Maximize number of days for which P chocolates can be distributed consecutively to N people

Given an integer, P denoting the number of chocolates and an array a[] where a_i denotes the type of i^thchocolate. There are N people who want to eat chocolate every day. Find the maximum number of consecutive days for which N people can eat chocolates considering the following conditions:

1. Each of the N people must eat exactly one chocolate on a particular day.
2. A single person can eat chocolate of the same type only for all days.

Examples:

Input: N = 4, P = 10, arr[] = {1, 5, 2, 1, 1, 1, 2, 5, 7, 2}
Output: 2
Explanation: Chocolates can be assigned in the following way:
Person 1: Type 1
Person 2: Type 1
Person 3: Type 2
Person 4: Type 5
In this way, there are sufficient chocolates for each person to eat one chocolate for two consecutive days. No other possible distribution of chocolates can make the people eat the chocolates for more than 2 days.

Input: N = 3, P = 10, arr[] = {1, 2, 2, 1, 1, 3, 3, 3, 2, 4}
Output: 3
Explanation: Chocolates can be distributed in the following way:
Person 1: Type 1
Person 2: Type 2
Person 3: Type 3
In this way, all the 3 people can eat their respective assigned type of chocolates for three days.

Approach: Follow the steps below to solve the problem:

• The minimum number of days for which it is possible to distribute the chocolates is 0 and the maximum number is P.
• So, for each number X in this range, check if it is possible to distribute chocolates to each person for X days.
• For all such Xs, find the maximum.
• Now, using Binary Search check for all the numbers in the range 0 to P.

Below is the implementation of the above approach:

C++

// C++ program to implement
// the above approach

#include <bits/stdc++.h>
using namespace std;

// Stores the frequency of
// each type of chocolate
map<int, int> mp;

int N, P;

// Function to check if chocolates
// can be eaten for 'mid' no. of days
bool helper(int mid)
{

    int cnt = 0;
    for (auto i : mp) {
        int temp = i.second;

        while (temp >= mid) {
            temp -= mid;
            cnt++;
        }
    }

    // If cnt exceeds N,
    // return true
    return cnt >= N;
}

// Function to find the maximum
// number of days for which
// chocolates can be eaten
int findMaximumDays(int arr[])
{

    // Store the frequency
    // of each type of chocolate
    for (int i = 0; i < P; i++) {
        mp[arr[i]]++;
    }

    // Initialize start and end
    // with 0 and P respectively
    int start = 0, end = P, ans = 0;
    while (start <= end) {

        // Calculate mid
        int mid = start
                  + ((end - start) / 2);

        // Check if chocolates can be
        // distributed for mid days
        if (mid != 0 and helper(mid)) {

            ans = mid;

            // Check if chocolates can
            // be distributed for more
            // than mid consecutive days
            start = mid + 1;
        }
        else if (mid == 0) {
            start = mid + 1;
        }
        else {
            end = mid - 1;
        }
    }

    return ans;
}

// Driver code
int main()
{

    N = 3, P = 10;
    int arr[] = { 1, 2, 2, 1, 1,
                  3, 3, 3, 2, 4 };

    // Function call
    cout << findMaximumDays(arr);

    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;

class GFG{

// Stores the frequency of
// each type of chocolate
static HashMap<Integer,
               Integer> mp = new HashMap<Integer,
                                         Integer>();

static int N, P;

// Function to check if chocolates
// can be eaten for 'mid' no. of days
static boolean helper(int mid)
{
    int cnt = 0;
    for(Map.Entry<Integer, Integer> i : mp.entrySet())
    {
        int temp = i.getValue();

        while (temp >= mid)
        {
            temp -= mid;
            cnt++;
        }
    }

    // If cnt exceeds N,
    // return true
    return cnt >= N;
}

// Function to find the maximum
// number of days for which
// chocolates can be eaten
static int findMaximumDays(int arr[])
{

    // Store the frequency
    // of each type of chocolate
    for(int i = 0; i < P; i++)
    {
        if (mp.containsKey(arr[i]))
        {
            mp.put(arr[i], mp.get(arr[i]) + 1);
        }
        else
        {
            mp.put(arr[i], 1);
        }
    }

    // Initialize start and end
    // with 0 and P respectively
    int start = 0, end = P, ans = 0;
    while (start <= end)
    {

        // Calculate mid
        int mid = start +
                  ((end - start) / 2);

        // Check if chocolates can be
        // distributed for mid days
        if (mid != 0 && helper(mid))
        {
            ans = mid;

            // Check if chocolates can
            // be distributed for more
            // than mid consecutive days
            start = mid + 1;
        }
        else if (mid == 0)
        {
            start = mid + 1;
        }
        else
        {
            end = mid - 1;
        }
    }
    return ans;
}

// Driver code
public static void main(String[] args)
{

    N = 3;
    P = 10;
    int arr[] = { 1, 2, 2, 1, 1,
                  3, 3, 3, 2, 4 };

