查找包含恰好 K 个唯一元音的所有子串

给定长度为N的字符串str包含大写和小写字母，以及一个整数K 。任务是找到包含恰好K个不同元音的所有子串。

例子：

Input:  str = “aeiou”, K = 2
Output: “ae”, “ei”, “io”, “ou”
Explanation: These are the substrings containing exactly 2 distinct vowels.

Input: str = “TrueGeek”, K = 3
Output: “”
Explanation: Though the string has more than 3 vowels but there is not three unique vowels.
So the answer is empty.

Input: str = “TrueGoik”, K = 3
Output: “TrueGo”, “rueGo”, “ueGo”, “eGoi”, “eGoik”

方法：这个问题可以通过贪心方法来解决。生成子字符串并检查每个子字符串。请按照以下步骤解决问题：

• 首先生成从0到N范围内的每个索引i开始的所有子字符串。
• 然后对于每个子字符串：
  • 保留一个哈希数组来存储唯一元音的出现。
  • 检查子字符串中的新字符是否为元音。
  • 如果它是元音，则增加其在哈希中的出现次数并保留找到的不同元音的计数
  • 现在对于每个子字符串，如果元音的不同计数是K ，则打印子字符串。
• 如果对于任何循环查找从i开始的子串，不同元音的计数超过K ，或者，子串长度已达到字符串长度，则中断循环并查找从i+1开始的子串。

下面是上述方法的实现。

C++

// C++ program to find all substrings with
// exactly k distinct vowels
#include 
using namespace std;
 
#define MAX 128
 
// Function to check whether
// a character is vowel or not
bool isVowel(char x)
{
    return (x == 'a' || x == 'e' || x == 'i'
            || x == 'o' || x == 'u' ||
            x == 'A' || x == 'E' || x == 'I'
            || x == 'O' || x == 'U');
}
 
int getIndex(char ch)
{
    return (ch - 'A' > 26 ?
            ch - 'a' : ch - 'A');
}
 
// Function to find all substrings
// with exactly k unique vowels
void countkDist(string str, int k)
{
    int n = str.length();
   
    // Consider all substrings
    // beginning with str[i]
    for (int i = 0; i < n; i++) {
        int dist_count = 0;
 
        // To store count of characters
        // from 'a' to 'z'
        vector cnt(26, 0);
 
        // Consider all substrings
        // between str[i..j]
        for (int j = i; j < n; j++) {
 
            // If this is a new vowel
            // for this substring,
            // increment dist_count.
            if (isVowel(str[j])
                && cnt[getIndex(str[j])] == 0) {
                dist_count++;
            }
             
            // Increment count of
            // current character
            cnt[getIndex(str[j])]++;
 
            // If distinct vowels count
            // becomes k, then print the
            // substring.
            if (dist_count == k) {
                cout << str.substr(i, j - i + 1) << endl;
            }
 
            if (dist_count > k)
                break;
        }
    }
}
 
// Driver code
int main()
{
    string str = "TrueGoik";
    int K = 3;
    countkDist(str, K);
    return 0;
}

Java

// Java program to find all subStrings with
// exactly k distinct vowels
import java.util.*;
 
class GFG{
 
  static final int MAX = 128;
 
  // Function to check whether
  // a character is vowel or not
  static boolean isVowel(char x)
  {
    return (x == 'a' || x == 'e' || x == 'i'
            || x == 'o' || x == 'u' ||
            x == 'A' || x == 'E' || x == 'I'
            || x == 'O' || x == 'U');
  }
 
  static int getIndex(char ch)
  {
    return (ch - 'A' > 26 ?
            ch - 'a' : ch - 'A');
  }
 
  // Function to find all subStrings
  // with exactly k unique vowels
  static void countkDist(String str, int k)
  {
    int n = str.length();
 
    // Consider all subStrings
    // beginning with str[i]
    for (int i = 0; i < n; i++) {
      int dist_count = 0;
 
      // To store count of characters
      // from 'a' to 'z'
      int[] cnt = new int[26];
 
      // Consider all subStrings
      // between str[i..j]
      for (int j = i; j < n; j++) {
        String print  = new String(str);
        // If this is a new vowel
        // for this subString,
        // increment dist_count.
        if (isVowel(str.charAt(j))
            && cnt[getIndex(str.charAt(j))] == 0) {
          dist_count++;
        }
 
        // Increment count of
        // current character
        cnt[getIndex(str.charAt(j))]++;
 
        // If distinct vowels count
        // becomes k, then print the
        // subString.
        if (dist_count == k) {
          System.out.print(print.substring(i, j +1) +"\n");
        }
 
        if (dist_count > k)
          break;
      }
    }
  }
 
  // Driver code
  public static void main(String[] args)
  {
    String str = "TrueGoik";
    int K = 3;
    countkDist(str, K);
  }
}
 
