Given a pattern and a string s, find if s follows the same pattern.
Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty word in s.
Example 1:
Input: pattern = "abba ", s = "dog cat cat dog " Output: true
Example 2:
Input: pattern = "abba ", s = "dog cat cat fish " Output: false
Example 3:
Input: pattern = "aaaa ", s = "dog cat cat dog " Output: false
Constraints:
1 <= pattern.length <= 300pattern contains only lower-case English letters.1 <= s.length <= 3000s contains only lowercase English letters and spaces ' '.s does not contain any leading or trailing spaces.s are separated by a single space.program main
implicit none
character(len=*), parameter :: pattern = "abba "
character(len=*), parameter :: s1 = "dog cat cat dog "
character(len=*), parameter :: s2 = "dog cat cat fish "
character(len=*), parameter :: s3 = "dog cat cat dog"
logical :: result1, result2, result3
result1 = follows_pattern(pattern, s1)
result2 = follows_pattern(pattern, s2)
result3 = follows_pattern(pattern, s3)
write (*,*) result1
write (*,*) result2
write (*,*) result3
contains
function follows_pattern(pattern, s) result(result)
implicit none
character(len=*), intent(in) :: pattern
character(len=*), intent(in) :: s
integer :: i, j, k
logical :: result
result = .false.
! Check if the length of the pattern and the input string are equal
if (len(trim(pattern)) /= len(trim(s))) return
! Check if the pattern and the input string have the same number of words
do i = 1, len(trim(pattern))
if (index(trim(pattern), ' ') /= index(trim(s), ' ')) return
end do
! Check if there is a bijection between each letter in the pattern and a non-empty word in the input string
do i = 1, len(trim(pattern))
k = index(trim(pattern), ' ')
do j = 1, len(trim(s))
if (trim(s)(j:j) == trim(pattern)(i:i)) then
if (trim(s)(k+1:) /= '') then
result = .true.
exit
end if
end if
end do
end do
end function follows_pattern
end program maintemp.f95:42:28:
42 | if (trim(s)(j:j) == trim(pattern)(i:i)) then
| 1
Error: Syntax error in IF-expression at (1)
temp.f95:43:32:
43 | if (trim(s)(k+1:) /= '') then
| 1
Error: Syntax error in IF-expression at (1)
temp.f95:46:23:
46 | end if
| 1
Error: Expecting END DO statement at (1)
temp.f95:47:19:
47 | end if
| 1
Error: Expecting END DO statement at (1)
! This script solves the problem of finding if a string follows a pattern.
! The pattern is given as a string, and the string to be checked is given as another string.
! The program should output "true" if the string follows the pattern, and "false" otherwise.
! Declare variables
integer :: i, j, k, l, m
character(len=300) :: pattern, s
character(len=3000) :: temp
logical :: found
! Read input from stdin
read(*,*) pattern
read(*,*) s
! Initialize variables
i = 1
j = 1
k = 1
l = 1
m = 1
found = .false.
! Check if the pattern is a palindrome
if (pattern(1:i) == pattern(len(pattern):-1:-1)) then
! The pattern is a palindrome, so we can use it to check if the string follows the pattern
do while (j <= len(s))
! Check if the current character in the string matches the current character in the pattern
if (s(j:j) == pattern(i:i)) then
! The current character in the string matches the current character in the pattern, so we move on to the next character in the string
j = j + 1
i = i + 1
else
! The current character in the string does not match the current character in the pattern, so we move on to the next character in the pattern
i = i + 1
end if
! Check if we have reached the end of the pattern
if (i > len(pattern)) then
! We have reached the end of the pattern, so we are done
found = .true.
exit
end if
end do
else
! The pattern is not a palindrome, so we need to use it to check if the string follows the pattern
do while (j <= len(s))
! Check if the current character in the string matches the current character in the pattern
if (s(j:j) == pattern(i:i)) then
! The current character in the string matches the current character in the pattern, so we move on to the next character in the string
j = j + 1
i = i + 1
else
! The current character in the string does not match the current character in the pattern, so we move on to the next character in the pattern
i = i + 1
end if
! Check if we have reached the end of the pattern
if (i > len(pattern)) then
! We have reached the end of the pattern, so we are done
found = .true.
exit
end if
end do
end if
! Output the result
if (found) then
write(*,*) "true"
else
write(*,*) "false"
end if
end
temp.f95:24:44:
24 | if (pattern(1:i) == pattern(len(pattern):-1:-1)) then
| 1
Error: Syntax error in SUBSTRING specification at (1)
temp.f95:43:4:
43 | else
| 1
Error: Unexpected ELSE statement at (1)
temp.f95:62:3:
62 | end if
| 1
Error: Expecting END PROGRAM statement at (1)
def word_pattern(pattern, s):
words = s.split()
if len(pattern) != len(words):
return False
char_map, word_map = {}, {}
for c, word in zip(pattern, words):
if c not in char_map and word not in word_map:
char_map[c] = word
word_map[word] = c
else:
if char_map.get(c) != word or word_map.get(word) != c:
return False
return True
1. First, split the string 's' into words array based on the space-delimited words.
#include <string>
#include <unordered_map>
#include <sstream>
using namespace std;
bool wordPattern(string pattern, string s) {
unordered_map<char, string> char_map;
unordered_map<string, char> word_map;
stringstream ss(s);
int i = 0;
for (string word; ss >> word; ++i) {
if (i == pattern.size()) return false;
char c = pattern[i];
if (char_map.count(c) == 0 && word_map.count(word) == 0) {
char_map[c] = word;
word_map[word] = c;
} else {
if (char_map[c] != word || word_map[word] != c) return false;
}
}
return i == pattern.size();
}
1. First, split the string 's' into words array based on the space-delimited words.