定的字符串str，任务是找到需要从字符串，使得字符串的每个字符的频率是独特删除的最少字符数。
例子：
Input: str = “ceabaacb”
Output: 2
Explanation:
The frequencies of each distinct character are as follows:
c —> 2
e —> 1
a —> 3
b —> 2
Possible ways to make frequency of each character unique by minimum number of moves are:
- Removing both occurrences of ‘c’ modifies str to “eabaab”
- Removing an occurrence of ‘c’ and ‘e’ modifies str to “abaacb”
Therefore, the minimum removals required is 2.
Input: S = “abbbcccd”
Output: 2
方法：可以使用贪婪技术解决问题。这个想法是使用映射和优先级队列。请按照以下步骤解决问题：
- 初始化一个映射，例如mp ，以存储字符串的每个不同字符的频率。
- 初始化变量，说cntChar存储需要被移除，以独特的字符串的每个字符的化妆频率的字符个数。
- 创建一个priority_queue(例如pq)以存储每个字符的频率，以使获得的最大频率出现在优先级队列pq的顶部。
- 遍历priority_queue直到PQ是空的，并检查是否最上面的PQ的元素的是等于或PQ不的第二顶端元件。如果发现是真的，则将pq的最顶部元素的值减1，并将cntChar的值减1 。
- 否则，弹出pq的最上层元素。
- 最后，打印cntChar的值。
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to find the minimum count of
// characters required to be deleted to make
// frequencies of all characters unique
int minCntCharDeletionsfrequency(string& str,
int N)
{
// Stores frequency of each
// distinct character of str
unordered_map mp;
// Store frequency of each distinct
// character such that the largest
// frequency is present at the top
priority_queue pq;
// Stores minimum count of characters
// required to be deleted to make
// frequency of each character unique
int cntChar = 0;
// Traverse the string
for (int i = 0; i < N; i++) {
// Update frequency of str[i]
mp[str[i]]++;
}
// Traverse the map
for (auto it : mp) {
// Insert current
// frequency into pq
pq.push(it.second);
}
// Traverse the priority_queue
while (!pq.empty()) {
// Stores topmost
// element of pq
int frequent
= pq.top();
// Pop the topmost element
pq.pop();
// If pq is empty
if (pq.empty()) {
// Return cntChar
return cntChar;
}
// If frequent and topmost
// element of pq are equal
if (frequent == pq.top()) {
// If frequency of the topmost
// element is greater than 1
if (frequent > 1) {
// Insert the decremented
// value of frequent
pq.push(frequent - 1);
}
// Update cntChar
cntChar++;
}
}
return cntChar;
}
// Driver Code
int main()
{
string str = "abbbcccd";
// Stores length of str
int N = str.length();
cout << minCntCharDeletionsfrequency(
str, N);
return 0;
} Java
// Java program to implement
// the above approach
import java.util.*;
class GFG{
// Function to find the minimum count of
// characters required to be deleted to make
// frequencies of all characters unique
static int minCntCharDeletionsfrequency(char[] str,
int N)
{
// Stores frequency of each
// distinct character of str
HashMap mp =
new HashMap<>();
// Store frequency of each distinct
// character such that the largest
// frequency is present at the top
PriorityQueue pq =
new PriorityQueue<>((x, y) ->
Integer.compare(y, x));
// Stores minimum count of characters
// required to be deleted to make
// frequency of each character unique
int cntChar = 0;
// Traverse the String
for (int i = 0; i < N; i++)
{
// Update frequency of str[i]
if(mp.containsKey(str[i]))
{
mp.put(str[i],
mp.get(str[i]) + 1);
}
else
{
mp.put(str[i], 1);
}
}
// Traverse the map
for (Map.Entry it :
mp.entrySet())
{
// Insert current
// frequency into pq
pq.add(it.getValue());
}
// Traverse the priority_queue
while (!pq.isEmpty())
{
// Stores topmost
// element of pq
int frequent = pq.peek();
// Pop the topmost element
pq.remove();
// If pq is empty
if (pq.isEmpty()) {
// Return cntChar
return cntChar;
}
// If frequent and topmost
// element of pq are equal
if (frequent == pq.peek())
{
// If frequency of the topmost