// This code is contributed by Rajput-Ji

Python3

# python3 program to find all substrings with
# exactly k distinct vowels
MAX = 128
 
# Function to check whether
# a character is vowel or not
def isVowel(x):
 
    return (x == 'a' or x == 'e' or x == 'i'
            or x == 'o' or x == 'u' or
            x == 'A' or x == 'E' or x == 'I'
            or x == 'O' or x == 'U')
 
def getIndex(ch):
    if ord(ch) - ord('A') > 26:
        return ord(ch) - ord('a')
    else:
        return ord(ch) - ord('A')
 
# Function to find all substrings
# with exactly k unique vowels
def countkDist(str, k):
    n = len(str)
 
    # Consider all substrings
    # beginning with str[i]
    for i in range(0, n):
        dist_count = 0
 
        # To store count of characters
        # from 'a' to 'z'
        cnt = [0 for _ in range(26)]
 
        # Consider all substrings
        # between str[i..j]
        for j in range(i, n):
 
            # If this is a new vowel
            # for this substring,
            # increment dist_count.
            if (isVowel(str[j])
                    and cnt[getIndex(str[j])] == 0):
                dist_count += 1
 
            # Increment count of
            # current character
            cnt[getIndex(str[j])] += 1
 
            # If distinct vowels count
            # becomes k, then print the
            # substring.
            if (dist_count == k):
                print(str[i:i+j - i + 1])
 
            if (dist_count > k):
                break
 
# Driver code
if __name__ == "__main__":
 
    str = "TrueGoik"
    K = 3
    countkDist(str, K)
 
    # This code is contributed by rakeshsahni

C#

// C# program to find all substrings with
// exactly k distinct vowels
using System;
class GFG {
 
  // Function to check whether
  // a character is vowel or not
  static bool isVowel(char x)
  {
    return (x == 'a' || x == 'e' || x == 'i' || x == 'o'
            || x == 'u' || x == 'A' || x == 'E'
            || x == 'I' || x == 'O' || x == 'U');
  }
 
  static int getIndex(char ch)
  {
    return (ch - 'A' > 26 ? ch - 'a' : ch - 'A');
  }
 
  // Function to find all substrings
  // with exactly k unique vowels
  static void countkDist(string str, int k)
  {
    int n = str.Length;
 
    // Consider all substrings
    // beginning with str[i]
    for (int i = 0; i < n; i++) {
      int dist_count = 0;
 
      // To store count of characters
      // from 'a' to 'z'
      int[] cnt = new int[26];
 
      // Consider all substrings
      // between str[i..j]
      for (int j = i; j < n; j++) {
 
        // If this is a new vowel
        // for this substring,
        // increment dist_count.
        if (isVowel(str[j])
            && cnt[getIndex(str[j])] == 0) {
          dist_count++;
        }
 
        // Increment count of
        // current character
        cnt[getIndex(str[j])]++;
 
        // If distinct vowels count
        // becomes k, then print the
        // substring.
        if (dist_count == k) {
          Console.WriteLine(
            str.Substring(i, j - i + 1));
        }
 
        if (dist_count > k)
          break;
      }
    }
  }
 
  // Driver code
  public static void Main()
  {
    string str = "TrueGoik";
    int K = 3;
    countkDist(str, K);
  }
}
 
// This code is contributed by ukasp.

Javascript

      // JavaScript code for the above approach
 
      // Function to check whether
      // a character is vowel or not
      function isVowel(x) {
          return (x == 'a' || x == 'e' || x == 'i' || x == 'o'
              || x == 'u' || x == 'A' || x == 'E' || x == 'I'
              || x == 'O' || x == 'U');
      }
 
      function getIndex(ch) {
          return ((ch.charCodeAt(0) - 'A'.charCodeAt(0)) > 26 ? (ch.charCodeAt(0) - 'a'.charCodeAt(0)) :
              (ch.charCodeAt(0) - 'A'.charCodeAt(0)));
      }
 
      // Function to find all substrings
      // with exactly k unique vowels
      function countkDist(str, k) {
          let n = str.length;
 
          // Consider all substrings
          // beginning with str[i]
          for (let i = 0; i < n; i++) {
              let dist_count = 0;
 
              // To store count of characters
              // from 'a' to 'z'
              let cnt = new Array(26).fill(0);
 
              // Consider all substrings
              // between str[i..j]
              for (let j = i; j < n; j++) {
 
                  // If this is a new vowel
                  // for this substring,
                  // increment dist_count.
                  if (isVowel(str[j])
                      && cnt[getIndex(str[j])] == 0) {
                      dist_count++;
                  }
 
                  // Increment count of
                  // current character
                  cnt[getIndex(str[j])]++;
 
                  // If distinct vowels count
                  // becomes k, then print the
                  // substring.
                  if (dist_count == k) {
                      document.write(str.slice(i, i + j - i + 1) + '<br>');
                  }
 
                  if (dist_count > k)
                      break;
              }
          }
      }
 
      // Driver code
 
      let s = "TrueGoik";
      let K = 3
 
      countkDist(s, K);
 
// This code is contributed by Potta Lokesh
  

输出

TrueGo
rueGo
ueGo
eGoi
eGoik

时间复杂度： O(N ² )
辅助空间： O(N ² ) 存储结果子串