문자열과 정수 k가 주어지면 모든 다른 문자가 정확히 k 번 나타나는 부분 문자열의 수를 찾습니다.
예:
Input : s = 'aabbcc' k = 2 Output : 6 The substrings are aa bb cc aabb bbcc and aabbcc. Input : s = 'aabccc' k = 2 Output : 3 There are three substrings aa cc and cc
순진한 접근 방식: 아이디어는 모든 하위 문자열을 탐색하는 것입니다. 선택한 지점부터 시작하여 모든 하위 문자열을 통과하는 시작점을 수정하고 모든 문자의 빈도를 계속 증가시킵니다. 모든 주파수가 k가 되면 결과가 증가합니다. 어떤 주파수의 개수가 k보다 커지면 중단하고 시작점을 변경합니다.
다음은 위의 접근 방식을 구현한 것입니다.
C++// C++ program to count number of substrings // with counts of distinct characters as k. #include using namespace std; const int MAX_CHAR = 26; // Returns true if all values // in freq[] are either 0 or k. bool check(int freq[] int k) { for (int i = 0; i < MAX_CHAR; i++) if (freq[i] && freq[i] != k) return false; return true; } // Returns count of substrings where frequency // of every present character is k int substrings(string s int k) { int res = 0; // Initialize result // Pick a starting point for (int i = 0; s[i]; i++) { // Initialize all frequencies as 0 // for this starting point int freq[MAX_CHAR] = { 0 }; // One by one pick ending points for (int j = i; s[j]; j++) { // Increment frequency of current char int index = s[j] - 'a'; freq[index]++; // If frequency becomes more than // k we can't have more substrings // starting with i if (freq[index] > k) break; // If frequency becomes k then check // other frequencies as well. else if (freq[index] == k && check(freq k) == true) res++; } } return res; } // Driver code int main() { string s = 'aabbcc'; int k = 2; cout << substrings(s k) << endl; s = 'aabbc'; k = 2; cout << substrings(s k) << endl; }
Java // Java program to count number of substrings // with counts of distinct characters as k. import java.io.*; class GFG { static int MAX_CHAR = 26; // Returns true if all values // in freq[] are either 0 or k. static boolean check(int freq[] int k) { for (int i = 0; i < MAX_CHAR; i++) if (freq[i] !=0 && freq[i] != k) return false; return true; } // Returns count of substrings where frequency // of every present character is k static int substrings(String s int k) { int res = 0; // Initialize result // Pick a starting point for (int i = 0; i< s.length(); i++) { // Initialize all frequencies as 0 // for this starting point int freq[] = new int[MAX_CHAR]; // One by one pick ending points for (int j = i; j<s.length(); j++) { // Increment frequency of current char int index = s.charAt(j) - 'a'; freq[index]++; // If frequency becomes more than // k we can't have more substrings // starting with i if (freq[index] > k) break; // If frequency becomes k then check // other frequencies as well. else if (freq[index] == k && check(freq k) == true) res++; } } return res; } // Driver code public static void main(String[] args) { String s = 'aabbcc'; int k = 2; System.out.println(substrings(s k)); s = 'aabbc'; k = 2; System.out.println(substrings(s k)); } } // This code has been contributed by 29AjayKumar
Python3 # Python3 program to count number of substrings # with counts of distinct characters as k. MAX_CHAR = 26 # Returns true if all values # in freq[] are either 0 or k. def check(freq k): for i in range(0 MAX_CHAR): if(freq[i] and freq[i] != k): return False return True # Returns count of substrings where # frequency of every present character is k def substrings(s k): res = 0 # Initialize result # Pick a starting point for i in range(0 len(s)): # Initialize all frequencies as 0 # for this starting point freq = [0] * MAX_CHAR # One by one pick ending points for j in range(i len(s)): # Increment frequency of current char index = ord(s[j]) - ord('a') freq[index] += 1 # If frequency becomes more than # k we can't have more substrings # starting with i if(freq[index] > k): break # If frequency becomes k then check # other frequencies as well elif(freq[index] == k and check(freq k) == True): res += 1 return res # Driver Code if __name__ == '__main__': s = 'aabbcc' k = 2 print(substrings(s k)) s = 'aabbc'; k = 2; print(substrings(s k)) # This code is contributed # by Sairahul Jella