// element is greater than 1
if (frequent > 1)
{
// Insert the decremented
// value of frequent
pq.add(frequent - 1);
}
// Update cntChar
cntChar++;
}
}
return cntChar;
}
// Driver Code
public static void main(String[] args)
{
String str = "abbbcccd";
// Stores length of str
int N = str.length();
System.out.print(minCntCharDeletionsfrequency(
str.toCharArray(), N));
}
}
// This code is contributed by Rajput-Ji Python3
# Python3 program to implement
# the above approach
# Function to find the minimum count of
# characters required to be deleted to make
# frequencies of all characters unique
def minCntCharDeletionsfrequency(str, N):
# Stores frequency of each
# distinct character of str
mp = {}
# Store frequency of each distinct
# character such that the largest
# frequency is present at the top
pq = []
# Stores minimum count of characters
# required to be deleted to make
# frequency of each character unique
cntChar = 0
# Traverse the string
for i in range(N):
# Update frequency of str[i]
mp[str[i]] = mp.get(str[i], 0) + 1
# Traverse the map
for it in mp:
# Insert current
# frequency into pq
pq.append(mp[it])
pq = sorted(pq)
# Traverse the priority_queue
while (len(pq) > 0):
# Stores topmost
# element of pq
frequent = pq[-1]
# Pop the topmost element
del pq[-1]
# If pq is empty
if (len(pq) == 0):
# Return cntChar
return cntChar
# If frequent and topmost
# element of pq are equal
if (frequent == pq[-1]):
# If frequency of the topmost
# element is greater than 1
if (frequent > 1):
# Insert the decremented
# value of frequent
pq.append(frequent - 1)
# Update cntChar
cntChar += 1
pq = sorted(pq)
return cntChar
# Driver Code
if __name__ == '__main__':
str = "abbbcccd"
# Stores length of str
N = len(str)
print(minCntCharDeletionsfrequency(str, N))
# This code is contributed by mohit kumar 29C#
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find the minimum count of
// characters required to be deleted to make
// frequencies of all characters unique
static int minCntCharDeletionsfrequency(char[] str,
int N)
{
// Stores frequency of each
// distinct character of str
Dictionary mp = new Dictionary();
// Store frequency of each distinct
// character such that the largest
// frequency is present at the top
List pq = new List();
// Stores minimum count of characters
// required to be deleted to make
// frequency of each character unique
int cntChar = 0;
// Traverse the String
for(int i = 0; i < N; i++)
{
// Update frequency of str[i]
if (mp.ContainsKey(str[i]))
{
mp[str[i]]++;
}
else
{
mp.Add(str[i], 1);
}
}
// Traverse the map
foreach(KeyValuePair it in mp)
{
// Insert current
// frequency into pq
pq.Add(it.Value);
}
pq.Sort();
pq.Reverse();
// Traverse the priority_queue
while (pq.Count != 0)
{
// Stores topmost
// element of pq
pq.Sort();
pq.Reverse();
int frequent = pq[0];
// Pop the topmost element
pq.RemoveAt(0);
// If pq is empty
if (pq.Count == 0)
{
// Return cntChar
return cntChar;
}
// If frequent and topmost
// element of pq are equal
if (frequent == pq[0])
{
// If frequency of the topmost
// element is greater than 1
if (frequent > 1)
{
// Insert the decremented
// value of frequent
pq.Add(frequent - 1);
pq.Sort();
// pq.Reverse();
}
// Update cntChar
cntChar++;
}
}
return cntChar;
}
// Driver Code
public static void Main(String[] args)
{
String str = "abbbcccd";
// Stores length of str
int N = str.Length;
Console.Write(minCntCharDeletionsfrequency(
str.ToCharArray(), N));
}
}
// This code is contributed by shikhasingrajput 输出：
2时间复杂度： O(N)
辅助空间： O(256)