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Poeilučių, kurių kiekvieno simbolio skaičius yra k, skaičius

Duota eilutė ir sveikasis skaičius k, raskite eilučių, kuriose visi skirtingi simboliai pasitaiko tiksliai k kartų, skaičių. 

Pavyzdžiai:  

  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

Naivus požiūris: Idėja yra pereiti per visas eilutes. Mes nustatome pradžios taško traversą per visas eilutes, pradedant nuo pasirinkto taško, ir nuolat didiname visų simbolių dažnį. Jei visi dažniai tampa k, mes padidiname rezultatą. Jei kurio nors dažnio skaičius tampa didesnis nei k, sulaužome ir keičiame pradžios tašką. 



Žemiau pateikiamas pirmiau minėto metodo įgyvendinimas:

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>

Išvestis
6 3

Laiko sudėtingumas: O(n*n) kur n yra įvesties eilutės ilgis. Function Check() vykdo pastovaus ilgio kilpą nuo 0 iki MAX_CHAR (ty; 26 visada), todėl ši funkcija check() veikia per O(MAX_CHAR) laiką, todėl laiko sudėtingumas yra O(MAX_CHAR*n*n)=O(n^2).
Pagalbinė erdvė: O(1) 

Efektyvus požiūris: Labai atidžiai stebime, kad pakanka patikrinti tą patį poeilėms, kurių ilgis Ktimes i forall iisin[1 D]                kur D              yra skirtingų simbolių, esančių nurodytoje eilutėje, skaičius.

Argumentas:
Apsvarstykite poeilelę S_{i+1}S_{i+2}taškai S_{i+p}, kurių ilgis 'p'. Jei ši poeilutė turi „m“ skirtingus simbolius ir kiekvienas atskiras simbolis pasitaiko tiksliai „K“ kartų, poeilutės „p“ ilgis pateikiamas taip, kad p = Ktimes m. Kadangi „p              “ visada yra „K“ kartotinis ir 1 le mle 26               duotąja eilute, pakanka kartoti eilutes, kurių ilgis dalijasi iš „K“ ir turintis m 1 le m le 26 skirtingus simbolius. Naudosime slankiojantį langą, kad kartotume fiksuoto ilgio eilutes.

Sprendimas:

  • Raskite skirtingų simbolių skaičių pateiktoje eilutėje. Tebūnie D.
  • Kiekvienam i 1le ile D atlikite šiuos veiksmus
    • Pakartokite eilutes, kurių ilgis $i padauginus K$, naudodami slankiojantį langą.
    • Patikrinkite, ar jie atitinka sąlygą – visi atskiri simboliai poeilėje atsiranda tiksliai K kartų.
    • Jei jie tenkina sąlygą, skaičių padidinkite.

Žemiau pateikiamas pirmiau minėto metodo įgyvendinimas:

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>

Išvestis
6 3

Laiko sudėtingumas: O (N * D) kur D yra skirtingų simbolių, esančių eilutėje, skaičius, o N yra eilutės ilgis.
Pagalbinė erdvė: O(N) 

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