C# // C# program to count number of substrings // with counts of distinct characters as k. using System; class GFG { static int MAX_CHAR = 26; // Returns true if all values // in freq[] are either 0 or k. static bool check(int []freq int k) { for (int i = 0; i < MAX_CHAR; i++) if (freq[i] != 0 && freq[i] != k) return false; return true; } // Returns count of substrings where frequency // of every present character is k static int substrings(String s int k) { int res = 0; // Initialize result // Pick a starting point for (int i = 0; i < s.Length; i++) { // Initialize all frequencies as 0 // for this starting point int []freq = new int[MAX_CHAR]; // One by one pick ending points for (int j = i; j < s.Length; j++) { // Increment frequency of current char int index = s[j] - 'a'; freq[index]++; // If frequency becomes more than // k we can't have more substrings // starting with i if (freq[index] > k) break; // If frequency becomes k then check // other frequencies as well. else if (freq[index] == k && check(freq k) == true) res++; } } return res; } // Driver code public static void Main(String[] args) { String s = 'aabbcc'; int k = 2; Console.WriteLine(substrings(s k)); s = 'aabbc'; k = 2; Console.WriteLine(substrings(s k)); } } /* This code contributed by PrinciRaj1992 */
PHP // PHP program to count number of substrings // with counts of distinct characters as k. $MAX_CHAR = 26; // Returns true if all values // in freq[] are either 0 or k. function check(&$freq $k) { global $MAX_CHAR; for ($i = 0; $i < $MAX_CHAR; $i++) if ($freq[$i] && $freq[$i] != $k) return false; return true; } // Returns count of substrings where frequency // of every present character is k function substrings($s $k) { global $MAX_CHAR; $res = 0; // Initialize result // Pick a starting point for ($i = 0; $i < strlen($s); $i++) { // Initialize all frequencies as 0 // for this starting point $freq = array_fill(0 $MAX_CHARNULL); // One by one pick ending points for ($j = $i; $j < strlen($s); $j++) { // Increment frequency of current char $index = ord($s[$j]) - ord('a'); $freq[$index]++; // If frequency becomes more than // k we can't have more substrings // starting with i if ($freq[$index] > $k) break; // If frequency becomes k then check // other frequencies as well. else if ($freq[$index] == $k && check($freq $k) == true) $res++; } } return $res; } // Driver code $s = 'aabbcc'; $k = 2; echo substrings($s $k).'n'; $s = 'aabbc'; $k = 2; echo substrings($s $k).'n'; // This code is contributed by Ita_c. ?> JavaScript <script> // Javascript program to count number of // substrings with counts of distinct // characters as k. let MAX_CHAR = 26; // Returns true if all values // in freq[] are either 0 or k. function check(freqk) { for(let i = 0; i < MAX_CHAR; i++) if (freq[i] != 0 && freq[i] != k) return false; return true; } // Returns count of substrings where frequency // of every present character is k function substrings(s k) { // Initialize result let res = 0; // Pick a starting point for(let i = 0; i< s.length; i++) { // Initialize all frequencies as 0 // for this starting point let freq = new Array(MAX_CHAR); for(let i = 0; i < freq.length ;i++) { freq[i] = 0; } // One by one pick ending points for(let j = i; j < s.length; j++) { // Increment frequency of current char let index = s[j].charCodeAt(0) - 'a'.charCodeAt(0); freq[index]++; // If frequency becomes more than // k we can't have more substrings // starting with i if (freq[index] > k) break; // If frequency becomes k then check // other frequencies as well. else if (freq[index] == k && check(freq k) == true) res++; } } return res; } // Driver code let s = 'aabbcc'; let k = 2; document.write(substrings(s k) + '
'); s = 'aabbc'; k = 2; document.write(substrings(s k) + '
'); // This code is contributed by avanitrachhadiya2155 </script>
산출
6 3
시간 복잡도: O(n*n) 여기서 n은 입력 문자열의 길이입니다. Check() 함수는 0에서 MAX_CHAR(즉, 항상 26)까지 일정한 길이의 루프를 실행하므로 이 함수 check()는 O(MAX_CHAR) 시간에 실행되므로 시간 복잡도는 O(MAX_CHAR*n*n)=O(n^2)입니다.
보조 공간: 오(1)
효율적인 접근 방식: 매우 주의 깊게 관찰하면 길이가 Ktimes i forall iisin[1 D] 여기서 D 는 주어진 문자열에 존재하는 고유 문자의 수인 부분 문자열에 대해 동일한 내용을 확인하는 것으로 충분하다는 것을 알 수 있습니다.
