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Contar substrings com k caracteres distintos

Dada uma string s consistindo apenas de letras minúsculas em inglês e um número inteiro k, conte o número total de substrings (não necessariamente distintas) de s que contêm exatamente k caracteres distintos.
Observação:

onclick javascript
  • Uma substring é uma sequência contígua de caracteres dentro de uma string.
  • Substrings que são idênticas, mas ocorrem em posições diferentes, devem ser contadas separadamente.

Exemplos:  



Entrada: s = 'abc' k = 2
Saída: 2
Explicação: Substrings possíveis são ['ab' 'bc']

Entrada: s = 'aba' k = 2
Saída: 3
Explicação: Substrings possíveis são ['ab' 'ba' 'aba']

Entrada: s = 'AA' k = 1
Saída: 3
Explicação: Substrings possíveis são ['a' 'a' 'aa']



vb e vb líquido

Índice

[Abordagem ingênua] Verificando todas as substrings - tempo O (n ^ 2) e espaço O (1)

A ideia é verificar todas as substrings possíveis iterando todas as posições iniciais (i) e finais (j) possíveis na string. Para cada substring, mantenha uma matriz booleana para rastrear caracteres distintos e um contador para o número de caracteres distintos. À medida que expande a substring da esquerda para a direita, ele atualiza a contagem de caracteres distintos, verificando se cada novo caractere foi visto antes. Sempre que o número de caracteres distintos corresponder exatamente ao k fornecido, ele incrementará a contagem de respostas.

C++ 
#include    #include  using namespace std; int countSubstr(string &s int k) {  int n = s.length();  int ans = 0;    for (int i=0; i<n; i++) {    // array to check if a character   // is present in substring i..j  vector<bool> map(26 0);  int distinctCnt = 0;    for (int j=i; j<n; j++) {    // if new character is present  // increment distinct count.  if (map[s[j] - 'a'] == false) {  map[s[j] - 'a'] = true;  distinctCnt++;  }    // if distinct count is equal to k.  if (distinctCnt == k) ans++;  }  }    return ans; } int main() {  string s = 'abc';  int k = 2;    cout << countSubstr(s k);  return 0; }
Java 
class GfG {  static int countSubstr(String s int k) {  int n = s.length();  int ans = 0;  for (int i = 0; i < n; i++) {  // array to check if a character   // is present in substring i..j  boolean[] map = new boolean[26];  int distinctCnt = 0;  for (int j = i; j < n; j++) {  // if new character is present  // increment distinct count.  if (!map[s.charAt(j) - 'a']) {  map[s.charAt(j) - 'a'] = true;  distinctCnt++;  }  // if distinct count is equal to k.  if (distinctCnt == k) ans++;  }  }  return ans;  }  public static void main(String[] args) {  String s = 'abc';  int k = 2;  System.out.println(countSubstr(s k));  } }
Python 
def countSubstr(s k): n = len(s) ans = 0 for i in range(n): # array to check if a character  # is present in substring i..j map = [False] * 26 distinctCnt = 0 for j in range(i n): # if new character is present # increment distinct count. if not map[ord(s[j]) - ord('a')]: map[ord(s[j]) - ord('a')] = True distinctCnt += 1 # if distinct count is equal to k. if distinctCnt == k: ans += 1 return ans if __name__ == '__main__': s = 'abc' k = 2 print(countSubstr(s k))
C# 
using System; class GfG {  static int countSubstr(string s int k) {  int n = s.Length;  int ans = 0;  for (int i = 0; i < n; i++) {  // array to check if a character   // is present in substring i..j  bool[] map = new bool[26];  int distinctCnt = 0;  for (int j = i; j < n; j++) {  // if new character is present  // increment distinct count.  if (!map[s[j] - 'a']) {  map[s[j] - 'a'] = true;  distinctCnt++;  }  // if distinct count is equal to k.  if (distinctCnt == k) ans++;  }  }  return ans;  }  static void Main() {  string s = 'abc';  int k = 2;  Console.WriteLine(countSubstr(s k));  } }
JavaScript 
function countSubstr(s k) {  let n = s.length;  let ans = 0;  for (let i = 0; i < n; i++) {  // array to check if a character   // is present in substring i..j  let map = new Array(26).fill(false);  let distinctCnt = 0;  for (let j = i; j < n; j++) {  // if new character is present  // increment distinct count.  if (!map[s.charCodeAt(j) - 'a'.charCodeAt(0)]) {  map[s.charCodeAt(j) - 'a'.charCodeAt(0)] = true;  distinctCnt++;  }  // if distinct count is equal to k.  if (distinctCnt === k) ans++;  }  }  return ans; } // Driver Code let s = 'abc'; let k = 2; console.log(countSubstr(s k));

Saída 
2

[Abordagem Eficiente] Usando o Método de Janela Deslizante - O(n) tempo e O(1) espaço

A ideia é usar janela deslizante técnica para contar substrings com eficiência com no máximo k caracteres distintos e, em seguida, subtrair a contagem de substrings com no máximo k-1 caracteres distintos para obter o número de substrings com exatamente k caracteres distintos.



hashmap em java

Implementação passo a passo:

