Dada una array de strings arr[] de tamaño N . Cada una de las strings contiene letras minúsculas o números en inglés solamente. La tarea es encontrar la frecuencia de strings alfabéticas y alfanuméricas en la array.
Nota: si alguna string tiene al menos un número, la string es una string alfanumérica.
Ejemplos:
Entrada: arr[] = {“abcd”, “mak87s”, “abcd”, “kakjdj”, “laojs7s6”}
Salida: 3 2
“abcd”: 2
“kakjdj”: 1
“mak87s”: 1
“laojs7s6”: 1
Explicación: De la entrada dada, las strings que solo tienen letras son
«abcd», «kakjdj» y las strings restantes contienen números. Así que 3 strings alfabéticas y dos alfanuméricas en total.
Estas dos strings tienen una frecuencia de 2 y 1 y las otras dos strings tienen una frecuencia de 1 cada una.Entrada: arr[] = {“defr3t”, “lsk4dk”, “njdcd”}
Salida: 1 2
“njdcd”: 1
“defr3t”: 1
“lsk4dk”: 1
Enfoque: El enfoque para resolver este problema se basa en la técnica hash . Siga los pasos que se mencionan a continuación para resolver el problema.
- Tome una función de mapa hash para contar la frecuencia
- Ahora itere a de 0 a N-1 y luego almacene.
- Ahora itere otro bucle for dentro del primer bucle for (bucle anidado) desde 0 hasta el tamaño de esa string.
- Ahora comprueba si cada elemento es un alfabeto o no.
- En caso afirmativo, incremente el recuento de strings alfabéticas y aumente la frecuencia de esa string en 1.
- De lo contrario, incremente el recuento de strings alfanuméricas y aumente la frecuencia de esa string en 1.
- Imprima el recuento total de strings alfanuméricas y alfabéticas y las frecuencias de cada string por separado.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ code to implement the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find frequency
void find_freq(string v[], int n)
{
// Take an hash map function
// to count frequency
map<string, int> mp1, mp2;
int count1 = 0, count2 = 0;
bool flag;
for (int i = 0; i < n; i++) {
flag = true;
for (int j = 0; j < v[i].size(); j++) {
// Check whether the string has
// numbers or not
if (v[i][j] >= '0'
&& v[i][j] <= '9') {
flag = false;
break;
}
}
// For alphabetic string
if (flag) {
count1++;
mp1[v[i]]++;
}
// For alphanumeric string
else {
count2++;
mp2[v[i]]++;
}
}
cout << count1 << " " << count2 << endl;
// Print the frequencies of
// alphabetic strings
for (auto it : mp1) {
cout << it.first << ": "
<< it.second << endl;
;
}
// Print the frequencies of
// alphanumeric strings
for (auto it : mp2) {
cout << it.first << ": "
<< it.second << endl;
;
}
}
// Drive code
int main()
{
int N = 5;
string arr[] = { "abcd", "mak87s", "abcd",
"kakjdj", "laojs7s6" };
// Function call
find_freq(arr, N);
return 0;
}
Java
// Java code for the above approach
import java.util.*;
class GFG {
// Function to find frequency
static void find_freq(String v[], int n)
{
// Take an hash map function
// to count frequency
Map<String, Integer> mp1 = new HashMap<String, Integer>();
Map<String, Integer> mp2 = new HashMap<String, Integer>();
int count1 = 0, count2 = 0;
Boolean flag;
for (int i = 0; i < n; i++) {
flag = true;
for (int j = 0; j < v[i].length(); j++) {
// Check whether the string has
// numbers or not
if (v[i].charAt(j) >= '0'
&& v[i].charAt(j) <= '9') {
flag = false;
break;
}
}
// For alphabetic string
if (flag) {
count1++;
if (mp1.containsKey(v[i]))
{
mp1.put(v[i], mp1.get(v[i]) + 1);
}
else
{
mp1.put(v[i], 1);
}
}
// For alphanumeric string
else {
count2++;
if (mp2.containsKey(v[i]))
{
mp2.put(v[i], mp2.get(v[i]) + 1);
}
else
{
mp2.put(v[i], 1);
}
}
}
System.out.println(count1 + " " + count2);
// Print the frequencies of
// alphabetic strings
for (Map.Entry<String, Integer> entry : mp1.entrySet())
{
System.out.println(entry.getKey() + ":" + entry.getValue());
}
// Print the frequencies of
// alphanumeric strings
for (Map.Entry<String, Integer> entry : mp2.entrySet())
{
System.out.println(entry.getKey() + ":" + entry.getValue());
}
}
// Drive code
public static void main (String[] args) {
int N = 5;
String arr[] = { "abcd", "mak87s", "abcd",
"kakjdj", "laojs7s6" };
// Function call
find_freq(arr, N);
}
}
// This code is contributed by hrithikgarg03188.
