将 P 块巧克力连续分发给 N 人的天数最大化

给定的整数，P表示巧克力的数目和阵列的[]其中a _i表示^第巧克力的类型i的。每天有N个人想吃巧克力。考虑以下条件，求N个人可以吃巧克力的最大连续天数：

1. 在特定的一天， N个人中的每一个人必须恰好吃一块巧克力。
2. 一个人只能整天吃同一类型的巧克力。

例子：

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.

处理方法：按照以下步骤解决问题：

• 可以分发巧克力的最小天数是0 ，最大天数是P 。
• 因此，对于此范围内的每个数字X ，请检查是否可以在X天内向每个人分发巧克力。
• 对于所有这样的 X，找出最大值。
• 现在，使用二分搜索检查 0 到 P 范围内的所有数字。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
 
#include 
using namespace std;
 
// Stores the frequency of
// each type of chocolate
map 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 mp = new HashMap();
                                          
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 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 mp = new Dictionary();                                        
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 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

 
// 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));
 

输出

3

时间复杂度： O(N * log N)
辅助空间： O(N)

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