Número de substrings con recuento de cada carácter como k

Dada una string y un entero k, encuentre el número de substrings en las que todos los caracteres diferentes aparecen exactamente k veces. 

Ejemplos: 

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

La idea es atravesar todas las substrings. Fijamos un punto de inicio, recorremos todas las substrings comenzando con el punto seleccionado, seguimos incrementando las frecuencias de todos los caracteres. Si todas las frecuencias se vuelven k, incrementamos el resultado. Si el conteo de cualquier frecuencia se vuelve más que k, rompemos y cambiamos el punto de partida. 

Implementación:

// C++ program to count number of substrings
// with counts of distinct characters as k.
#include <bits/stdc++.h>
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 program to count number of substrings
// with counts of distinct characters as k.
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 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# 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_CHAR,NULL);
 
        // 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.
?>

<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(freq,k)
{
    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) + "<br>");
  
s = "aabbc";
k = 2;
document.write(substrings(s, k) + "<br>");
 
// This code is contributed by avanitrachhadiya2155
     
</script>

6
3

Complejidad de tiempo: O(n*n) donde n es la longitud de la string de entrada. La función Check() ejecuta un bucle de longitud constante de 0 a MAX_CHAR (es decir, 26 siempre), por lo que esta función check() se ejecuta en tiempo O(MAX_CHAR), por lo que la complejidad del tiempo es O(MAX_CHAR*n*n)=O( n^2).
Espacio Auxiliar: O(1) 

Enfoque óptimo:

En una observación muy cuidadosa, podemos ver que es suficiente verificar lo mismo para las substrings de longitud   donde   está el número de caracteres distintos presentes en la string dada.

Argumento:

Considere una substring S_{i+1}S_{i+2}\dots S_{i+p} de longitud ‘p’. Si esta substring tiene ‘m’ caracteres distintos y cada carácter distinto aparece exactamente ‘K’ veces, entonces la longitud de la substring, ‘p’, viene dada por p = K\veces m. Dado que ‘ ‘ es siempre un múltiplo de ‘K’ y   para la string dada, es suficiente iterar sobre las substrings cuya longitud es divisible por ‘K’ y que tienen m, 1 \le m \le 26 caracteres distintos. Usaremos la ventana deslizante para iterar sobre las substrings de longitud fija.

Solución:

• Encuentre el número de caracteres distintos presentes en la string dada. Que sea D.
• Para cada i, 1\le i\le D, haz lo siguiente
  • Iterar sobre las substrings de longitud $i \times K$, usando una ventana deslizante.
  • Compruebe si cumplen la condición: todos los caracteres distintos de la substring aparecen exactamente K veces.
  • Si cumplen la condición, incremente el conteo.

Implementación:

#include <iostream>
#include <map>
#include <set>
#include <string>
 
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;
}

#include <stdbool.h>
#include <stdio.h>
#include <string.h>
 
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("%d\n", count_substrings(s, 6, k));
    s = "aabbc";
    k = 2;
    printf("%d\n", count_substrings(s, 5, k));
    return 0;
}

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));
    }
}

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))

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

<script>
 
function have_same_frequency(freq,k)
{
    for (let i = 0; i < 26; i++) {
            if (freq[i] != 0 && freq[i] != k) {
                return false;
            }
        }
        return true;
}
 
function count_substrings(s,k)
{
    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)+"<br>");
s = "aabbc";
k = 2;
document.write(count_substrings(s, k)+"<br>");
 
 
// This code is contributed by rag2127
 
</script>

6
3

Complejidad de tiempo: O(N * D) donde D es el número de caracteres distintos presentes en la string y N es la longitud de la string.
Espacio Auxiliar: O(N) 

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