给定长度为N的字符串S ，任务是找到需要删除的最小字符数，以使字符串中的每个不同字符出现相同的次数。
例子：
Input: S = “abcbccdddd”
Output: 4
Explanation: Removing an occurrence of the characters ‘a’, ‘c’ and two occurrences of ‘d’ modifies the string to “bbccdd”. Therefore, every character in the string ocuurs same number of times.
Input : S = “geeksforgeeks”
Output : 5
Explanation: Removing an occurrence of ‘r’, ‘f’, ‘and ‘o’ and removing two occurrences of ‘e’ modifies S to “geeksgks”.
方法：想法是使用多集和映射。请按照以下步骤解决问题：
- 初始化一个map
表示countMultiset来存储每个字符的频率。 - 将变量ans初始化为INT_MAX，以存储要删除的最小字符数。
- 在countMap中遍历字符串S和S [i]的递增计数。
- 遍历地图countMap并将字符的频率插入countMultiset中。
- 找到多集countMultiset的大小，并将其存储在变量m中。
- 遍历多集countMultiset并将ans更新为ans = min(ans，(N-(mi)*(* it))))，然后将i递增1 。
- 最后，完成上述步骤后，将答案打印为ans。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find minimum number of
// character removals required to make
// frequency of all distinct characters the same
int minimumDeletion(string s, int n)
{
// Stores the frequency
// of each character
map countMap;
// Traverse the string
for (int i = 0; i < n; i++) {
countMap[s[i]]++;
}
// Stores the frequency of each
// charachter in sorted order
multiset countMultiset;
// Traverse the Map
for (auto it : countMap) {
// Insert the frequency in multiset
countMultiset.insert(it.second);
}
// Stores the count of elements
// required to be removed
int ans = INT_MAX;
int i = 0;
// Stores the size of multiset
int m = countMultiset.size();
// Traverse the multiset
for (auto j : countMultiset) {
// Update the ans
ans = min(ans, n - (m - i) * j);
// Increment i by 1
i++;
}
// Return
return ans;
}
// Driver Code
int main()
{
// Input
string S = "geeksforgeeks";
int N = S.length();
cout << minimumDeletion(S, N);
return 0;
} Java
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG{
// Function to find minimum number of
// character removals required to make
// frequency of all distinct characters the same
static int minimumDeletion(String s, int n)
{
// Stores the frequency
// of each character
HashMap countMap = new HashMap<>();
// Traverse the string
for(int i = 0; i < n; i++)
{
char ch = s.charAt(i);
countMap.put(ch,
countMap.getOrDefault(ch, 0) + 1);
}
// Stores the frequency of each
// charachter in sorted order
ArrayList countMultiset = new ArrayList<>(
countMap.values());
Collections.sort(countMultiset);
// Stores the count of elements
// required to be removed
int ans = Integer.MAX_VALUE;
int i = 0;
// Stores the size of multiset
int m = countMultiset.size();
// Traverse the multiset
for(int j : countMultiset)
{
// Update the ans
ans = Math.min(ans, n - (m - i) * j);
// Increment i by 1
i++;
}
// Return
return ans;
}
// Driver Code
public static void main(String[] args)
{
// Input
String S = "geeksforgeeks";
int N = S.length();
System.out.println(minimumDeletion(S, N));
}
}
// This code is contributed by Kingash Python3
# Python3 program for the above approach
import sys
# Function to find minimum number of
# character removals required to make
# frequency of all distinct characters the same
def minimumDeletion(s, n):
# Stores the frequency
# of each character
countMap = {}
# Traverse the string
for i in s:
countMap[i] = countMap.get(i, 0) + 1
# Stores the frequency of each
# charachter in sorted order
countMultiset = []
# Traverse the Map
for it in countMap:
# Insert the frequency in multiset
countMultiset.append(countMap[it])
# Stores the count of elements
# required to be removed
ans = sys.maxsize + 1
i = 0
# Stores the size of multiset
m = len(countMultiset)
# Traverse the multiset
for j in sorted(countMultiset):
# Update the ans
ans = min(ans, n - (m - i) * j)
# Increment i by 1
i += 1
# Return
return ans
# Driver Code
if __name__ == '__main__':
# Input
S = "geeksforgeeks"
N = len(S)
print (minimumDeletion(S, N))
# This code is contributed by mohit kumar 29C#
// C# program for the above approach
using System.Collections.Generic;
using System;
class GFG{
// Function to find minimum number of
// character removals required to make
// frequency of all distinct characters the same
static int minimumDeletion(string s, int n)
{
// Stores the frequency
// of each character
Dictionary countMap = new Dictionary();
// Traverse the string
for(int i = 0; i < n; i++)
{
if (countMap.ContainsKey(s[i]) == true)
{
countMap[s[i]] += 1;
}
else
{
countMap[s[i]] = 1;
}
}
// Stores the frequency of each
// charachter in sorted order
List countMultiset = new List();
foreach(var values in countMap.Values)
{
countMultiset.Add(values);
}
countMultiset.Sort();
// Stores the count of elements
// required to be removed
int ans = 100000000;
int index = 0;
// Stores the size of multiset
int m = countMultiset.Count;
// Traverse the multiset
foreach(var j in countMultiset)
{
// Update the ans
ans = Math.Min(ans, n - (m - index) * j);
// Increment i by 1
index++;
}
// Return
return ans;
}
// Driver Code
public static void Main(String[] args)
{
// Input
string S = "geeksforgeeks";
int N = S.Length;
Console.WriteLine(minimumDeletion(S, N));
}
}
// This code is contributed by Stream_Cipher 输出：
5时间复杂度： O(N * log(N))
辅助空间： O(1)