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📜  所有字符至少出现 K 次的最长子串 |设置 3

📅  最后修改于: 2021-09-06 06:40:02             🧑  作者: Mango

给定一个字符串str和一个整数K,任务是找到最长子串的长度S,使得S中的每一个字符出现至少K倍。

例子:

朴素方法:解决给定问题的最简单方法在第 1 组中讨论。
时间复杂度: O(N 2 )
辅助空间: O(26)

分治法:针对给定问题的分治法在第 2 组中讨论。
时间复杂度: O(N*log N)
辅助空间: O(26)

高效的方法:可以通过使用滑动窗口技术进一步优化上述两种方法。请按照以下步骤解决问题:

  • 将字符串str中唯一字符的数量存储在一个变量中,比如unique
  • 使用{0}初始化大小为26的数组freq[]并将每个字符的频率存储在该数组中。
  • 使用变量curr_unique迭代范围[1, unique] 。在每次迭代中, curr_unique是窗口中必须存在的唯一字符的最大数量。
    • 使用{0}重新初始化数组freq[]以存储此窗口中每个字符的频率。
    • startend初始化为0 ,分别存储窗口的起点和终点。
    • 使用两个变量CNT,用于存储唯一的字符和countK数,用于存储字符的数目与至少K的当前窗口重复字符。
    • 现在,在end < N 时迭代一个循环,并执行以下操作:
      • 如果cnt的值小于或等于curr_unique,则通过在窗口末尾添加一个字符从右侧扩展窗口。并将其频率在freq[] 中增加1
      • 否则,通过从开始处删除一个字符并将其频率在freq[] 中减少1来从左侧减少窗口。
      • 在每一步,更新cntcountK的值。
      • 如果cnt的值与curr_unique相同,并且每个字符至少出现K 次,则更新整体最大长度并将其存储在ans 中
  • 完成以上步骤后,打印ans的值作为结果。

下面是上述方法的实现:

C++
// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the length of
// the longest substring
int longestSubstring(string s, int k)
{
    // Store the required answer
    int ans = 0;
 
    // Create a frequency map of the
    // characters of the string
    int freq[26] = { 0 };
 
    // Store the length of the string
    int n = s.size();
 
    // Traverse the string, s
    for (int i = 0; i < n; i++)
 
        // Increment the frequency of
        // the current character by 1
        freq[s[i] - 'a']++;
 
    // Stores count of unique characters
    int unique = 0;
 
    // Find the number of unique
    // characters in string
    for (int i = 0; i < 26; i++)
        if (freq[i] != 0)
            unique++;
 
    // Iterate in range [1, unique]
    for (int curr_unique = 1;
         curr_unique <= unique;
         curr_unique++) {
 
        // Initialize frequency of all
        // characters as 0
        memset(freq, 0, sizeof(freq));
 
        // Stores the start and the
        // end of the window
        int start = 0, end = 0;
 
        // Stores the current number of
        // unique characters and characters
        // occuring atleast K times
        int cnt = 0, count_k = 0;
 
        while (end < n) {
            if (cnt <= curr_unique) {
                int ind = s[end] - 'a';
 
                // New unique character
                if (freq[ind] == 0)
                    cnt++;
 
                freq[ind]++;
 
                // New character which
                // occurs atleast k times
                if (freq[ind] == k)
                    count_k++;
 
                // Expand window by
                // incrementing end by 1
                end++;
            }
            else {
                int ind = s[start] - 'a';
 
                // Check if this character
                // is present atleast k times
                if (freq[ind] == k)
                    count_k--;
 
                freq[ind]--;
 
                // Check if this character
                // is unique
                if (freq[ind] == 0)
                    cnt--;
 
                // Shrink the window by
                // incrementing start by 1
                start++;
            }
 
            // If there are curr_unique
            // characters and each character
            // is atleast k times
            if (cnt == curr_unique
                && count_k == curr_unique)
 
                // Update the overall
                // maximum length
                ans = max(ans, end - start);
        }
    }
 
    // Print the answer
    cout << ans;
}
 
// Driver Code
int main()
{
    string S = "aabbba";
    int K = 3;
    longestSubstring(S, K);
 
    return 0;
}


Java
// Java program for the above approach
import java.util.*;
class GFG
{
 
// Function to find the length of
// the longest subString
static void longestSubString(char[] s, int k)
{
   
    // Store the required answer
    int ans = 0;
 
    // Create a frequency map of the
    // characters of the String
    int freq[] = new int[26];
 
    // Store the length of the String
    int n = s.length;
 
    // Traverse the String, s
    for (int i = 0; i < n; i++)
 
        // Increment the frequency of
        // the current character by 1
        freq[s[i] - 'a']++;
 
    // Stores count of unique characters
    int unique = 0;
 
    // Find the number of unique
    // characters in String
    for (int i = 0; i < 26; i++)
        if (freq[i] != 0)
            unique++;
 
    // Iterate in range [1, unique]
    for (int curr_unique = 1;
         curr_unique <= unique;
         curr_unique++)
    {
 
        // Initialize frequency of all
        // characters as 0
        Arrays.fill(freq, 0);
 
        // Stores the start and the
        // end of the window
        int start = 0, end = 0;
 
        // Stores the current number of
        // unique characters and characters
        // occuring atleast K times
        int cnt = 0, count_k = 0;
        while (end < n)
        {
            if (cnt <= curr_unique)
            {
                int ind = s[end] - 'a';
 
                // New unique character
                if (freq[ind] == 0)
                    cnt++;
                freq[ind]++;
 
                // New character which
                // occurs atleast k times
                if (freq[ind] == k)
                    count_k++;
 
