频率为 K 的给定字符串每个不同字符的最大索引

给定一个由小写英文字母和整数K组成的字符串S ，任务是为 S 中的每个不同字符找到恰好包含该字符K次的最大索引。如果不存在这样的字符，则打印 -1。按字典顺序打印结果。
注意：考虑 S 中基于 0 的索引。

例子：

Input: S = “cbaabaacbcd”, K = 2 
Output: { {a 4}, {b 7}, {c 8} } 
Explanation: 
For ‘a’, the largest index having 2 a’s is “cbaab”. 
For ‘b’, the largest index having 2 b’s is “cbaabaac”. 
For ‘c’, the largest index having 2 c’s is “cbaabaacb”. 
For ‘d’, the is no index up to which we have 2 d’s

Input: P = “acbacbacbaba”, K = 3 
Output: { {a 8}, {b 9}, {c 11} }

方法：思路是先找出字符串S 中所有不同的字符，然后对于每个小写英文字符，检查它是否存在于 S 中，并从 S 的开头运行一个 for 循环并保持该字符的计数发生到现在。当计数变得等于 K 时，相应地更新索引答案。最后，在向量结果中附加这个字符及其对应的索引。
下面是上述方法的实现。

C++

// C++ implementation of the approach
 
#include 
using namespace std;
 
// Function to find largest index for each
// distinct character occuring exactly K times.
void maxSubstring(string& S, int K, int N)
{
 
    // finding all characters present in S
    int freq[26];
    memset(freq, 0, sizeof freq);
 
    // Finding all distinct characters in S
    for (int i = 0; i < N; ++i) {
        freq[S[i] - 'a'] = 1;
    }
 
    // vector to store result for each character
    vector > answer;
 
    // loop through each lower case English character
    for (int i = 0; i < 26; ++i) {
 
        // if current character is absent in s
        if (freq[i] == 0)
            continue;
 
        // getting current character
        char ch = i + 97;
 
        // finding count of character ch in S
        int count = 0;
 
        // to store max Index encountred so far
        int index = -1;
 
        for (int j = 0; j < N; ++j) {
            if (S[j] == ch)
                count++;
 
            if (count == K)
                index = j;
        }
 
        answer.push_back({ ch, index });
    }
 
    int flag = 0;
 
    // printing required result
    for (int i = 0; i < (int)answer.size(); ++i) {
 
        if (answer[i].second > -1) {
            flag = 1;
            cout << answer[i].first << " "
                << answer[i].second << endl;
        }
    }
 
    // If no such character exists, print -1
    if (flag == 0)
        cout << "-1" << endl;
}
 
// Driver code
int main()
{
    string S = "cbaabaacbcd";
    int K = 2;
    int N = S.length();
 
    maxSubstring(S, K, N);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
class GFG{
    static class pair
    {     char first;
        int  second;
        public pair(char first, int second) 
        {
            this.first = first;
            this.second = second;
        }   
}
   
// Function to find largest
//  index for each distinct
// character occuring exactly K times.
static void maxSubString(char [] S,
                         int K, int N)
{
    // finding all characters present in S
    int []freq = new int[26];   
 
    // Finding all distinct characters in S
    for (int i = 0; i < N; ++i)
    {
        freq[S[i] - 'a'] = 1;
    }
 
    // vector to store result for each character
    Vector answer = new Vector();
 
    // loop through each lower case English character
    for (int i = 0; i < 26; ++i)
    {
        // if current character is absent in s
        if (freq[i] == 0)
            continue;
 
        // getting current character
        char ch = (char) (i + 97);
 
        // finding count of character ch in S
        int count = 0;
 
        // to store max Index encountred so far
        int index = -1;
 
        for (int j = 0; j < N; ++j)
        {
            if (S[j] == ch)
                count++;
            if (count == K)
                index = j;
        }
        answer.add(new pair(ch, index ));
    }
 
    int flag = 0;
 
    // printing required result
    for (int i = 0; i < (int)answer.size(); ++i)
    {
        if (answer.get(i).second > -1)
        {
            flag = 1;
            System.out.print(answer.get(i).first + " " +
                             answer.get(i).second + "\n");
        }
    }
 
    // If no such character exists, print -1
    if (flag == 0)
        System.out.print("-1" + "\n");
}
 
// Driver code
public static void main(String[] args)
{
    String S = "cbaabaacbcd";
    int K = 2;
    int N = S.length();
    maxSubString(S.toCharArray(), K, N);
}
}
 
// This code is contributed by shikhasingrajput

Python3

# Python3 implementation of the approach
# Function to find largest index for each
# distinct character occuring exactly K times.
def maxSubstring(S, K, N):
     
