Dada la string str de tamaño N que contiene letras mayúsculas y minúsculas y un número entero K . La tarea es encontrar el recuento de substrings que contengan exactamente K vocales distintas.
Ejemplos:
Entrada: str = “aeiou”, K = 2
Salida: 4
Explicación: Las substrings que tienen dos vocales distintas son “ae”, “ei”, “io” y “ou”.Entrada: str = “TrueGoik”, K = 3
Salida: 5
Explicación: Las substrings son “TrueGo”, “rueGo”, “ueGo”, “eGoi” y “eGoik”.
Enfoque: el problema se puede resolver generando todas las substrings. De las substrings generadas, cuente las que tienen K vocales distintas. Siga los pasos que se mencionan a continuación para implementar el enfoque:
- Primero genere todas las substrings a partir de cada índice i en el rango [0, N]
- Luego, para cada substring, siga los pasos:
- Mantenga una array hash para almacenar las ocurrencias 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 recuento distinto de vocales es K , incremente el recuento final .
- Si para cualquier ciclo para encontrar substrings que comiencen desde i , el recuento de vocales distintas excede K , o si la longitud de la substring ha alcanzado la longitud de la string, rompa el ciclo y busque substrings que comiencen desde i+1 .
- Cuando se hayan considerado todas las substrings, imprima el recuento final.
A continuación se muestra la implementación del enfoque anterior.
C++
// C++ program to count number of substrings
// with exactly k distinct vowels
#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 count number of substrings
// with exactly k unique vowels
int countkDist(string str, int k)
{
int n = str.length();
// Initialize result
int res = 0;
// Consider all substrings
// beginning with str[i]
for (int i = 0; i < n; 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 < n; 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 distinct vowels count
// becomes k then increment result
if (dist_count == k)
res++;
if (dist_count > k)
break;
}
}
return res;
}
// Driver code
int main()
{
string str = "TrueGoik";
int K = 3;
cout << countkDist(str, K) << endl;
return 0;
}
Java
// Java program to count number of substrings
// with exactly k distinct vowels
import java.util.*;
public class GFG
{
// 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 count number of substrings
// with exactly k unique vowels
static int countkDist(String str, int k)
{
int n = str.length();
// Initialize result
int res = 0;
// Consider all substrings
// beginning with str[i]
for (int i = 0; i < n; i++) {
int dist_count = 0;
// To store count of characters
// from 'a' to 'z'
int cnt[] = new int[26];
for(int t = 0; t < 26; t++) {
cnt[t] = 0;
}
// Consider all substrings
// between str[i..j]
for (int j = i; j < n; 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 distinct vowels count
// becomes k then increment result
if (dist_count == k)
res++;
if (dist_count > k)
break;
}
}
return res;
}
// Driver code
public static void main(String args[])
{
String str = "TrueGoik";
int K = 3;
System.out.println(countkDist(str, K));
}
}
// This code is contributed by Samim Hossain Mondal.
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 count number of substrings
# with exactly k unique vowels
def countkDist(str, k):
n = len(str)
# Initialize result
res = 0
# Consider all substrings
# beginning with str[i]
for i in range(n):
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, n):
# 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 distinct vowels count
# becomes k then increment result
if (dist_count == k):
res += 1
if (dist_count > k):
break
return res
# Driver code
s = "TrueGoik"
K = 3
print(countkDist(s, K))
# This code is contributed by Saurabh Jaiswal
C#
// C# program to count number of substrings
// with exactly k distinct vowels
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 count number of substrings
// with exactly k unique vowels
static int countkDist(string str, int k)
{
int n = str.Length;
// Initialize result
int res = 0;
// Consider all substrings
// beginning with str[i]
for (int i = 0; i < n; i++) {
int dist_count = 0;
// To store count of characters
// from 'a' to 'z'
int []cnt = new int[26];
for(int t = 0; t < 26; t++) {
cnt[t] = 0;
}
// Consider all substrings
// between str[i..j]
for (int j = i; j < n; 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 distinct vowels count
// becomes k then increment result
if (dist_count == k)
res++;
if (dist_count > k)
break;
}
}
return res;
}
// Driver code
public static void Main()
{
string str = "TrueGoik";
int K = 3;
Console.Write(countkDist(str, K));
}
}
// This code is contributed by Samim Hossain Mondal.
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 count number of substrings
// with exactly k unique vowels
function countkDist(str, k)
{
let n = str.length;
// Initialize result
let res = 0;
// Consider all substrings
// beginning with str[i]
for (let i = 0; i < n; 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 < n; 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 distinct vowels count
// becomes k then increment result
if (dist_count == k)
res++;
if (dist_count > k)
break;
}
}
return res;
}
// Driver code
let s = "TrueGoik";
let K = 3
document.write(countkDist(s, K));
// This code is contributed by Potta Lokesh
</script>
Producción
5
Tiempo Complejidad: O(N 2 )
Espacio Auxiliar: O(N 2 )