    // Function call
    System.out.print(findMaximumDays(arr));
}
}

// This code is contributed by Amit Katiyar

Python3

# Python3 program to implement
# the above approach

# Stores the frequency of
# each type of chocolate
mp = {}

N, P = 0, 0

# Function to check if chocolates
# can be eaten for 'mid' no. of days
def helper(mid):

    cnt = 0;
    for i in mp:
        temp = mp[i]

        while (temp >= mid):
            temp -= mid
            cnt += 1

    # If cnt exceeds N,
    # return true
    return cnt >= N

# Function to find the maximum
# number of days for which
# chocolates can be eaten
def findMaximumDays(arr):

    # Store the frequency
    # of each type of chocolate
    for i in range(P):
        mp[arr[i]] = mp.get(arr[i], 0) + 1

    # Initialize start and end
    # with 0 and P respectively
    start = 0
    end = P
    ans = 0

    while (start <= end):

        # Calculate mid
        mid = start + ((end - start) // 2)

        # Check if chocolates can be
        # distributed for mid days
        if (mid != 0 and helper(mid)):
            ans = mid

            # Check if chocolates can
            # be distributed for more
            # than mid consecutive days
            start = mid + 1
        elif (mid == 0):
            start = mid + 1
        else:
            end = mid - 1

    return ans

# Driver code
if __name__ == '__main__':

    N = 3
    P = 10

    arr = [ 1, 2, 2, 1, 1,
            3, 3, 3, 2, 4 ]

    # Function call
    print(findMaximumDays(arr))

# This code is contributed by mohit kumar 29

C#

// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{

// Stores the frequency of
// each type of chocolate
static Dictionary<int,
                  int> mp = new Dictionary<int,
                                           int>();
static int N, P;

// Function to check if
// chocolates can be eaten
// for 'mid' no. of days
static bool helper(int mid)
{
  int cnt = 0;
  foreach(KeyValuePair<int,
                       int> i in mp)
  {
    int temp = i.Value;

    while (temp >= mid)
    {
      temp -= mid;
      cnt++;
    }
  }

  // If cnt exceeds N,
  // return true
  return cnt >= N;
}

// Function to find the maximum
// number of days for which
// chocolates can be eaten
static int findMaximumDays(int []arr)
{
  // Store the frequency
  // of each type of chocolate
  for(int i = 0; i < P; i++)
  {
    if (mp.ContainsKey(arr[i]))
    {
      mp[arr[i]] =  mp[arr[i]] + 1;
    }
    else
    {
      mp.Add(arr[i], 1);
    }
  }

  // Initialize start and end
  // with 0 and P respectively
  int start = 0, end = P, ans = 0;
  while (start <= end)
  {
    // Calculate mid
    int mid = start +
              ((end - start) / 2);

    // Check if chocolates can be
    // distributed for mid days
    if (mid != 0 && helper(mid))
    {
      ans = mid;

      // Check if chocolates can
      // be distributed for more
      // than mid consecutive days
      start = mid + 1;
    }
    else if (mid == 0)
    {
      start = mid + 1;
    }
    else
    {
      end = mid - 1;
    }
  }
  return ans;
}

// Driver code
public static void Main(String[] args)
{
  N = 3;
  P = 10;
  int []arr = {1, 2, 2, 1, 1,
               3, 3, 3, 2, 4};

  // Function call
  Console.Write(findMaximumDays(arr));
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// Javascript program to implement
// the above approach

// Stores the frequency of
// each type of chocolate
var mp = new Map();

var N, P;

// Function to check if chocolates
// can be eaten for 'mid' no. of days
function helper(mid)
{

    var cnt = 0;
    mp.forEach((value,) => {
        var temp = value;

        while (temp >= mid) {
            temp -= mid;
            cnt++;
        }
    });

    // If cnt exceeds N,
    // return true
    return cnt >= N;
}

// Function to find the maximum
// number of days for which
// chocolates can be eaten
function findMaximumDays(arr)
{

    // Store the frequency
    // of each type of chocolate
    for (var i = 0; i < P; i++) {
        if(mp.has(arr[i]))
            mp.set(arr[i], mp.get(arr[i])+1)
        else
            mp.set(arr[i], 1);
    }

    // Initialize start and end
    // with 0 and P respectively
    var start = 0, end = P, ans = 0;
    while (start <= end) {

        // Calculate mid
        var mid = start
                  + parseInt((end - start) / 2);

        // Check if chocolates can be
        // distributed for mid days
        if (mid != 0 && helper(mid)) {

            ans = mid;

            // Check if chocolates can
            // be distributed for more
            // than mid consecutive days
            start = mid + 1;
        }
        else if (mid == 0) {
            start = mid + 1;
        }
        else {
            end = mid - 1;
        }
    }

    return ans;
}

// Driver code
N = 3, P = 10;
var arr = [1, 2, 2, 1, 1,
              3, 3, 3, 2, 4 ];
// Function call
document.write( findMaximumDays(arr));

</script>

Output

3

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