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：
Input: str = “aabbba”, K = 3
Output: 6
Explanation: In substring “aabbba”, each character repeats at least k times and its length is 6.
Input: str = “ababacb”, K = 3
Output: 0
Explanation: There is no substring where each character repeats at least k times.
天真的方法：在解决方案1中讨论了解决给定问题的Simpelst方法。
时间复杂度： O(N 2 )
辅助空间： O(26)
分而治之的方法：给定问题的分而治之的方法在第2集中进行了讨论。
时间复杂度： O(N * log N)
辅助空间： O(26)
高效方法：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，例如unique 。
- 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此数组中。
- 使用变量curr_unique迭代[1，unique]范围。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。
- 用{0}重新初始化数组freq []以在此窗口中存储每个字符的频率。
- 将start和end初始化为0 ，以分别存储窗口的起点和终点。
- 使用两个变量cnt (用于存储唯一字符的数量)和countK (用于存储当前窗口中具有至少K个重复字符的字符的数量)。
- 现在，在end
- 如果cnt的值小于或等于curr_unique，则通过在窗口的末尾添加一个字符，从右侧扩展窗口。并以freq []的形式将其频率增加1 。
- 否则，通过从开头删除一个字符并在freq []中将其频率减1来减小左侧窗口的大小。
- 在每一步中，更新cnt和countK的值。
- 如果cnt的值与curr_unique相同，并且每个字符至少出现K次，则更新总的最大长度并将其存储在ans中。
- 完成上述步骤后，输出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 29AjayKumarPython3
# 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.输出：
6时间复杂度： O(N)
辅助空间： O(1)