    # Finding all characters present in S
    freq = [0 for i in range(26)]
     
    # Finding all distinct characters in S
    for i in range(N):
        freq[ord(S[i]) - 97] = 1
 
    # To store result for each character
    answer = []
 
    # Loop through each lower
    # case English character
    for i in range(26):
         
        # If current character is absent in s
        if (freq[i] == 0):
            continue
 
        # Getting current character
        ch = chr(i + 97)
 
        # Finding count of character ch in S
        count = 0
 
        # To store max Index encountred so far
        index = -1
 
        for j in range(N):
            if (S[j] == ch):
                count += 1
 
            if (count == K):
                index = j
 
        answer.append([ch, index])
 
    flag = 0
 
    # Printing required result
    for i in range(len(answer)):
        if (answer[i][1] > -1):
            flag = 1
            print(answer[i][0],
                  answer[i][1])
 
    # If no such character exists,
    # print -1
    if (flag == 0):
        print("-1")
 
# Driver code
if __name__ == '__main__':
     
    S = "cbaabaacbcd"
    K = 2
    N = len(S)
 
    maxSubstring(S, K, N)
 
# This code is contributed by Surendra_Gangwar

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG{
     
class pair
{    
    public char first;
    public int second;
     
    public pair(char first, int second)
    {
        this.first = first;
        this.second = second;
    }
}
 
// Function to find largest
// index for each distinct
// character occuring exactly K times.
static void maxSubString(char [] S,
                         int K, int N)
{
     
    // Finding all characters present in S
    int []freq = new int[26];
 
    // Finding all distinct characters in S
    for(int i = 0; i < N; ++i)
    {
        freq[S[i] - 'a'] = 1;
    }
 
    // To store result for each character
    List answer = new List();
 
    // Loop through each lower case
    // English character
    for(int i = 0; i < 26; ++i)
    {
         
        // If current character is absent in s
        if (freq[i] == 0)
            continue;
 
        // Getting current character
        char ch = (char)(i + 97);
 
        // Finding count of character ch in S
        int count = 0;
 
        // To store max Index encountred so far
        int index = -1;
 
        for(int j = 0; j < N; ++j)
        {
            if (S[j] == ch)
                count++;
            if (count == K)
                index = j;
        }
        answer.Add(new pair(ch, index));
    }
 
    int flag = 0;
 
    // Printing required result
    for(int i = 0; i < (int)answer.Count; ++i)
    {
        if (answer[i].second > -1)
        {
            flag = 1;
            Console.Write(answer[i].first + " " +
                          answer[i].second + "\n");
        }
    }
 
    // If no such character exists, print -1
    if (flag == 0)
        Console.Write("-1" + "\n");
}
 
// Driver code
public static void Main(String[] args)
{
    String S = "cbaabaacbcd";
    int K = 2;
    int N = S.Length;
     
    maxSubString(S.ToCharArray(), K, N);
}
}
 
// This code is contributed by Amit Katiyar

Javascript

      // JavaScript implementation of the approach
      // Function to find largest
      // index for each distinct
      // character occuring exactly K times.
      function maxSubString(S, K, N) {
        // Finding all characters present in S
        var freq = new Array(26).fill(0);
 
        // Finding all distinct characters in S
        for (var i = 0; i < N; ++i) {
          freq[S[i].charCodeAt(0) - "a".charCodeAt(0)] = 1;
        }
 
        // To store result for each character
        var answer = [];
 
        // Loop through each lower case
        // English character
        for (var i = 0; i < 26; ++i) {
          // If current character is absent in s
          if (freq[i] === 0)
              continue;
 
          // Getting current character
          var ch = String.fromCharCode(i + 97);
 
          // Finding count of character ch in S
          var count = 0;
 
          // To store max Index encountred so far
          var index = -1;
 
          for (var j = 0; j < N; ++j) {
            if (S[j] === ch) count++;
            if (count === K) index = j;
          }
          answer.push([ch, index]);
        }
 
        var flag = 0;
 
        // Printing required result
        for (var i = 0; i < answer.length; ++i) {
          if (answer[i][1] > -1) {
            flag = 1;
            document.write(answer[i][0] + " " + answer[i][1] + "<br>");
          }
        }
 
        // If no such character exists, print -1
        if (flag === 0)
            document.write("-1" + "<br>");
      }
 
      // Driver code
      var S = "cbaabaacbcd";
      var K = 2;
      var N = S.length;
 
      maxSubString(S.split(""), K, N);

输出：

a 4
b 7
c 8

时间复杂度： O(26 * N)

如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。