Python3
# Python code to implement the above approach
# Function to find frequency
def find_freq(v, n):
# Take an hash map function
# to count frequency
mp1 = {}
mp2 = {}
count1 = 0
count2 = 0
flag = False
for i in range(n):
flag = True
for j in range(len(v[i])):
# Check whether the string has
# numbers or not
if(v[i][j] >= '0' and v[i][j] <= '9'):
flag = False
break
# For alphabetic string
if (flag):
count1 = count1+1
if v[i] in mp1:
mp1[v[i]] = mp1[v[i]]+1
else :
mp1[v[i]] = 1
# For alphanumeric string
else :
count2 = count2+1
if v[i] in mp1:
mp2[v[i]] = mp2[v[i]]+1
else :
mp2[v[i]] = 1
print(f"{count1} {count2}")
# Print the frequencies of
# alphabetic strings
for key,value in mp1.items():
print(f"{key} : {value}")
# Print the frequencies of
# alphanumeric strings
for key,value in mp2.items():
print(f"{key} : {value}")
# Driver code
N = 5
arr = [ "abcd", "mak87s", "abcd", "kakjdj", "laojs7s6" ];
# Function call
find_freq(arr, N);
# This code is contributed by shinjanpatra.
C#
// C# program to implement above approach
using System;
using System.Collections;
using System.Collections.Generic;
class GFG
{
// Function to find frequency
static void find_freq(String[] v, int n)
{
// Take an hash map function
// to count frequency
SortedDictionary<String, int> mp1 = new SortedDictionary<String, int>();
SortedDictionary<String, int> mp2 = new SortedDictionary<String, int>();
int count1 = 0, count2 = 0;
bool flag;
for (int i = 0; i < n; i++) {
flag = true;
for (int j = 0 ; j < v[i].Length ; j++) {
// Check whether the string has
// numbers or not
if (v[i][j] >= '0' && v[i][j] <= '9') {
flag = false;
break;
}
}
// For alphabetic string
if (flag) {
count1++;
if (mp1.ContainsKey(v[i]))
{
mp1[v[i]] += 1;
}
else
{
mp1.Add(v[i], 1);
}
}
// For alphanumeric string
else {
count2++;
if (mp2.ContainsKey(v[i]))
{
mp2[v[i]] += 1;
}
else
{
mp2.Add(v[i], 1);
}
}
}
Console.WriteLine(count1 + " " + count2);
// Print the frequencies of
// alphabetic strings
foreach (KeyValuePair<String, int> entry in mp1)
{
Console.WriteLine(entry.Key + ": " + entry.Value);
}
// Print the frequencies of
// alphanumeric strings
foreach (KeyValuePair<String, int> entry in mp2)
{
Console.WriteLine(entry.Key + ": " + entry.Value);
}
}
// Driver code
public static void Main(string[] args)
{
int N = 5;
String[] arr = new String[]{ "abcd", "mak87s", "abcd", "kakjdj", "laojs7s6" };
// Function call
find_freq(arr, N);
}
}
// This code is contributed by subhamgoyal2014.
Javascript
<script>
// JavaScript code to implement the above approach
// Function to find frequency
const find_freq = (v, n) =>
// Take an hash map function
// to count frequency
let mp1 = {}, mp2 = {};
let count1 = 0, count2 = 0;
let flag = false;
for (let i = 0; i < n; i++) {
flag = true;
for (let j = 0; j < v[i].length; j++) {
// Check whether the string has
// numbers or not
if (v[i][j] >= '0'
&& v[i][j] <= '9') {
flag = false;
break;
}
}
// For alphabetic string
if (flag) {
count1++;
if (v[i] in mp1) mp1[v[i]] += 1
else mp1[v[i]] = 1;
}
// For alphanumeric string
else {
count2++;
if (v[i] in mp2) mp2[v[i]] += 1;
else mp2[v[i]] = 1;
}
}
document.write(`${count1} ${count2}<br/>`);
// Print the frequencies of
// alphabetic strings
for (let it in mp1) {
document.write(`${it}: ${mp1[it]}<br/>`);
}
// Print the frequencies of
// alphanumeric strings
for (let it in mp2) {
document.write(`${it}: ${mp2[it]}<br/>`);
}
}
// Drive code
let N = 5;
let arr = ["abcd", "mak87s", "abcd",
"kakjdj", "laojs7s6"];
// Function call
find_freq(arr, N);
// This code is contributed by rakeshsahni
</script>
Producción
3 2 abcd: 2 kakjdj: 1 laojs7s6: 1 mak87s: 1
Complejidad de tiempo: O(N * M) donde M es la longitud máxima de una string
Espacio auxiliar: O(N)
Publicación traducida automáticamente
Artículo escrito por bunny09262002 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA