Dada una string str de alfabetos ingleses en minúsculas y un entero m . La tarea es contar cuántas posiciones hay en la string de modo que si divide la string en dos substrings no vacías, hay al menos m caracteres con la misma frecuencia en ambas substrings.
Los caracteres deben estar presentes en la string str .
Ejemplos:
Entrada: str = “aabbccaa”, m = 2
Salida: 2
La string tiene una longitud de 8, por lo que hay 7 posiciones disponibles para realizar la partición.
es decir, a|a|b|b|c|c|a|a
Sólo son posibles dos particiones que satisfagan las restricciones dadas.
aab|bccaa: en la mitad izquierda del separador, ‘a’ tiene la frecuencia 2 y ‘b’ tiene la frecuencia 1,
que es la misma que la de la mitad derecha.
aabbc|caa: en la mitad izquierda del separador, ‘a’ tiene una frecuencia de 2 y ‘c’ tiene una frecuencia de 1,
que es la misma que la de la mitad derecha.
Entrada: str = “aabbaa”, m = 2
Salida: 1
Enfoque: para cada posición de partición, calcule las frecuencias de cada uno de los caracteres de la string en ambas particiones. Luego calcule la cantidad de caracteres que tienen la misma frecuencia en ambas particiones. Si el recuento de dichos caracteres es al menos m, agregue 1 al recuento requerido de particiones.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the number of ways
// to partition the given so that the
// given condition is satisfied
int countWays(string str, int m)
{
// Hashset to store unique characters
// in the given string
set<char> s;
for (int i = 0; i < str.length(); i++)
s.insert(str[i]);
// To store the number of ways
// to partition the string
int result = 0;
for (int i = 1; i < str.length(); i++)
{
// Hashmaps to store frequency of characters
// of both the partitions
map<char, int> first_map, second_map;
// Iterate in the first partition
for (int j = 0; j < i; j++)
// If character already exists in the hashmap
// then increase it's frequency
first_map[str[j]]++;
// Iterate in the second partition
for (int k = 0; k < str.length(); k++)
// If character already exists in the hashmap
// then increase it's frequency
second_map[str[k]]++;
// Iterator for HashSet
set<char>::iterator itr = s.begin();
// To store the count of characters that have
// equal frequencies in both the partitions
int total_count = 0;
while (++itr != s.end())
{
// first_count and second_count keeps track
// of the frequencies of each character
int first_count = 0, second_count = 0;
char ch = *(itr);
// Frequency of the character
// in the first partition
if (first_map.find(ch) != first_map.end())
first_count = first_map[ch];
// Frequency of the character
// in the second partition
if (second_map.find(ch) != second_map.end())
second_count = second_map[ch];
// Check if frequency is same
// in both the partitions
if (first_count == second_count &&
first_count != 0)
total_count += 1;
}
// Check if the condition is satisfied
// for the current partition
if (total_count >= m)
result += 1;
}
return result;
}
// Driver code
int main(int argc, char const *argv[])
{
string str = "aabbccaa";
int m = 2;
cout << countWays(str, m) << endl;
return 0;
}
// This code is contributed by
// sanjeev2552
Java
// Java implementation of the approach
import java.util.*;
class GFG {
// Function to return the number of ways
// to partition the given so that the
// given condition is satisfied
static int countWays(String str, int m)
{
// Hashset to store unique characters
// in the given string
HashSet<Character> set = new HashSet<Character>();
for (int i = 0; i < str.length(); i++)
set.add(str.charAt(i));
// To store the number of ways
// to partition the string
int result = 0;
for (int i = 1; i < str.length(); i++) {
// Hashmaps to store frequency of characters
// of both the partitions
HashMap<Character, Integer> first_map
= new HashMap<Character, Integer>();
HashMap<Character, Integer> second_map
= new HashMap<Character, Integer>();
// Iterate in the first partition
for (int j = 0; j < i; j++) {
// If character already exists in the hashmap
// then increase it's frequency
if (first_map.containsKey(str.charAt(j)))
first_map.put(str.charAt(j),
(first_map.get(str.charAt(j)) + 1));
// Else create an entry for it in the Hashmap
else
first_map.put(str.charAt(j), 1);
}
// Iterate in the second partition
for (int k = i; k < str.length(); k++) {
// If character already exists in the hashmap
// then increase it's frequency
if (second_map.containsKey(str.charAt(k)))
second_map.put(str.charAt(k),
(second_map.get(str.charAt(k)) + 1));
// Else create an entry for it in the Hashmap
else
second_map.put(str.charAt(k), 1);
}
// Iterator for HashSet
Iterator itr = set.iterator();
// To store the count of characters that have
// equal frequencies in both the partitions
int total_count = 0;
while (itr.hasNext()) {
// first_count and second_count keeps track
// of the frequencies of each character
int first_count = 0, second_count = 0;
char ch = (char)itr.next();
// Frequency of the character
// in the first partition
if (first_map.containsKey(ch))
first_count = first_map.get(ch);
// Frequency of the character
// in the second partition
if (second_map.containsKey(ch))
second_count = second_map.get(ch);
// Check if frequency is same in both the partitions
if (first_count == second_count && first_count != 0)
total_count += 1;
}
// Check if the condition is satisfied
// for the current partition
if (total_count >= m)
result += 1;
}
return result;
}
// Driver code
public static void main(String[] args)
{
String str = "aabbccaa";
int m = 2;
System.out.println(countWays(str, m));
}
}
Python3
