I can't seem to figure out where in my loop I don't come back to referencing the same indexed values to validate the palindrome.
I tried the code and should expect to see "This is a palindrome" but it returns the opposite
3 Answers
An easier way would be to check that all elements of your string match all elements of the reversed string:
is_palindrome <- function(s) {
x <- strsplit(s, "")[[1]]
all(x == rev(x))
}
is_palindrome("radar")
#> [1] TRUE
is_palindrome("reader")
#> [1] FALSE
If for some reason you need to use a loop, you can do:
is_palindrome <- function(s) {
x <- strsplit(s, "")[[1]]
for(i in 1:length(x)) {
if(x[i] != x[length(x) + 1 - i]) {
return(paste0("'", s, "' is not a palindrome"))
}
}
return(paste0("'", s, "' is a palindrome"))
}
is_palindrome("radar")
#> [1] "'radar' is a palindrome"
is_palindrome("reader")
#> [1] "'reader' is not a palindrome"
Created on 2022-10-29 with reprex v2.0.2
Comments
Use seq_along
Edit
Also there is an error at indexing x: must be len-i+1
Finally i do some modifications
s <-"radar"
x <- strsplit(s, "")[[1]]
len <-length(x)
#find the middle
middle <- ceiling(length(x)/2)
### Loop to the middle
answer <- "The word is a palindrome"
for (i in seq_along(x)){
if (i < middle){
#check for index position to match the other side
if (x[i] != x[len-i+1]){
answer <- "The word is not a palindrome"
break;
}
}
}
print(answer)
1 Comment
toplevel5
is there a way to get around using seq_along?
You can try the code below using rev + utf8ToInt
isPalindrome <- function(s) s == intToUtf8(rev(utf8ToInt(s)))
Comments