                // Expand window by
                // incrementing end by 1
                end++;
            }
            else
            {
                int ind = s[start] - 'a';
 
                // Check if this character
                // is present atleast k times
                if (freq[ind] == k)
                    count_k--;
                freq[ind]--;
 
                // Check if this character
                // is unique
                if (freq[ind] == 0)
                    cnt--;
 
                // Shrink the window by
                // incrementing start by 1
                start++;
            }
 
            // If there are curr_unique
            // characters and each character
            // is atleast k times
            if (cnt == curr_unique
                && count_k == curr_unique)
 
                // Update the overall
                // maximum length
                ans = Math.max(ans, end - start);
        }
    }
 
    // Print the answer
    System.out.print(ans);
}
 
// Driver Code
public static void main(String[] args)
{
    String S = "aabbba";
    int K = 3;
    longestSubString(S.toCharArray(), K);
}
}
 
// This code is contributed by 29AjayKumar


Python3
# Python3 program for the above approach
 
# Function to find the length of
# the longest substring
def longestSubstring(s, k) :
 
    # Store the required answer
    ans = 0
 
    # Create a frequency map of the
    # characters of the string
    freq = [0]*26
 
    # Store the length of the string
    n = len(s)
 
    # Traverse the string, s
    for i in range(n) :
 
        # Increment the frequency of
        # the current character by 1
        freq[ord(s[i]) - ord('a')] += 1
 
    # Stores count of unique characters
    unique = 0
 
    # Find the number of unique
    # characters in string
    for i in range(26) :
        if (freq[i] != 0) :
            unique += 1
 
    # Iterate in range [1, unique]
    for curr_unique in range(1, unique + 1) :
 
        # Initialize frequency of all
        # characters as 0
        Freq = [0]*26
 
        # Stores the start and the
        # end of the window
        start, end = 0, 0
 
        # Stores the current number of
        # unique characters and characters
        # occuring atleast K times
        cnt, count_k = 0, 0
 
        while (end < n) :
            if (cnt <= curr_unique) :
                ind = ord(s[end]) - ord('a')
 
                # New unique character
                if (Freq[ind] == 0) :
                    cnt += 1
 
                Freq[ind] += 1
 
                # New character which
                # occurs atleast k times
                if (Freq[ind] == k) :
                    count_k += 1
 
                # Expand window by
                # incrementing end by 1
                end += 1
             
            else :
                ind = ord(s[start]) - ord('a')
 
                # Check if this character
                # is present atleast k times
                if (Freq[ind] == k) :
                    count_k -= 1
 
                Freq[ind] -= 1
 
                # Check if this character
                # is unique
                if (Freq[ind] == 0) :
                    cnt -= 1
 
                # Shrink the window by
                # incrementing start by 1
                start += 1
 
            # If there are curr_unique
            # characters and each character
            # is atleast k times
            if ((cnt == curr_unique) and (count_k == curr_unique)) :
 
                # Update the overall
                # maximum length
                ans = max(ans, end - start)
 
    # Print the answer
    print(ans)
 
S = "aabbba"
K = 3
longestSubstring(S, K)
 
# This code is contributed by divyesh072019.


C#
// C# program to implement
// the above approach
using System;
 
class GFG
{
   
// Function to find the length of
// the longest substring
static void longestSubstring(string s, int k)
{
   
    // Store the required answer
    int ans = 0;
 
    // Create a frequency map of the
    // characters of the string
    int[] freq = new int[26];
 
    // Store the length of the string
    int n = s.Length;
 
    // Traverse the string, s
    for (int i = 0; i < n; i++)
 
        // Increment the frequency of
        // the current character by 1
        freq[s[i] - 'a']++;
 
    // Stores count of unique characters
    int unique = 0;
 
    // Find the number of unique
    // characters in string
    for (int i = 0; i < 26; i++)
        if (freq[i] != 0)
            unique++;
 
    // Iterate in range [1, unique]
    for (int curr_unique = 1;
         curr_unique <= unique;
         curr_unique++)
    {
 
        // Initialize frequency of all
        // characters as 0
        for (int i = 0; i < freq.Length; i++)
        {
            freq[i] = 0;
        }
         
        // Stores the start and the
        // end of the window
        int start = 0, end = 0;
 
        // Stores the current number of
        // unique characters and characters
        // occuring atleast K times
        int cnt = 0, count_k = 0;
        while (end < n)
        {
            if (cnt <= curr_unique)
            {
                int ind = s[end] - 'a';
 
                // New unique character
                if (freq[ind] == 0)
                    cnt++;
                freq[ind]++;
 
                // New character which
                // occurs atleast k times
                if (freq[ind] == k)
                    count_k++;
 
                // Expand window by
                // incrementing end by 1
                end++;
            }
            else
            {
                int ind = s[start] - 'a';
 
                // Check if this character
                // is present atleast k times
                if (freq[ind] == k)
                    count_k--;
                freq[ind]--;
 
                // Check if this character
                // is unique
                if (freq[ind] == 0)
                    cnt--;
 
                // Shrink the window by
                // incrementing start by 1
                start++;
            }
 
            // If there are curr_unique
            // characters and each character
            // is atleast k times
            if (cnt == curr_unique
                && count_k == curr_unique)
 
                // Update the overall
                // maximum length
                ans = Math.Max(ans, end - start);
        }
    }
 
    // Print the answer
    Console.Write(ans);
}
 
// Driver Code
public static void Main()
{
   string S = "aabbba";
    int K = 3;
    longestSubstring(S, K);
}
}
 
// This code is contributed by splevel62.


Javascript


输出:
6

时间复杂度: O(N)
辅助空间: O(1)

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