给定一个整数， P表示巧克力的数量，数组a []其中a i表示第i个巧克力的类型。每天有N个人想吃巧克力。考虑以下条件，找到N个人可以连续吃巧克力的最大天数：
- N个人每个人在特定的一天必须只吃一种巧克力。
- 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.
方法：请按照以下步骤解决问题：
- 可以分发巧克力的最小天数为0 ，最大数目为P。
- 因此，对于此范围内的每个数字X ，请检查是否有可能在X天内向每个人分发巧克力。
- 对于所有这样的X，找到最大值。
- 现在，使用“二进制搜索”检查范围在0到P之间的所有数字。
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 29C#
// 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 输出
3
时间复杂度： O(N * log N)
辅助空间： O(N)