C", "F");
        dataSet.put("D", "E");
        dataSet.put("E", "F");
        dataSet.put("F", "F");
 
        populateResult(dataSet);
        System.out.println("result = " + result);
    }
 
    // This function populates 'result' for given input 'dataset'
    private static void populateResult(Map<String, String> dataSet)
    {
        // To store reverse of original map, each key will have 0
        // to multiple values
        Map<String, List<String>> mngrEmpMap =
                                  new HashMap<String, List<String>>();
 
        // To fill mngrEmpMap, iterate through the given map
        for (Map.Entry<String,String> entry: dataSet.entrySet())
        {
            String emp = entry.getKey();
            String mngr = entry.getValue();
            if (!emp.equals(mngr)) // excluding emp-emp entry
            {
                // Get the previous list of direct reports under
                // current 'mgr' and add the current 'emp' to the list
                List<String> directReportList = mngrEmpMap.get(mngr);
 
                // If 'emp' is the first employee under 'mgr'
                if (directReportList == null)
                {
                    directReportList = new ArrayList<String>();
                    // add a new entry for the mngr with empty directReportList
                    mngrEmpMap.put(mngr, directReportList);
                }
                directReportList.add(emp);
            }
        }
 
        // Now use manager-Emp map built above to populate result
        // with use of populateResultUtil()
 
        // note- we are iterating over original emp-manager map and
        // will use mngr-emp map in helper to get the count
        for (String mngr: dataSet.keySet())
            populateResultUtil(mngr, mngrEmpMap);
    }
 
    // This is a recursive function to fill count for 'mgr' using
    // mngrEmpMap.  This function uses memoization to avoid re-
    // computations of subproblems.
    private static int populateResultUtil(String mngr,
                               Map<String, List<String>> mngrEmpMap)
    {
        int count = 0;
 
        // means employee is not a manager of any other employee
        if (!mngrEmpMap.containsKey(mngr))
        {
            result.put(mngr, 0);
            return 0;
        }
 
        // this employee count has already been done by this
        // method, so avoid re-computation
        else if (result.containsKey(mngr))
            count = result.get(mngr);
 
        else
        {
            List<String> directReportEmpList = mngrEmpMap.get(mngr);
            count = directReportEmpList.size();
            for (String directReportEmp: directReportEmpList)
               count +=  populateResultUtil(directReportEmp, mngrEmpMap);
 
            result.put(mngr, count);
        }
        return count;
    }
}

Python3




class Solution():
    def __init__(self):
        pass
 
    def assignAndPrint(self,t):
        #We will directily permute over t. Access 2nd element(i.e. manager) of certain tuple and assign the relation in
        # dictionary. We will assign list of employees to a particular manager so that with iterations, we can append
        # more employees to that list and list grows.
        d = dict()
        for pair in t:
            if(pair[0]==pair[1]):  # because we dont want to assign self managing role
                continue
            if pair[0] not in d:  # assign employee a empty list of employees
                d[pair[0]] = []
            # for managers -
            if pair[1] not in d:
                d[pair[1]] = [pair[0]]
            else:
                d[pair[1]].append(pair[0])
        #print(d)
        # now we know how many employees are directly under a particular manager.
        # now lets count the total number of employees under a particular manager.
        c = dict()   # store    manager:count of employee    as key value
        for manager in d:
            c[manager] = len(d[manager])
            for employee in d[manager]:
                c[manager] += len(d[employee])
            print("{} : {}".format(manager,c[manager]))     # prints which manager has total how many employees
        # Note : Employees having no employees under are also considered as managed with 0 employees.
 
 
if __name__=="__main__":
    # t is tuple containing employee and boss pair.
    t = (("A", "C"),("B", "C"),("C", "F"),("D", "E"),("E", "F"),("F", "F"))
    obj = Solution()
    obj.assignAndPrint(t)
Output
result = {A=0, B=0, C=2, D=0, E=1, F=5}

This article is contributed by Chandan Prakash. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
 

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