Dota virkne un vesels skaitlis k atrodiet apakšvirkņu skaitu, kurās visas dažādās rakstzīmes parādās tieši k reizes.
Piemēri:
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
Naiva pieeja: Ideja ir šķērsot visas apakšvirknes. Mēs nofiksējam sākuma punkta traversu cauri visām apakšvirknēm, sākot ar izvēlēto punktu, mēs turpinām palielināt visu rakstzīmju biežumu. Ja visas frekvences kļūst par k, mēs palielinām rezultātu. Ja jebkuras frekvences skaits kļūst lielāks par k, mēs pārtraucam un mainām sākuma punktu.
Tālāk ir aprakstīta iepriekš minētās pieejas īstenošana.
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>
Izvade
6 3
Laika sarežģītība: O(n*n) kur n ir ievades virknes garums. Funkciju pārbaude () izpilda konstanta garuma cilpu no 0 līdz MAX_CHAR (ti, vienmēr 26), tāpēc šī funkcijas pārbaude () darbojas O(MAX_CHAR) laikā, tāpēc laika sarežģītība ir O(MAX_CHAR*n*n)=O(n^2).
Palīgtelpa: O(1)
Efektīva pieeja: Ļoti rūpīgi novērojot, mēs varam redzēt, ka pietiek pārbaudīt to pašu apakšvirknēm ar garumu Ktimes i forall iisin[1 D] kur D ir norādītajā virknē esošo atšķirīgo rakstzīmju skaits.
Arguments:
Apsveriet apakšvirkni S_{i+1}S_{i+2}punktus S_{i+p} ar garumu “p”. Ja šai apakšvirknei ir “m” atšķirīgas rakstzīmes un katra atšķirīgā rakstzīme parādās tieši “K” reizes, tad apakšvirknes “p” garumu nosaka ar p = Ktimes m. Tā kā “p ” vienmēr ir skaitļa “K” un 1 mle 26 reizināts dotajai virknei, pietiek ar to, lai atkārtotu apakšvirknes, kuru garums dalās ar “K” un kurām ir m 1 le mle 26 atšķirīgas rakstzīmes. Mēs izmantosim bīdāmo logu, lai atkārtotu noteikta garuma apakšvirknes.
Risinājums:
- Atrodiet norādītajā virknē esošo atšķirīgo rakstzīmju skaitu. Lai tas būtu D.
- Katrai i 1le ile D rīkojieties šādi
- Atkārtojiet apakšvirknes, kuru garums ir $i reizes K$, izmantojot bīdāmo logu.
- Pārbaudiet, vai tās atbilst nosacījumam — visas apakšvirknes atšķirīgās rakstzīmes parādās tieši K reizes.
- Ja tie atbilst nosacījumam, palieliniet skaitu.
Tālāk ir aprakstīta iepriekš minētās pieejas īstenošana.
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>
Izvade
6 3
Laika sarežģītība: O(N * D) kur D ir virknē esošo atšķirīgo rakstzīmju skaits un N ir virknes garums.
Palīgtelpa: O(N)
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