清空管柱所需的操作
给定一个字符串 str ，任务是用给定的操作将字符串清空。在单个操作中，您可以挑选字符串中的一些字符(每个挑选的字符应该具有相同的频率)并将其从字符串中移除。打印使字符串为空所需的全部操作。 例:
输入: str = "aabbccc" 输出: 2 在一次操作中，字符‘a’和‘b’可以被移除，因为两者具有相同的频率。 第二次操作可以删除频率为 3 的字符‘c’。 共需要 2 次操作。 输入:str = " geeks forgeeks " 输出: 3
方法:找到字符串字符的唯一频率。唯一频率的总计数将是使字符串为空所需的操作数。 对于str = " aaabbcccc "，唯一的频率是 3 和 4 。唯一频率总数为 2 。 HashMap 可用于存储字符及其频率，然后 HashSet 可用于查找唯一频率的计数，即所需操作的数量。 以下是上述方法的实施:
C++
// CPP implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the count of operations required
int totalOperations(string str, int len)
{
// HashMap to store characters and their frequencies
unordered_map<char, int> h;
for (int i = 0; i < len; i++)
// If already contains the character then
// increment its frequency by 1
h[str[i]]++;
// HashSet to store unique frequency
unordered_set<int> hs;
// Insert frequencies into HashSet
for (auto i : h)
hs.insert(i.second);
// Count of unique frequencies
return hs.size();
}
// Driver Code
int main()
{
string str = "geeksforgeeks";
int len = str.length();
cout << totalOperations(str, len) << endl;
return 0;
}
// This code is contributed by
// sanjeev2552
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
import java.util.*;
class GFG {
// Function to return the count of operations required
static int totalOperations(String str, int len)
{
// HashMap to store characters and their frequencies
HashMap<Character, Integer> h = new HashMap<Character, Integer>();
for (int i = 0; i < len; i++) {
// If already contains the character then
// increment its frequency by 1
if (h.containsKey(str.charAt(i)))
h.put(str.charAt(i), h.get(str.charAt(i)) + 1);
// Else add the character to the HashMap with frequency 1
else
h.put(str.charAt(i), 1);
}
// Set to iterate over HashMap
Set<Map.Entry<Character, Integer> > set = h.entrySet();
// HashSet to store unique frequency
HashSet<Integer> hs = new HashSet<Integer>();
// Insert frequencies into HashSet
for (Map.Entry<Character, Integer> me : set)
hs.add(me.getValue());
// Count of unique frequencies
return hs.size();
}
// Driver code
public static void main(String[] args)
{
String str = "geeksforgeeks";
int len = str.length();
System.out.println(totalOperations(str, len));
}
}
Python 3
# Python implementation of the approach
# Function to return the count of operations required
def totalOperations(st, length):
# Dictionary to store characters and their frequencies
d = {}
for i in range(length):
# If already contains the character then
# increment its frequency by 1
if st[i] in d:
d[st[i]] += 1
# Else add the character to the HashMap with frequency 1
else:
d[st[i]] = 1
# Set to Store unique frequency
valueSet = set()
# Insert frequencies into HashSet
for key in d.keys():
valueSet.add(d[key])
# Count of unique frequencies
return len(valueSet)
# Driver Code
st = "geeksforgeeks"
l = len(st)
print(totalOperations(st, l))
# This code is contributed by Vivekkumar Singh
C
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to return
// the count of operations required
static int totalOperations(String str, int len)
{
// HashMap to store characters
// and their frequencies
Dictionary<char,
int> h = new Dictionary<char,
int>();
for (int i = 0; i < len; i++)
{
// If already contains the character then
// increment its frequency by 1
if (h.ContainsKey(str[i]))
h[str[i]] = h[str[i]] + 1;
// Else add the character
// to the HashMap with frequency 1
else
h.Add(str[i], 1);
}
// Set to iterate over HashMap
// HashSet to store unique frequency
HashSet<int> hs = new HashSet<int>();
// Insert frequencies into HashSet
foreach(KeyValuePair<char, int> me in h)
hs.Add(me.Value);
// Count of unique frequencies
return hs.Count;
}
// Driver code
public static void Main(String[] args)
{
String str = "geeksforgeeks";
int len = str.Length;
Console.WriteLine(totalOperations(str, len));
}
}
// This code is contributed by Rajput-Ji
java 描述语言
<script>
// Javascript implementation of the approach
// Function to return the count of operations required
function totalOperations(str, len)
{
// HashMap to store characters and their frequencies
var h = new Map();
for (var i = 0; i < len; i++)
// If already contains the character then
// increment its frequency by 1
if(h.has(str[i]))
h.set(str[i], h.get(str[i])+1)
else
h.set(str[i], 1)
// HashSet to store unique frequency
var hs = new Set();
// Insert frequencies into HashSet
h.forEach((value, key) => {
hs.add(value);
});
// Count of unique frequencies
return hs.size;
}
// Driver Code
var str = "geeksforgeeks";
var len = str.length;
document.write( totalOperations(str, len));
// This code is contributed by itsok.
</script>
Output:
3
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