Given a string s, find the first non-repeating character in it and return its index. If it does not exist, return -1.
Example 1:
Input: s = "leetcode" Output: 0
Example 2:
Input: s = "loveleetcode" Output: 2
Example 3:
Input: s = "aabb" Output: -1
Constraints:
1 <= s.length <= 105s consists of only lowercase English letters.program main
implicit none
integer :: i, j, k, n, m
character(len=100) :: s
logical :: found
! Example 1
s = 'leetcode'
print *, find_first_non_repeating_char(s)
! Example 2
s = 'loveleetcode'
print *, find_first_non_repeating_char(s)
! Example 3
s = 'aabb'
print *, find_first_non_repeating_char(s)
contains
function find_first_non_repeating_char(str) result(index)
implicit none
character(len=*), intent(in) :: str
integer :: index
logical :: found
index = -1
do i = 1, len(str)
found = .false.
do j = 1, len(str)
if (i /= j) then
if (str(i:i) == str(j:j)) then
found = .true.
exit
end if
end if
end do
if (.not. found) then
index = i
exit
end if
end do
end function find_first_non_repeating_char
end program main 1
3
-1
! Declare variables
integer :: i, j, len, first_index
character(len=105) :: s
! Read input
read(*,*) s
! Initialize variables
len = len(s)
first_index = -1
! Loop through the string
do i = 1, len
! Check if the current character is already in the string
do j = 1, i - 1
if (s(i:i) == s(j:j)) then
cycle
end if
end do
! If the current character is not in the string, set the first index
if (first_index == -1) then
first_index = i
end if
end do
! Output the result
write(*,*) first_index
end
temp.f95:9:1:
9 | len = len(s)
| 1
Error: Unclassifiable statement at (1)
def firstUniqChar(s):
char_count = {}
for c in s:
char_count[c] = char_count.get(c, 0) + 1
for i in range(len(s)):
if char_count[s[i]] == 1:
return i
return -1
The algorithm first creates a dictionary to store the count of each character in the input string 's'. Then, it iterates through the string, counting the occurrences of each character and updating the dictionary accordingly.
After that, the algorithm iterates through the input string 's' again, and it checks if the count of the current character in the dictionary is 1. If it is, the algorithm returns the index of the current character as the answer. If no such character is found, the algorithm returns -1 as the answer.
The time complexity of this algorithm is O(n), where n is the length of the input string 's'.
int firstUniqChar(std::string s) {
std::unordered_map<char, int> charCount;
for (char c : s)
charCount[c]++;
for (int i = 0; i < s.length(); i++)
if (charCount[s[i]] == 1)
return i;
return -1;
}
The algorithm first creates a dictionary to store the count of each character in the input string 's'. Then, it iterates through the string, counting the occurrences of each character and updating the dictionary accordingly.
After that, the algorithm iterates through the input string 's' again, and it checks if the count of the current character in the dictionary is 1. If it is, the algorithm returns the index of the current character as the answer. If no such character is found, the algorithm returns -1 as the answer.
The time complexity of this algorithm is O(n), where n is the length of the input string 's'.