Dada una string str y un entero K , la tarea es encontrar la longitud de la substring más larga S’ tal que cada carácter en S’ aparezca al menos K veces.
Entrada: s = “xyxyyz”, k = 2
Salida: 5
“xyxyy” es la substring más larga donde
cada carácter aparece al menos dos veces.
Entrada: s = «geeksforgeeks», k = 2
Salida: 2
Enfoque: considere todas las substrings posibles y, para cada substring, calcule la frecuencia de cada uno de sus caracteres y verifique si todos los caracteres aparecen al menos K veces. Para todas esas substrings válidas, encuentre la mayor longitud posible.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define MAX 26
// Function that return true if frequency
// of all the present characters is at least k
bool atLeastK(int freq[], int k)
{
for (int i = 0; i < MAX; i++) {
// If the character is present and
// its frequency is less than k
if (freq[i] != 0 && freq[i] < k)
return false;
}
return true;
}
// Function to set every frequency to zero
void setZero(int freq[])
{
for (int i = 0; i < MAX; i++)
freq[i] = 0;
}
// Function to return the length of the longest
// sub-string such that it contains every
// character at least k times
int findlength(string str, int n, int k)
{
// To store the required maximum length
int maxLen = 0;
int freq[MAX];
// Starting index of the sub-string
for (int i = 0; i < n; i++) {
setZero(freq);
// Ending index of the sub-string
for (int j = i; j < n; j++) {
freq[str[j] - 'a']++;
// If the frequency of every character
// of the current sub-string is at least k
if (atLeastK(freq, k)) {
// Update the maximum possible length
maxLen = max(maxLen, j - i + 1);
}
}
}
return maxLen;
}
// Driver code
int main()
{
string str = "xyxyyz";
int n = str.length();
int k = 2;
cout << findlength(str, n, k);
return 0;
}
Java
// Java Implementation of the above approach
class GFG
{
static final int MAX = 26;
// Function that return true if frequency
// of all the present characters is at least k
static boolean atLeastK(int freq[], int k)
{
for (int i = 0; i < MAX; i++)
{
// If the character is present and
// its frequency is less than k
if (freq[i] != 0 && freq[i] < k)
return false;
}
return true;
}
// Function to set every frequency to zero
static void setZero(int freq[])
{
for (int i = 0; i < MAX; i++)
freq[i] = 0;
}
// Function to return the length of the longest
// sub-string such that it contains every
// character at least k times
static int findlength(String str, int n, int k)
{
// To store the required maximum length
int maxLen = 0;
int freq[] = new int[MAX];
// Starting index of the sub-string
for (int i = 0; i < n; i++)
{
setZero(freq);
// Ending index of the sub-string
for (int j = i; j < n; j++)
{
freq[str.charAt(j) - 'a']++;
// If the frequency of every character
// of the current sub-string is at least k
if (atLeastK(freq, k))
{
// Update the maximum possible length
maxLen = Math.max(maxLen, j - i + 1);
}
}
}
return maxLen;
}
// Driver code
public static void main(String args[])
{
String str = "xyxyyz";
int n = str.length();
int k = 2;
System.out.println(findlength(str, n, k));
}
}
// This code has been contributed by 29AjayKumar
Python3
# Python3 implementation of the approach
MAX = 26
# Function that return true if frequency
# of all the present characters is at least k
def atLeastK(freq, k) :
for i in range(MAX) :
# If the character is present and
# its frequency is less than k
if (freq[i] != 0 and freq[i] < k) :
return False;
return True;
# Function to set every frequency to zero
def setZero(freq) :
for i in range(MAX) :
freq[i] = 0;
# Function to return the length of the longest
# sub-string such that it contains every
# character at least k times
def findlength(string, n, k) :
# To store the required maximum length
maxLen = 0;
freq = [0]*MAX;
# Starting index of the sub-string
for i in range(n) :
setZero(freq);
# Ending index of the sub-string
for j in range(i,n) :
freq[ord(string[j]) - ord('a')] += 1;
# If the frequency of every character
# of the current sub-string is at least k
if (atLeastK(freq, k)) :
# Update the maximum possible length
maxLen = max(maxLen, j - i + 1);
return maxLen;
# Driver code
if __name__ == "__main__" :
string = "xyxyyz";
n = len(string);
k = 2;
print(findlength(string, n, k));
# This code is contributed by AnkitRai01
C#
// C# Implementation of the above approach
using System;
class GFG
{
static int MAX = 26;
// Function that return true if frequency
// of all the present characters is at least k
static Boolean atLeastK(int []freq, int k)
{
for (int i = 0; i < MAX; i++)
{
// If the character is present and
// its frequency is less than k
if (freq[i] != 0 && freq[i] < k)
return false;
}
return true;
}
// Function to set every frequency to zero
static void setZero(int []freq)
{
for (int i = 0; i < MAX; i++)
freq[i] = 0;
}
// Function to return the length of the longest
// sub-string such that it contains every
// character at least k times
static int findlength(String str, int n, int k)
{
// To store the required maximum length
int maxLen = 0;
int []freq = new int[MAX];
// Starting index of the sub-string
for (int i = 0; i < n; i++)
{
setZero(freq);
// Ending index of the sub-string
for (int j = i; j < n; j++)
{
freq[str[j] - 'a']++;
// If the frequency of every character
// of the current sub-string is at least k
if (atLeastK(freq, k))
{
// Update the maximum possible length
maxLen = Math.Max(maxLen, j - i + 1);
}
}
}
return maxLen;
}
// Driver code
public static void Main(String []args)
{
String str = "xyxyyz";
int n = str.Length;
int k = 2;
Console.WriteLine(findlength(str, n, k));
}
}
// This code is contributed by Princi Singh
Javascript
<script>
// Javascript implementation of the approach
var MAX = 26;
// Function that return true if frequency
// of all the present characters is at least k
function atLeastK(freq, k)
{
for (var i = 0; i < MAX; i++) {
// If the character is present and
// its frequency is less than k
if (freq[i] != 0 && freq[i] < k)
return false;
}
return true;
}
// Function to set every frequency to zero
function setZero(freq)
{
for (var i = 0; i < MAX; i++)
freq[i] = 0;
}
// Function to return the length of the longest
// sub-string such that it contains every
// character at least k times
function findlength(str, n, k)
{
// To store the required maximum length
var maxLen = 0;
var freq = Array(MAX).fill(0);
// Starting index of the sub-string
for (var i = 0; i < n; i++) {
setZero(freq);
// Ending index of the sub-string
for (var j = i; j < n; j++) {
freq[str[j].charCodeAt(0) - 'a'.charCodeAt(0)]++;
// If the frequency of every character
// of the current sub-string is at least k
if (atLeastK(freq, k)) {
// Update the maximum possible length
maxLen = Math.max(maxLen, j - i + 1);
}
}
}
return maxLen;
}
// Driver code
var str = "xyxyyz";
var n = str.length;
var k = 2;
document.write( findlength(str, n, k));
</script>
Producción:
5
Complejidad de tiempo: O(n 2 )
Espacio auxiliar: O(MAX), donde MAX es el número máximo de caracteres únicos en la string.
Publicación traducida automáticamente
Artículo escrito por everythingispossible y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA