Dada una string, encuentre las palabras de longitud mínima y máxima en ella.
Ejemplos:
Input : "This is a test string"
Output : Minimum length word: a
Maximum length word: string
Input : "GeeksforGeeks A computer Science portal for Geeks"
Output : Minimum length word: A
Maximum length word: GeeksforGeeks
Acercarse
La idea es mantener un índice inicial si y un índice final ei .
- si apunta al comienzo de una nueva palabra y recorremos la string usando ei.
- Cada vez que se encuentra un espacio o un carácter ‘\0’, calculamos la longitud de la palabra actual usando (ei – si) y la comparamos con la longitud mínima y máxima hasta el momento.
- Si es menor, actualice min_length y min_start_index (que apunta al comienzo de la palabra de longitud mínima).
- Si es mayor, actualice max_length y max_start_index (que apunta al inicio de la palabra de longitud máxima).
- Finalmente, actualice minWord y maxWord, que son strings de salida que se han enviado por referencia con las substrings que comienzan en min_start_index y max_start_index de longitud min_length y max_length respectivamente.
Implementación:
C++
// CPP Program to find Smallest and
// Largest Word in a String
#include<iostream>
#include<cstring>
using namespace std;
void minMaxLengthWords(string input, string &minWord, string &maxWord)
{
// minWord and maxWord are received by reference
// and not by value
// will be used to store and return output
int len = input.length();
int si = 0, ei = 0;
int min_length = len, min_start_index = 0, max_length = 0, max_start_index = 0;
// Loop while input string is not empty
while (ei <= len)
{
if (ei < len && input[ei] != ' ')
ei++;
else
{
// end of a word
// find curr word length
int curr_length = ei - si;
if (curr_length < min_length)
{
min_length = curr_length;
min_start_index = si;
}
if (curr_length > max_length)
{
max_length = curr_length;
max_start_index = si;
}
ei++;
si = ei;
}
}
// store minimum and maximum length words
minWord = input.substr(min_start_index, min_length);
maxWord = input.substr(max_start_index, max_length);
}
// Driver code
int main()
{
string a = "GeeksforGeeks A Computer Science portal for Geeks";
string minWord, maxWord;
minMaxLengthWords(a, minWord, maxWord);
// to take input in string use getline(cin, a);
cout << "Minimum length word: "
<< minWord << endl
<< "Maximum length word: "
<< maxWord << endl;
}
Java
// Java Program to find Smallest and
// Largest Word in a String
class GFG
{
static String minWord = "", maxWord = "";
static void minMaxLengthWords(String input)
{
input=input.trim();//Triming any space before the String else space at start would be consider as smallest word
// minWord and maxWord are received by reference
// and not by value
// will be used to store and return output
int len = input.length();
int si = 0, ei = 0;
int min_length = len, min_start_index = 0,
max_length = 0, max_start_index = 0;
// Loop while input string is not empty
while (ei <= len)
{
if (ei < len && input.charAt(ei) != ' ')
{
ei++;
}
else
{
// end of a word
// find curr word length
int curr_length = ei - si;
if (curr_length < min_length)
{
min_length = curr_length;
min_start_index = si;
}
if (curr_length > max_length)
{
max_length = curr_length;
max_start_index = si;
}
ei++;
si = ei;
}
}
// store minimum and maximum length words
minWord = input.substring(min_start_index, min_start_index + min_length);
maxWord = input.substring(max_start_index, max_start_index+max_length);//Earlier code was not working if the largests word is inbetween String
}
// Driver code
public static void main(String[] args)
{
String a = "GeeksforGeeks A Computer Science portal for Geeks";
minMaxLengthWords(a);
// to take input in string use getline(cin, a);
System.out.print("Minimum length word: "
+ minWord
+ "\nMaximum length word: "
+ maxWord);
}
}
// This code contributed by Rajput-Ji
Python 3
# Python3 program to find Smallest and
# Largest Word in a String
# defining the method to find the longest
# word and the shortest word
def minMaxLengthWords(inp):
length = len(inp)
si = ei = 0
min_length = length
min_start_index = max_length = max_start_index = 0
# loop to find the length and stating index
# of both longest and shortest words
while ei <= length:
if (ei < length) and (inp[ei] != " "):
ei += 1
else:
curr_length = ei - si
# condition checking for the shortest word
if curr_length < min_length:
min_length = curr_length
min_start_index = si
# condition for the longest word
if curr_length > max_length:
max_length = curr_length
max_start_index = si
ei += 1
si = ei
# extracting the shortest word using
# it's starting index and length
minWord = inp[min_start_index :
min_start_index + min_length]
# extracting the longest word using
# it's starting index and length
maxWord = inp[max_start_index : max_length]
# printing the final result
print("Minimum length word: ", minWord)
print ("Maximum length word: ", maxWord)
# Driver Code
# Using this string to test our code
a = "GeeksforGeeks A Computer Science portal for Geeks"
minMaxLengthWords(a)
# This code is contributed by Animesh_Gupta
C#
// C# Program to find Smallest and
// Largest Word in a String
using System;
class GFG
{
static String minWord = "", maxWord = "";
static void minMaxLengthWords(String input)
{
// minWord and maxWord are received by reference
// and not by value
// will be used to store and return output
int len = input.Length;
int si = 0, ei = 0;
int min_length = len, min_start_index = 0,
max_length = 0, max_start_index = 0;
// Loop while input string is not empty
while (ei <= len)
{
if (ei < len && input[ei] != ' ')
{
ei++;
}
else
{
// end of a word
// find curr word length
int curr_length = ei - si;
if (curr_length < min_length)
{
min_length = curr_length;
min_start_index = si;
}
if (curr_length > max_length)
{
max_length = curr_length;
max_start_index = si;
}
ei++;
si = ei;
}
}
// store minimum and maximum length words
minWord = input.Substring(min_start_index, min_length);
maxWord = input.Substring(max_start_index, max_length);
}
// Driver code
public static void Main(String[] args)
{
String a = "GeeksforGeeks A Computer Science portal for Geeks";
minMaxLengthWords(a);
// to take input in string use getline(cin, a);
Console.Write("Minimum length word: "
+ minWord
+ "\nMaximum length word: "
+ maxWord);
}
}
// This code has been contributed by 29AjayKumar
Javascript
<script>
// JavaScript Program to find Smallest and
// Largest Word in a String
let minWord = "";
let maxWord = "";
function minMaxLengthWords(input)
{
// minWord and maxWord are received by reference
// and not by value
// will be used to store and return output
let len = input.length;
let si = 0, ei = 0;
let min_length = len;
let min_start_index = 0;
let max_length = 0;
let max_start_index = 0;
// Loop while input string is not empty
while (ei <= len)
{
if (ei < len && input[ei] != ' ')
{
ei++;
}
else
{
// end of a word
// find curr word length
let curr_length = ei - si;
if (curr_length < min_length)
{
min_length = curr_length;
min_start_index = si;
}
if (curr_length > max_length)
{
max_length = curr_length;
max_start_index = si;
}
ei++;
si = ei;
}
}
// store minimum and maximum length words
minWord =
input.substring(min_start_index,min_start_index + min_length);
maxWord =
input.substring(max_start_index, max_length);
}
// Driver code
let a = "GeeksforGeeks A Computer Science portal for Geeks";
minMaxLengthWords(a);
// to take input in string use getline(cin, a);
document.write("Minimum length word: "
+ minWord+"<br>"
+ "Maximum length word: "
+ maxWord);
</script>
Producción
Minimum length word: A Maximum length word: GeeksforGeeks
Complejidad de tiempo: O(n), donde n es la longitud de la string.
Espacio auxiliar: O(n), donde n es la longitud de la string.
Esto se debe a que cuando se pasa una string en la función, crea una copia de sí mismo en la pila.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA