Dada una string de palabras separadas por espacios. La tarea es encontrar la primera palabra par de longitud máxima de la string. Ej.: “Te dan una array de n números” La respuesta sería “una” y no “de” porque “una” viene antes de “de”.
Ejemplos:
Input: "this is a test string" Output: string Even length words are this, is, test, string. Even maximum length word is string. Input: "geeksforgeeks is a platform for geeks" Output: platform Only even length word is platform.
Enfoque: La idea es recorrer la string de entrada y encontrar la longitud de cada palabra. Compruebe si la longitud de la palabra es par o no. Si es par, compare la longitud con la longitud máxima encontrada hasta ahora. Si la longitud es estrictamente mayor que la longitud máxima, almacene la palabra actual como string requerida.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to find maximum length even word
#include <bits/stdc++.h>
using namespace std;
// Function to find maximum length even word
string findMaxLenEven(string str)
{
int n = str.length();
int i = 0;
// To store length of current word.
int currlen = 0;
// To store length of maximum length word.
int maxlen = 0;
// To store starting index of maximum
// length word.
int st = -1;
while (i < n) {
// If current character is space then
// word has ended. Check if it is even
// length word or not. If yes then
// compare length with maximum length
// found so far.
if (str[i] == ' ') {
if (currlen % 2 == 0) {
if (maxlen < currlen) {
maxlen = currlen;
st = i - currlen;
}
}
// Set currlen to zero for next word.
currlen = 0;
}
else {
// Update length of current word.
currlen++;
}
i++;
}
// Check length of last word.
if (currlen % 2 == 0) {
if (maxlen < currlen) {
maxlen = currlen;
st = i - currlen;
}
}
// If no even length word is present
// then return -1.
if (st == -1)
return "-1";
return str.substr(st, maxlen);
}
// Driver code
int main()
{
string str = "this is a test string";
cout << findMaxLenEven(str);
return 0;
}
Java
// Java program to find maximum length even word
class GFG
{
// Function to find maximum length even word
static String findMaxLenEven(String str)
{
int n = str.length();
int i = 0;
// To store length of current word.
int currlen = 0;
// To store length of maximum length word.
int maxlen = 0;
// To store starting index of maximum
// length word.
int st = -1;
while (i < n)
{
// If current character is space then
// word has ended. Check if it is even
// length word or not. If yes then
// compare length with maximum length
// found so far.
if (str.charAt(i) == ' ')
{
if (currlen % 2 == 0)
{
if (maxlen < currlen)
{
maxlen = currlen;
st = i - currlen;
}
}
// Set currlen to zero for next word.
currlen = 0;
}
else
{
// Update length of current word.
currlen++;
}
i++;
}
// Check length of last word.
if (currlen % 2 == 0)
{
if (maxlen < currlen)
{
maxlen = currlen;
st = i - currlen;
}
}
// If no even length word is present
// then return -1.
if (st == -1)
return "-1";
return str.substring(st, st + maxlen);
}
// Driver code
public static void main(String args[])
{
String str = "this is a test string";
System.out.println( findMaxLenEven(str));
}
}
// This code is contributed by Arnab Kundu
Python 3
# Python3 program to find maximum
# length even word
# Function to find maximum length
# even word
def findMaxLenEven(str):
n = len(str)
i = 0
# To store length of current word.
currlen = 0
# To store length of maximum length word.
maxlen = 0
# To store starting index of maximum
# length word.
st = -1
while (i < n):
# If current character is space then
# word has ended. Check if it is even
# length word or not. If yes then
# compare length with maximum length
# found so far.
if (str[i] == ' '):
if (currlen % 2 == 0):
if (maxlen < currlen):
maxlen = currlen
st = i - currlen
# Set currlen to zero for next word.
currlen = 0
else :
# Update length of current word.
currlen += 1
i += 1
# Check length of last word.
if (currlen % 2 == 0):
if (maxlen < currlen):
maxlen = currlen
st = i - currlen
# If no even length word is present
# then return -1.
if (st == -1):
print("trie")
return "-1"
return str[st: st + maxlen]
# Driver code
if __name__ == "__main__":
str = "this is a test string"
print(findMaxLenEven(str))
# This code is contributed by Ita_c
C#
// C# program to find maximum length even word
using System;
class GFG
{
// Function to find maximum length even word
static String findMaxLenEven(string str)
{
int n = str.Length;
int i = 0;
// To store length of current word.
int currlen = 0;
// To store length of maximum length word.
int maxlen = 0;
// To store starting index of maximum
// length word.
int st = -1;
while (i < n)
{
// If current character is space then
// word has ended. Check if it is even
// length word or not. If yes then
// compare length with maximum length
// found so far.
if (str[i] == ' ')
{
if (currlen % 2 == 0)
{
if (maxlen < currlen)
{
maxlen = currlen;
st = i - currlen;
}
}
// Set currlen to zero for next word.
currlen = 0;
}
else
{
// Update length of current word.
currlen++;
}
i++;
}
// Check length of last word.
if (currlen % 2 == 0)
{
if (maxlen < currlen)
{
maxlen = currlen;
st = i - currlen;
}
}
// If no even length word is present
// then return -1.
if (st == -1)
return "-1";
return str.Substring(st, maxlen);
}
// Driver code
public static void Main()
{
string str = "this is a test string";
Console.WriteLine(findMaxLenEven(str));
}
// This code is contributed by Ryuga
}
Javascript
<script>
// Javascript program to find maximum length even word
// Function to find maximum length even word
function findMaxLenEven(str)
{
var n = str.length;
var i = 0;
// To store length of current word.
var currlen = 0;
// To store length of maximum length word.
var maxlen = 0;
// To store starting index of maximum
// length word.
var st = -1;
while (i < n) {
// If current character is space then
// word has ended. Check if it is even
// length word or not. If yes then
// compare length with maximum length
// found so far.
if (str[i] == ' ') {
if (currlen % 2 == 0) {
if (maxlen < currlen) {
maxlen = currlen;
st = i - currlen;
}
}
// Set currlen to zero for next word.
currlen = 0;
}
else {
// Update length of current word.
currlen++;
}
i++;
}
// Check length of last word.
if (currlen % 2 == 0) {
if (maxlen < currlen) {
maxlen = currlen;
st = i - currlen;
}
}
// If no even length word is present
// then return -1.
if (st == -1)
return "-1";
return str.substr(st, maxlen);
}
// Driver code
var str = "this is a test string";
document.write( findMaxLenEven(str));
// This code is contributed by noob2000.
</script>
Producción:
string
Complejidad de tiempo: O(N), donde N es la longitud de la string.
Espacio Auxiliar: O(1)