논쟁:
길이가 'p'인 하위 문자열 S_{i+1}S_{i+2}점 S_{i+p}를 고려하세요. 이 하위 문자열에 'm'개의 고유 문자가 있고 각 고유 문자가 정확히 'K'회 발생하는 경우 하위 문자열 'p'의 길이는 p = Ktimes m으로 지정됩니다. 'p '는 항상 'K'의 배수이고 1le mle 26 주어진 문자열에 대해 길이가 'K'로 나누어지고 m 1 le m le 26개의 개별 문자가 있는 하위 문자열을 반복하는 것으로 충분합니다. 고정 길이의 하위 문자열을 반복하기 위해 슬라이딩 윈도우를 사용할 것입니다.
해결책:
- 주어진 문자열에 존재하는 고유 문자 수를 찾습니다. D라고 하자.
- 각 i 1le ile D에 대해 다음을 수행합니다.
- 슬라이딩 윈도우를 사용하여 $i 곱하기 K$ 길이의 하위 문자열을 반복합니다.
- 조건을 충족하는지 확인하십시오. 하위 문자열의 모든 고유 문자는 정확히 K 번 발생합니다.
- 조건을 만족하면 개수가 증가합니다.
다음은 위의 접근 방식을 구현한 것입니다.
C++#include #include #include #include int min(int a int b) { return a < b ? a : b; } using namespace std; bool have_same_frequency(map<char int>& freq int k) { for (auto& pair : freq) { if (pair.second != k && pair.second != 0) { return false; } } return true; } int count_substrings(string s int k) { int count = 0; int distinct = (set<char>(s.begin() s.end())).size(); for (int length = 1; length <= distinct; length++) { int window_length = length * k; map<char int> freq; int window_start = 0; int window_end = window_start + window_length - 1; for (int i = window_start; i <= min(window_end s.length() - 1); i++) { freq[s[i]]++; } while (window_end < s.length()) { if (have_same_frequency(freq k)) { count++; } freq[s[window_start]]--; window_start++; window_end++; if (window_length < s.length()) { freq[s[window_end]]++; } } } return count; } int main() { string s = 'aabbcc'; int k = 2; cout << count_substrings(s k) << endl; s = 'aabbc'; k = 2; cout << count_substrings(s k) << endl; return 0; }
C #include #include #include int min(int a int b) { return a < b ? a : b; } bool have_same_frequency(int freq[] int k) { for (int i = 0; i < 26; i++) { if (freq[i] != 0 && freq[i] != k) { return false; } } return true; } int count_substrings(char* s int n int k) { int count = 0; int distinct = 0; bool have[26] = { false }; for (int i = 0; i < n; i++) { have[s[i] - 'a'] = true; } for (int i = 0; i < 26; i++) { if (have[i]) { distinct++; } } for (int length = 1; length <= distinct; length++) { int window_length = length * k; int freq[26] = { 0 }; int window_start = 0; int window_end = window_start + window_length - 1; for (int i = window_start; i <= min(window_end n - 1); i++) { freq[s[i] - 'a']++; } while (window_end < n) { if (have_same_frequency(freq k)) { count++; } freq[s[window_start] - 'a']--; window_start++; window_end++; if (window_end < n) { freq[s[window_end] - 'a']++; } } } return count; } int main() { char* s = 'aabbcc'; int k = 2; printf('%dn' count_substrings(s 6 k)); s = 'aabbc'; k = 2; printf('%dn' count_substrings(s 5 k)); return 0; }
Java import java.util.*; class GFG { static boolean have_same_frequency(int[] freq int k) { for (int i = 0; i < 26; i++) { if (freq[i] != 0 && freq[i] != k) { return false; } } return true; } static int count_substrings(String s int k) { int count = 0; int distinct = 0; boolean[] have = new boolean[26]; Arrays.fill(have false); for (int i = 0; i < s.length(); i++) { have[((int)(s.charAt(i) - 'a'))] = true; } for (int i = 0; i < 26; i++) { if (have[i]) { distinct++; } } for (int length = 1; length <= distinct; length++) { int window_length = length * k; int[] freq = new int[26]; Arrays.fill(freq 0); int window_start = 0; int window_end = window_start + window_length - 1; for (int i = window_start; i <= Math.min(window_end s.length() - 1); i++) { freq[((int)(s.charAt(i) - 'a'))]++; } while (window_end < s.length()) { if (have_same_frequency(freq k)) { count++; } freq[( (int)(s.charAt(window_start) - 'a'))]--; window_start++; window_end++; if (window_end < s.length()) { freq[((int)(s.charAt(window_end) - 'a'))]++; } } } return count; } public static void main(String[] args) { String s = 'aabbcc'; int k = 2; System.out.println(count_substrings(s k)); s = 'aabbc'; k = 2; System.out.println(count_substrings(s k)); } }
Python3 from collections import defaultdict def have_same_frequency(freq: defaultdict k: int): return all([freq[i] == k or freq[i] == 0 for i in freq]) def count_substrings(s: str k: int) -> int: count = 0 distinct = len(set([i for i in s])) for length in range(1 distinct + 1): window_length = length * k freq = defaultdict(int) window_start = 0 window_end = window_start + window_length - 1 for i in range(window_start min(window_end + 1 len(s))): freq[s[i]] += 1 while window_end < len(s): if have_same_frequency(freq k): count += 1 freq[s[window_start]] -= 1 window_start += 1 window_end += 1 if window_end < len(s): freq[s[window_end]] += 1 return count if __name__ == '__main__': s = 'aabbcc' k = 2 print(count_substrings(s k)) s = 'aabbc' k = 2 print(count_substrings(s k))
C# using System; class GFG{ static bool have_same_frequency(int[] freq int k) { for(int i = 0; i < 26; i++) { if (freq[i] != 0 && freq[i] != k) { return false; } } return true; } static int count_substrings(string s int k) { int count = 0; int distinct = 0; bool[] have = new bool[26]; Array.Fill(have false); for(int i = 0; i < s.Length; i++) { have[((int)(s[i] - 'a'))] = true; } for(int i = 0; i < 26; i++) { if (have[i]) { distinct++; } } for(int length = 1; length <= distinct; length++) { int window_length = length * k; int[] freq = new int[26]; Array.Fill(freq 0); int window_start = 0; int window_end = window_start + window_length - 1; for(int i = window_start; i <= Math.Min(window_end s.Length - 1); i++) { freq[((int)(s[i] - 'a'))]++; } while (window_end < s.Length) { if (have_same_frequency(freq k)) { count++; } freq[((int)(s[window_start] - 'a'))]--; window_start++; window_end++; if (window_end < s.Length) { freq[((int)(s[window_end] - 'a'))]++; } } } return count; } // Driver code public static void Main(string[] args) { string s = 'aabbcc'; int k = 2; Console.WriteLine(count_substrings(s k)); s = 'aabbc'; k = 2; Console.WriteLine(count_substrings(s k)); } } // This code is contributed by gaurav01
JavaScript <script> function have_same_frequency(freqk) { for (let i = 0; i < 26; i++) { if (freq[i] != 0 && freq[i] != k) { return false; } } return true; } function count_substrings(sk) { let count = 0; let distinct = 0; let have = new Array(26); for(let i=0;i<26;i++) { have[i]=false; } for (let i = 0; i < s.length; i++) { have[((s[i].charCodeAt(0) - 'a'.charCodeAt(0)))] = true; } for (let i = 0; i < 26; i++) { if (have[i]) { distinct++; } } for (let length = 1; length <= distinct; length++) { let window_length = length * k; let freq = new Array(26); for(let i=0;i<26;i++) freq[i]=0; let window_start = 0; let window_end = window_start + window_length - 1; for (let i = window_start; i <= Math.min(window_end s.length - 1); i++) { freq[((s[i].charCodeAt(0) - 'a'.charCodeAt(0)))]++; } while (window_end < s.length) { if (have_same_frequency(freq k)) { count++; } freq[( (s[window_start].charCodeAt(0) - 'a'.charCodeAt(0)))]--; window_start++; window_end++; if (window_end < s.length) { freq[(s[window_end].charCodeAt(0) - 'a'.charCodeAt(0))]++; } } } return count; } let s = 'aabbcc'; let k = 2; document.write(count_substrings(s k)+'
'); s = 'aabbc'; k = 2; document.write(count_substrings(s k)+'
'); // This code is contributed by rag2127 </script>
산출
6 3
시간 복잡도: O(N * D) 여기서 D는 문자열에 존재하는 고유 문자의 수이고 N은 문자열의 길이입니다.
보조 공간: 에)
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