检查字符串是否包含超过 K 个活动字符的任何索引

给定一个包含小写英文字母的字符串S和一个整数K ，任务是找到包含超过K 个活动字符的字符串的任何索引。如果找到，则打印Yes 。否则，打印No 。

Count of active characters for any index is the number of characters having previous occurrences before or at the current index and last occurrence at or after the current index.

例子：

Input: S = “aabbcd”, K = 1 
Output: No 
Explanation: 
Index 1: Active character: a 
Index 2: Active character: a 
Index 3: Active character: b 
Index 4: Active character: b 
Index 5: Active character: c 
Index 6: Active character: d 
There are no more than one active character at any index in the string.
Input: S = “aabbcdca”, K = 2 
Output: Yes 
Explanation: 
Index 1: Active character : a 
Index 2: Active character : a 
Index 3: Active characters : a, b 
Index 4: Active characters : a, b 
Index 5: Active characters : a, c 
Index 6: Active characters : a, c, d 
Therefore, there exists an index with more than 2 active characters.

方法：
请按照以下步骤解决问题：

• 这个想法是将字符串中出现的每个字符的最后一次出现存储在 Map 中。
• 遍历字符串并保持将活动字母存储在一个 Set 中。
• 如果在任何索引处， Set的大小超过K ，则打印“是” 。
• 否则，检查当前索引是否是当前字符的最后一次出现。如果是，请从Set 中删除该字符。
• 最后，如果没有找到超过K 个活动字符的索引，则打印“No” 。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
#include 
using namespace std;
 
// Function to check if any index
// contains more than K active characters
string checkString(string s, int K)
{
 
    int n = s.length();
 
    // Store the last occurrence of
    // each character in the map.
    unordered_map mp;
 
    for (int i = 0; i < n; i++) {
        mp[s[i]] = i;
    }
 
    int cnt = 0, f = 0;
 
    // Stores the active
    // characters
    unordered_set st;
 
    for (int i = 0; i < n; i++) {
 
        // Insert the character
        st.insert(s[i]);
 
        // If the size of set
        // exceeds K
        if (st.size() > K) {
            f = 1;
            break;
        }
 
        // Remove the character from
        // set if i is the last index
        // of the current character
        if (mp[s[i]] == i)
            st.erase(s[i]);
    }
 
    return (f == 1 ? "Yes" : "No");
}
 
// Driver Code
int main()
{
 
    string s = "aabbcdca";
    int k = 2;
    cout << checkString(s, k);
    return 0;
}

Java

// Java program to implement the
// above approach
import java.util.*;
 
class GFG{
 
// Function to check if any index
// contains more than K active characters
static String checkString(String s, int K)
{
    int n = s.length();
 
    // Store the last occurrence of
    // each character in the map.
    Map mp = new HashMap<>();
 
    for(int i = 0; i < n; i++)
    {
        mp.put(s.charAt(i), i);
    }
 
    int cnt = 0, f = 0;
 
    // Stores the active
    // characters
    Set st = new HashSet<>();
 
    for(int i = 0; i < n; i++)
    {
         
        // Insert the character
        st.add(s.charAt(i));
 
        // If the size of set
        // exceeds K
        if (st.size() > K)
        {
            f = 1;
            break;
        }
 
        // Remove the character from
        // set if i is the last index
        // of the current character
        if (mp.get(s.charAt(i)) == i)
            st.remove(s.charAt(i));
    }
    return (f == 1 ? "Yes" : "No");
}
 
// Driver code
public static void main(String[] args)
{
    String s = "aabbcdca";
    int k = 2;
     
    System.out.println(checkString(s, k));
}
}
 
// This code is contributed by offbeat

Python3

# Python3 program to implement
# the above approach
 
# Function to check if any index
# contains more than K active characters
def checkString(s, K):
     
    n = len(s)
     
    # Store the last occurrence of
    # each character 
    dict = {}
 
    for i in range(n):
        dict[s[i]] = i;
         
    # Stores the active
    # characters
    st = set()
 
    for i in range(n):
         
        # Insert the character
        st.add(s[i])
         
        # If the size of set
        # exceeds K
        if len(st) > K:
            print("Yes")
            return
         
        # Remove the character from
        # set if i is the last index
        # of the current character
        if dict[s[i]] == i:
            st.remove(s[i])
 
    print("No")
 
# Driver code
s = "aabbcdca"
K = 2
 
checkString(s, K)
 
# This code is contributed by vashisthamadhur2

C#

// C# program to implement the
// above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to check if any index
// contains more than K active characters
static String checkString(String s, int K)
{
    int n = s.Length;
 
    // Store the last occurrence of
    // each character in the map.
    Dictionary mp = new Dictionary();
 
    for(int i = 0; i < n; i++)
    {
        if(mp.ContainsKey(s[i]))
            mp[s[i]] = i;
        else
            mp.Add(s[i], i);
    }
 
    int f = 0;
 
    // Stores the active
    // characters
    HashSet st = new HashSet();
 
    for(int i = 0; i < n; i++)
    {
         
        // Insert the character
        st.Add(s[i]);
 
        // If the size of set
        // exceeds K
        if (st.Count > K)
        {
            f = 1;
            break;
        }
 
        // Remove the character from
        // set if i is the last index
        // of the current character
        if (mp[s[i]] == i)
            st.Remove(s[i]);
    }
    return (f == 1 ? "Yes" : "No");
}
 
// Driver code
public static void Main(String[] args)
{
    String s = "aabbcdca";
    int k = 2;
     
    Console.WriteLine(checkString(s, k));
}
}
 
// This code is contributed by shikhasingrajput

Javascript

 
// Javascript Program to implement
// the above approach
 
// Function to check if any index
// contains more than K active characters
function checkString(s, K)
{
 
    var n = s.length;
 
    // Store the last occurrence of
    // each character in the map.
    var mp = new Map();
 
    for (var i = 0; i < n; i++) {
        if(mp.has(s[i]))
        {
            mp.set(s[i], mp.get(s[i])+1);
        }
        else
            mp.set(s[i], 1);
    }
 
    var cnt = 0, f = 0;
 
    // Stores the active
    // characters
    var st = new Set();
 
    for (var i = 0; i < n; i++) {
 
        // Insert the character
        st.add(s[i]);
 
        // If the size of set
        // exceeds K
        if (st.size > K) {
            f = 1;
            break;
        }
 
        // Remove the character from
        // set if i is the last index
        // of the current character
        if (mp.get(s[i]) == i)
            st.delete(s[i]);
    }
 
    return (f == 1 ? "Yes" : "No");
}
 
// Driver Code
var s = "aabbcdca";
var k = 2;
document.write( checkString(s, k));
 
// This code is contributed by importantly.

输出：

Yes

时间复杂度： O(N)
辅助空间： O(N)

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