  • Use uma janela deslizante com uma matriz de tamanho 26 para rastrear as frequências dos caracteres.
  • Expanda a janela à direita adicionando caracteres.
  • Reduza a janela da esquerda quando caracteres distintos excederem k.
  • Conte todas as substrings válidas na janela.
  • Subtraia substrings com k-1 caracteres distintos de k caracteres distintos.
C++ 
#include    #include  using namespace std; // function which finds the number of  // substrings with atmost k Distinct // characters. int count(string &s int k) {  int n = s.length();  int ans = 0;    // use sliding window technique  vector<int> freq(26 0);  int distinctCnt = 0;  int i = 0;    for (int j = 0; j < n; j++) {    // expand window and add character  freq[s[j] - 'a']++;  if (freq[s[j] - 'a'] == 1) distinctCnt++;    // shrink window if distinct characters exceed k  while (distinctCnt > k) {  freq[s[i] - 'a']--;  if (freq[s[i] - 'a'] == 0) distinctCnt--;  i++;  }    // add number of valid substrings ending at j  ans += j - i + 1;  }    return ans; } // function to find the number of substrings // with exactly k Distinct characters. int countSubstr(string &s int k) {  int n = s.length();  int ans = 0;    // subtract substrings with at most   // k-1 distinct characters from substrings  // with at most k distinct characters  ans = count(s k) - count(s k-1);    return ans; } int main() {  string s = 'abc';  int k = 2;  cout << countSubstr(s k);  return 0; }
Java 
class GfG {  // function which finds the number of   // substrings with atmost k Distinct  // characters.  static int count(String s int k) {  int n = s.length();  int ans = 0;  // use sliding window technique  int[] freq = new int[26];  int distinctCnt = 0;  int i = 0;  for (int j = 0; j < n; j++) {  // expand window and add character  freq[s.charAt(j) - 'a']++;  if (freq[s.charAt(j) - 'a'] == 1) distinctCnt++;  // shrink window if distinct characters exceed k  while (distinctCnt > k) {  freq[s.charAt(i) - 'a']--;  if (freq[s.charAt(i) - 'a'] == 0) distinctCnt--;  i++;  }  // add number of valid substrings ending at j  ans += j - i + 1;  }  return ans;  }  // function to find the number of substrings  // with exactly k Distinct characters.  static int countSubstr(String s int k) {  int n = s.length();  int ans = 0;  // Subtract substrings with at most   // k-1 distinct characters from substrings  // with at most k distinct characters  ans = count(s k) - count(s k - 1);  return ans;  }  public static void main(String[] args) {  String s = 'abc';  int k = 2;  System.out.println(countSubstr(s k));  } }
Python 
# function which finds the number of  # substrings with atmost k Distinct # characters. def count(s k): n = len(s) ans = 0 # ese sliding window technique freq = [0] * 26 distinctCnt = 0 i = 0 for j in range(n): # expand window and add character freq[ord(s[j]) - ord('a')] += 1 if freq[ord(s[j]) - ord('a')] == 1: distinctCnt += 1 # shrink window if distinct characters exceed k while distinctCnt > k: freq[ord(s[i]) - ord('a')] -= 1 if freq[ord(s[i]) - ord('a')] == 0: distinctCnt -= 1 i += 1 # add number of valid substrings ending at j ans += j - i + 1 return ans # function to find the number of substrings # with exactly k Distinct characters. def countSubstr(s k): n = len(s) ans = 0 # subtract substrings with at most  # k-1 distinct characters from substrings # with at most k distinct characters ans = count(s k) - count(s k - 1) return ans if __name__ == '__main__': s = 'abc' k = 2 print(countSubstr(s k))
C# 
using System; class GfG {  // function which finds the number of   // substrings with atmost k Distinct  // characters.  static int count(string s int k) {  int n = s.Length;  int ans = 0;  // use sliding window technique  int[] freq = new int[26];  int distinctCnt = 0;  int i = 0;  for (int j = 0; j < n; j++) {  // expand window and add character  freq[s[j] - 'a']++;  if (freq[s[j] - 'a'] == 1) distinctCnt++;  // shrink window if distinct characters exceed k  while (distinctCnt > k) {  freq[s[i] - 'a']--;  if (freq[s[i] - 'a'] == 0) distinctCnt--;  i++;  }  // add number of valid substrings ending at j  ans += j - i + 1;  }  return ans;  }  // function to find the number of substrings  // with exactly k Distinct characters.  static int countSubstr(string s int k) {  int n = s.Length;  int ans = 0;  // subtract substrings with at most   // k-1 distinct characters from substrings  // with at most k distinct characters  ans = count(s k) - count(s k - 1);  return ans;  }  static void Main() {  string s = 'abc';  int k = 2;  Console.WriteLine(countSubstr(s k));  } }
JavaScript 
// function which finds the number of  // substrings with atmost k Distinct // characters. function count(s k) {  let n = s.length;  let ans = 0;  // use sliding window technique  let freq = new Array(26).fill(0);  let distinctCnt = 0;  let i = 0;  for (let j = 0; j < n; j++) {  // expand window and add character  freq[s.charCodeAt(j) - 'a'.charCodeAt(0)]++;  if (freq[s.charCodeAt(j) - 'a'.charCodeAt(0)] === 1)  distinctCnt++;  // shrink window if distinct characters exceed k  while (distinctCnt > k) {  freq[s.charCodeAt(i) - 'a'.charCodeAt(0)]--;  if (freq[s.charCodeAt(i) - 'a'.charCodeAt(0)] === 0)  distinctCnt--;  i++;  }  // add number of valid substrings ending at j  ans += j - i + 1;  }  return ans; } // sunction to find the number of substrings // with exactly k Distinct characters. function countSubstr(s k) {  let n = s.length;  let ans = 0;  // subtract substrings with at most   // k-1 distinct characters from substrings  // with at most k distinct characters  ans = count(s k) - count(s k - 1);  return ans; } // Driver Code let s = 'abc'; let k = 2; console.log(countSubstr(s k));

Saída 
2