# Python3 implementation of the approach from collections import defaultdict # Function to return the number of ways # to partition the given so that the # given condition is satisfied def countWays(string, m): # Hashset to store unique # characters in the given string Set = set() for i in range(0, len(string)): Set.add(string[i]) # To store the number of ways # to partition the string result = 0 for i in range(1, len(string)): # Hashmaps to store frequency of # characters of both the partitions first_map = defaultdict(lambda:0) second_map = defaultdict(lambda:0) # Iterate in the first partition for j in range(0, i): first_map[string[j]] += 1 # Iterate in the second partition for k in range(i, len(string)): second_map[string[k]] += 1 # To store the count of characters that have # equal frequencies in both the partitions total_count = 0 for ch in Set: # first_count and second_count keeps track # of the frequencies of each character first_count, second_count = 0, 0 # Frequency of the character # in the first partition if ch in first_map: first_count = first_map[ch] # Frequency of the character # in the second partition if ch in second_map: second_count = second_map[ch] # Check if frequency is same in both the partitions if first_count == second_count and first_count != 0: total_count += 1 # Check if the condition is satisfied # for the current partition if total_count >= m: result += 1 return result # Driver code if __name__ == "__main__": string = "aabbccaa" m = 2 print(countWays(string, m)) # This code is contributed by Rituraj Jain
C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
public class GFG {
// Function to return the number of ways
// to partition the given so that the
// given condition is satisfied
static int countWays(String str, int m)
{
// Hashset to store unique characters
// in the given string
HashSet<char> set = new HashSet<char>();
for (int i = 0; i < str.Length; i++)
set.Add(str[i]);
// To store the number of ways
// to partition the string
int result = 0;
for (int i = 1; i < str.Length; i++) {
// Hashmaps to store frequency of characters
// of both the partitions
Dictionary<char, int> first_map
= new Dictionary<char, int>();
Dictionary<char, int> second_map
= new Dictionary<char, int>();
// Iterate in the first partition
for (int j = 0; j < i; j++) {
// If character already exists in the hashmap
// then increase it's frequency
if (first_map.ContainsKey(str[j]))
first_map[str[j]] =
(first_map[str[j]] + 1);
// Else create an entry for it in the Hashmap
else
first_map.Add(str[j], 1);
}
// Iterate in the second partition
for (int k = i; k < str.Length; k++) {
// If character already exists in the hashmap
// then increase it's frequency
if (second_map.ContainsKey(str[k]))
second_map[str[k]] =
(second_map[str[k]] + 1);
// Else create an entry for it in the Hashmap
else
second_map.Add(str[k], 1);
}
// To store the count of characters that have
// equal frequencies in both the partitions
int total_count = 0;
// Iterator for HashSet
foreach (int itr in set) {
// first_count and second_count keeps track
// of the frequencies of each character
int first_count = 0, second_count = 0;
char ch = (char)itr;
// Frequency of the character
// in the first partition
if (first_map.ContainsKey(ch))
first_count = first_map[ch];
// Frequency of the character
// in the second partition
if (second_map.ContainsKey(ch))
second_count = second_map[ch];
// Check if frequency is same in both the partitions
if (first_count == second_count && first_count != 0)
total_count += 1;
}
// Check if the condition is satisfied
// for the current partition
if (total_count >= m)
result += 1;
}
return result;
}
// Driver code
public static void Main(String[] args)
{
String str = "aabbccaa";
int m = 2;
Console.WriteLine(countWays(str, m));
}
}
// This code is contributed by Rajput-Ji
Javascript
<script>
// Javascript implementation of the approach
// Function to return the number of ways
// to partition the given so that the
// given condition is satisfied
function countWays(str, m)
{
// Hashset to store unique characters
// in the given string
var s = new Set();
for (var i = 0; i < str.length; i++)
s.add(str[i]);
// To store the number of ways
// to partition the string
var result = 0;
for (var i = 1; i < str.length; i++)
{
// Hashmaps to store frequency of characters
// of both the partitions
var first_map = new Map(), second_map = new Map();
// Iterate in the first partition
for (var j = 0; j < i; j++)
// If character already exists in the hashmap
// then increase it's frequency
if(first_map.has(str[j]))
first_map.set(str[j], first_map.get(str[j])+1)
else
first_map.set(str[j], 1)
// Iterate in the second partition
for (var k = 0; k < str.length; k++)
// If character already exists in the hashmap
// then increase it's frequency
if(second_map.has(str[k]))
second_map.set(str[k], second_map.get(str[k])+1)
else
second_map.set(str[k], 1)
// To store the count of characters that have
// equal frequencies in both the partitions
var total_count = 0;
s.forEach(itr => {
// first_count and second_count keeps track
// of the frequencies of each character
var first_count = 0, second_count = 0;
var ch = itr;
// Frequency of the character
// in the first partition
if (first_map.has(ch))
first_count = first_map.get(ch);
// Frequency of the character
// in the second partition
if (second_map.has(ch))
second_count = second_map.get(ch);
// Check if frequency is same
// in both the partitions
if (first_count == second_count &&
first_count != 0)
total_count += 1;
});
// Check if the condition is satisfied
// for the current partition
if (total_count >= m)
result += 1;
}
return result;
}
// Driver code
var str = "aabbccaa";
var m = 2;
document.write( countWays(str, m));
// This code is contributed by itsok.
</script>
Producción:
2
Complejidad de tiempo: O(n*n*log(n)), ya que los bucles anidados se utilizan para la iteración
Espacio auxiliar: O(n), ya que el espacio adicional de tamaño n se utiliza para crear un conjunto y un mapa
Publicación traducida automáticamente
Artículo escrito por Divyank_Sheth y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA