给定一个字符串S ，其中包含长度为N 的小写英文字母和一个整数K ，使得K ≤ N 。任务是从字符串删除K 个字符后找到最大数量的非重复字符。
例子：
Input: S = “geeksforgeeks”, K = 3
Output: 6
Explanation:
Remove 1 occurrences of each g, k and s so the final string is “geeksforee” and the 6 distinct elements are g, k, s, f, o and r
Input: S = “aabbccddeeffgghh”, k = 1
Output: 1
Explanation:
Remove 1 occurrences of any character we will have only one character which will non repeating.
朴素的方法：朴素的想法是删除给定字符串所有可能的K 个字符，然后在所有形成的字符串找到不重复的字符。打印所有非重复字符数中的最大值。
时间复杂度： O(N!)，其中 N 是给定字符串的长度。
辅助空间： O(NK)
高效的方法：为了优化上述方法，
The idea is to delete K characters in increasing order of frequency whose frequency is at least 2 to get the count of maximum non-repeating characters.
以下是步骤：
- 创建一个哈希表来存储每个元素的频率。
- 将每个元素的频率插入向量V 中，并按升序对向量V进行排序。
- 对于向量V 的每个元素（比如currentElement ），找到K和currentElement – 1 中的最小值，并将K和V[i]减少两者中的最小值。
- 重复上述步骤，直到K不为零。
- 向量V中1的计数给出了删除K 个字符后的最大非重复字符数。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find maximum distinct
// character after removing K character
int maxDistinctChar(string s, int n, int k)
{
// Freq implemented as hash table to
// store frequency of each character
unordered_map freq;
// Store frequency of each character
for (int i = 0; i < n; i++)
freq[s[i]]++;
vector v;
// Insert each frequency in v
for (auto it = freq.begin();
it != freq.end(); it++) {
v.push_back(it->second);
}
// Sort the freq of character in
// non-decresing order
sort(v.begin(), v.end());
// Traverse the vector
for (int i = 0; i < v.size(); i++) {
int mn = min(v[i] - 1, k);
// Update v[i] and k
v[i] -= mn;
k -= mn;
}
// If K is still not 0
if (k > 0) {
for (int i = 0; i < v.size(); i++) {
int mn = min(v[i], k);
v[i] -= mn;
k -= mn;
}
}
// Store the final answer
int res = 0;
for (int i = 0; i < v.size(); i++)
// Count this character if freq 1
if (v[i] == 1)
res++;
// Return count of distinct characters
return res;
}
// Driver Code
int main()
{
// Given string
string S = "geeksforgeeks";
int N = S.size();
// Given k
int K = 1;
// Function Call
cout << maxDistinctChar(S, N, K);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to find maximum distinct
// character after removing K character
static int maxDistinctChar(char []s, int n, int k)
{
// Freq implemented as hash table to
// store frequency of each character
HashMap freq = new HashMap();
// Store frequency of each character
for (int i = 0; i < n; i++)
{
if(freq.containsKey((int)s[i]))
{
freq.put((int)s[i],
freq.get((int)s[i]) + 1);
}
else
{
freq.put((int)s[i], 1);
}
}
Vector v = new Vector();
// Insert each frequency in v
for (Map.Entry it : freq.entrySet())
{
v.add(it.getValue());
}
// Sort the freq of character in
// non-decresing order
Collections.sort(v);
// Traverse the vector
for (int i = 0; i < v.size(); i++)
{
int mn = Math.min(v.get(i) - 1, k);
// Update v[i] and k
v.set(i, v.get(i) - mn);
k -= mn;
}
// If K is still not 0
if (k > 0)
{
for (int i = 0; i < v.size(); i++)
{
int mn = Math.min(v.get(i), k);
v.set(i, v.get(i) - mn);
k -= mn;
}
}
// Store the final answer
int res = 0;
for (int i = 0; i < v.size(); i++)
// Count this character if freq 1
if (v.get(i) == 1)
res++;
// Return count of distinct characters
return res;
}
// Driver Code
public static void main(String[] args)
{
// Given String
String S = "geeksforgeeks";
int N = S.length();
// Given k
int K = 1;
// Function Call
System.out.print(maxDistinctChar(S.toCharArray(),
N, K));
}
}
// This code is contributed by shikhasingrajput Python3
# Python3 program for the above approach
from collections import defaultdict
# Function to find maximum distinct
# character after removing K character
def maxDistinctChar(s, n, k):
# Freq implemented as hash table to
# store frequency of each character
freq = defaultdict (int)
# Store frequency of each character
for i in range (n):
freq[s[i]] += 1
v = []
# Insert each frequency in v
for it in freq.values():
v.append(it)
# Sort the freq of character in
# non-decresing order
v.sort()
# Traverse the vector
for i in range (len(v)):
mn = min(v[i] - 1, k)
# Update v[i] and k
v[i] -= mn
k -= mn
# If K is still not 0
if (k > 0):
for i in range (len(v)):
mn = min(v[i], k);
v[i] -= mn
k -= mn
# Store the final answer
res = 0
for i in range (len(v)):
# Count this character if freq 1
if (v[i] == 1):
res += 1
# Return count of distinct characters
return res
# Driver Code
if __name__ == "__main__":
# Given string
S = "geeksforgeeks"
N = len(S)
# Given k
K = 1
# Function Call
print(maxDistinctChar(S, N, K))
# This code is contributed by ChitranayalC#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find maximum distinct
// character after removing K character
static int maxDistinctChar(char []s, int n, int k)
{
// Freq implemented as hash table to
// store frequency of each character
Dictionary freq = new Dictionary();
// Store frequency of each character
for(int i = 0; i < n; i++)
{
if(freq.ContainsKey((int)s[i]))
{
freq[(int)s[i]] = freq[(int)s[i]] + 1;
}
else
{
freq.Add((int)s[i], 1);
}
}
List v = new List();
// Insert each frequency in v
foreach(KeyValuePair it in freq)
{
v.Add(it.Value);
}
// Sort the freq of character in
// non-decresing order
v.Sort();
// Traverse the vector
for(int i = 0; i < v.Count; i++)
{
int mn = Math.Min(v[i] - 1, k);
// Update v[i] and k
v[i] = v[i] - mn;
k -= mn;
}
// If K is still not 0
if (k > 0)
{
for(int i = 0; i < v.Count; i++)
{
int mn = Math.Min(v[i], k);
v[i] = v[i] - mn;
k -= mn;
}
}
// Store the readonly answer
int res = 0;
for(int i = 0; i < v.Count; i++)
// Count this character if freq 1
if (v[i] == 1)
res++;
// Return count of distinct characters
return res;
}
// Driver Code
public static void Main(String[] args)
{
// Given String
String S = "geeksforgeeks";
int N = S.Length;
// Given k
int K = 1;
// Function call
Console.Write(maxDistinctChar(S.ToCharArray(),
N, K));
}
}
// This code is contributed by Amit Katiyar Javascript
输出：
4时间复杂度： O(N)
辅助空间： O(26)
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