Dada la string str de tamaño N que contiene letras mayúsculas y minúsculas, y dos enteros K y X . La tarea es encontrar el recuento de substrings de tamaño K que contengan como máximo X vocales distintas.
Ejemplos:
Entrada: str = «TrueGoik», K = 3, X = 2
Salida: 6
Explicación: Las cinco substrings son «Tru», «rue», «ueG», «eGo», «Goi» y «oik».Entrada: str = “GeeksForGeeks”, K = 2, X = 2
Salida: 12
Explicación: “Ge”, “ee”, “ek”, “ks”, “sf”, “fo”, “or”, “rG ”, “Ge”, “ee”, “ek” y “ks”.
Enfoque: Para resolver el problema, primero hay que generar todas las substrings de longitud K. Luego, en cada substring, verifique si el número de vocales distintas es menor que X o no. Siga los pasos que se mencionan a continuación.
- Primero genere todas las substrings de longitud K a partir de cada índice i en [0, N – K].
- Luego, para cada substring de longitud K , haga lo siguiente:
- Mantenga un hash para almacenar las apariciones de vocales únicas.
- Compruebe si un nuevo carácter en la substring es una vocal o no.
- Si es una vocal, incremente su ocurrencia en el hash y mantenga un conteo de vocales distintas encontradas
- Ahora, para cada substring, si el conteo distinto de vocales es menor o igual a X , incremente el conteo final.
- Cuando se hayan considerado todas las substrings, imprima el recuento final.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ code to implement above approach
#include <bits/stdc++.h>
using namespace std;
#define MAX 128
// Function to check whether
// a character is vowel or not
bool isVowel(char x)
{
return (x == 'a' || x == 'e' || x == 'i'
|| x == 'o' || x == 'u' ||
x == 'A' || x == 'E' || x == 'I'
|| x == 'O' || x == 'U');
}
int getIndex(char ch)
{
return (ch - 'A' > 26 ? ch - 'a' :
ch - 'A');
}
// Function to find the count of K length
// substring with at most x distinct vowels
int get(string str, int k, int x)
{
int n = str.length();
// Initialize result
int res = 0;
// Consider all substrings
// beginning with str[i]
for (int i = 0; i <= n - k; i++) {
int dist_count = 0;
// To store count of characters
// from 'a' to 'z'
vector<int> cnt(26, 0);
// Consider all substrings
// between str[i..j]
for (int j = i; j < i + k; j++) {
// If this is a new vowels
// for this substring,
// increment dist_count.
if (isVowel(str[j])
&& cnt[getIndex(str[j])]
== 0) {
dist_count++;
}
// Increment count of
// current character
cnt[getIndex(str[j])]++;
}
// If count of distinct vowels
// in current substring
// of length k is less than
// equal to x, then increment result.
if (dist_count <= x)
res++;
}
return res;
}
// Driver code
int main(void)
{
string s = "TrueGoik";
int K = 3, X = 2;
cout << get(s, K, X);
return 0;
}
Java
// Java code to implement above approach
import java.util.Arrays;
class GFG {
static int MAX = 128;
// Function to check whether
// a character is vowel or not
static boolean isVowel(char x) {
return (x == 'a' || x == 'e' || x == 'i'
|| x == 'o' || x == 'u' ||
x == 'A' || x == 'E' || x == 'I'
|| x == 'O' || x == 'U');
}
static int getIndex(char ch) {
return (ch - 'A' > 26 ? ch - 'a' : ch - 'A');
}
// Function to find the count of K length
// substring with at most x distinct vowels
static int get(String str, int k, int x) {
int n = str.length();
// Initialize result
int res = 0;
// Consider all substrings
// beginning with str[i]
for (int i = 0; i <= n - k; i++) {
int dist_count = 0;
// To store count of characters
// from 'a' to 'z'
int[] cnt = new int[26];
Arrays.fill(cnt, 0);
// Consider all substrings
// between str[i..j]
for (int j = i; j < i + k; j++) {
// If this is a new vowels
// for this substring,
// increment dist_count.
if (isVowel(str.charAt(j))
&& cnt[getIndex(str.charAt(j))] == 0) {
dist_count++;
}
// Increment count of
// current character
cnt[getIndex(str.charAt(j))]++;
}
// If count of distinct vowels
// in current substring
// of length k is less than
// equal to x, then increment result.
if (dist_count <= x)
res++;
}
return res;
}
// Driver code
public static void main(String args[]) {
String s = "TrueGoik";
int K = 3, X = 2;
System.out.println(get(s, K, X));
}
}
// This code is contributed by saurabh_jaiswal.
Python3
# Python code for the above approach
# Function to check whether
# a character is vowel or not
def isVowel(x):
return (x == 'a' or x == 'e' or x == 'i' or x == 'o'
or x == 'u' or x == 'A' or x == 'E' or x == 'I'
or x == 'O' or x == 'U')
def getIndex(ch):
return (ord(ch) - ord('a')) if (ord(ch) - ord('A')) > 26 else (ord(ch) - ord('A'))
# Function to find the count of K length
# substring with at most x distinct vowels
def get(str, k, x):
n = len(str)
# Initialize result
res = 0
# Consider all substrings
# beginning with str[i]
for i in range(n - k + 1):
dist_count = 0
# To store count of characters
# from 'a' to 'z'
cnt = [0] * 26
# Consider all substrings
# between str[i..j]
for j in range(i, i + k):
# If this is a new vowels
# for this substring,
# increment dist_count.
if (isVowel(str[j]) and cnt[getIndex(str[j])] == 0):
dist_count += 1
# Increment count of
# current character
cnt[getIndex(str[j])] += 1
# If count of distinct vowels
# in current substring
# of length k is less than
# equal to x, then increment result.
if (dist_count <= x):
res += 1
return res
# Driver code
s = "TrueGoik"
K = 3
X = 2
print(get(s, K, X))
# This code is contributed by Saurabh Jaiswal
C#
// C# code to implement above approach
using System;
class GFG {
// Function to check whether
// a character is vowel or not
static bool isVowel(char x)
{
return (x == 'a' || x == 'e' || x == 'i' || x == 'o'
|| x == 'u' || x == 'A' || x == 'E'
|| x == 'I' || x == 'O' || x == 'U');
}
static int getIndex(char ch)
{
return (ch - 'A' > 26 ? ch - 'a' : ch - 'A');
}
// Function to find the count of K length
// substring with at most x distinct vowels
static int get(string str, int k, int x)
{
int n = str.Length;
// Initialize result
int res = 0;
// Consider all substrings
// beginning with str[i]
for (int i = 0; i <= n - k; i++) {
int dist_count = 0;
// To store count of characters
// from 'a' to 'z'
int[] cnt = new int[26];
// Arrays.fill(cnt, 0);
// Consider all substrings
// between str[i..j]
for (int j = i; j < i + k; j++) {
// If this is a new vowels
// for this substring,
// increment dist_count.
if (isVowel(str[j])
&& cnt[getIndex(str[j])] == 0) {
dist_count++;
}
// Increment count of
// current character
cnt[getIndex(str[j])]++;
}
// If count of distinct vowels
// in current substring
// of length k is less than
// equal to x, then increment result.
if (dist_count <= x)
res++;
}
return res;
}
// Driver code
public static void Main()
{
string s = "TrueGoik";
int K = 3, X = 2;
Console.WriteLine(get(s, K, X));
}
}
// This code is contributed by ukasp.
Javascript
<script>
// JavaScript code for the above approach
// Function to check whether
// a character is vowel or not
function isVowel(x) {
return (x == 'a' || x == 'e' || x == 'i' || x == 'o'
|| x == 'u' || x == 'A' || x == 'E' || x == 'I'
|| x == 'O' || x == 'U');
}
function getIndex(ch) {
return ((ch.charCodeAt(0) - 'A'.charCodeAt(0)) > 26 ? (ch.charCodeAt(0) - 'a'.charCodeAt(0)) :
(ch.charCodeAt(0) - 'A'.charCodeAt(0)));
}
// Function to find the count of K length
// substring with at most x distinct vowels
function get(str, k, x) {
let n = str.length;
// Initialize result
let res = 0;
// Consider all substrings
// beginning with str[i]
for (let i = 0; i <= n - k; i++) {
let dist_count = 0;
// To store count of characters
// from 'a' to 'z'
let cnt = new Array(26).fill(0);
// Consider all substrings
// between str[i..j]
for (let j = i; j < i + k; j++) {
// If this is a new vowels
// for this substring,
// increment dist_count.
if (isVowel(str[j])
&& cnt[getIndex(str[j])]
== 0) {
dist_count++;
}
// Increment count of
// current character
cnt[getIndex(str[j])]++;
}
// If count of distinct vowels
// in current substring
// of length k is less than
// equal to x, then increment result.
if (dist_count <= x)
res++;
}
return res;
}
// Driver code
let s = "TrueGoik";
let K = 3, X = 2;
document.write(get(s, K, X));
// This code is contributed by Potta Lokesh
</script>
Producción
6
Complejidad temporal: O(N * K)
Espacio auxiliar: O(N * K)