Given a string s consisting of only the characters 'a' and 'b', return true if every 'a' appears before every 'b' in the string. Otherwise, return false.
Example 1:
Input: s = "aaabbb " Output: true Explanation: The 'a's are at indices 0, 1, and 2, while the 'b's are at indices 3, 4, and 5. Hence, every 'a' appears before every 'b' and we return true.
Example 2:
Input: s = "abab " Output: false Explanation: There is an 'a' at index 2 and a 'b' at index 1. Hence, not every 'a' appears before every 'b' and we return false.
Example 3:
Input: s = "bbb " Output: true Explanation: There are no 'a's, hence, every 'a' appears before every 'b' and we return true.
Constraints:
1 <= s.length <= 100s[i] is either 'a' or 'b'.program main
implicit none
character(len=100) :: s
logical :: result
! Example 1
s = "aaabbb"
result = check_order(s)
if (result) then
print *, "Every 'a' appears before every 'b' in the string."
else
print *, "Not every 'a' appears before every 'b' in the string."
end if
! Example 2
s = "abab"
result = check_order(s)
if (result) then
print *, "Every 'a' appears before every 'b' in the string."
else
print *, "Not every 'a' appears before every 'b' in the string."
end if
! Example 3
s = "bbb"
result = check_order(s)
if (result) then
print *, "Every 'a' appears before every 'b' in the string."
else
print *, "Not every 'a' appears before every 'b' in the string."
end if
contains
function check_order(s) result(result)
implicit none
character(len=*), intent(in) :: s
integer :: i, j
logical :: result
do i = 1, len(s)-1
if (s(i:i) == 'a') then
do j = i+1, len(s)
if (s(j:j) == 'a') then
exit
else if (s(j:j) == 'b') then
result = .false.
return
end if
end do
else if (s(i:i) == 'b') then
do j = i+1, len(s)
if (s(j:j) == 'b') then
exit
else if (s(j:j) == 'a') then
result = .false.
return
end if
end do
end if
end do
result = .true.
end function check_order
end program mainNot every 'a' appears before every 'b' in the string. Not every 'a' appears before every 'b' in the string. Every 'a' appears before every 'b' in the string.
! Declare the input string and a flag to indicate if the condition is met
character(len=*), intent(in) :: s
logical :: is_sorted
! Initialize the flag to false
is_sorted = .false.
! Loop through the string and check if every 'a' appears before every 'b'
do i = 1, len(s)
if (s(i:i) == 'a') then
if (i < len(s) .and. s(i+1:i+1) == 'b') then
exit
else
is_sorted = .true.
end if
else if (s(i:i) == 'b') then
exit
end if
end do
! If the flag is still false, then the condition is not met
if (.not. is_sorted) then
write (*,*) "false"
else
write (*,*) "true"
end if
end
temp.f95:2:33:
2 | character(len=*), intent(in) :: s
| 1
Error: Symbol at (1) is not a DUMMY variable
def firstDayBeenInAllRooms(nextVisit):
n = len(nextVisit)
ans = 0
MOD = 10**9 + 7
dp = [0] * n
for i in range(1, n):
dp[i] = (dp[i - 1] + 1 + dp[i - 1] - dp[nextVisit[i - 1]] + MOD) % MOD
ans = (ans + dp[i]) % MOD
return ans
We can solve this problem using dynamic programming. We create an array dp of size n and initialize it with 0's. The dp[i] should represent the number of days it takes to go from room i to room i+1.
Starting from the second room (i=1), we loop through all rooms and update each dp[i] value. To calculate dp[i], we need to perform a cumulative summation of the days it has taken to visit the previous room (dp[i - 1]) plus 1 extra day (day i) and subtract the value at the index stored in nextVisit for the previous room (dp[nextVisit[i - 1]]).
Finally, we add the calculated value modulo 10^9 + 7 to the answer ans. Once we exit the loop, we return ans as the result.
int firstDayBeenInAllRooms(vector<int>& nextVisit) {
int n = nextVisit.size();
long long ans = 0, MOD = 1e9 + 7;
vector<long long> dp(n, 0);
for (int i = 1; i < n; ++i) {
dp[i] = (dp[i - 1] + 1 + dp[i - 1] - dp[nextVisit[i - 1]] + MOD) % MOD;
ans = (ans + dp[i]) % MOD;
}
return ans;
}
We can solve this problem using dynamic programming. We create an array dp of size n and initialize it with 0's. The dp[i] should represent the number of days it takes to go from room i to room i+1.
Starting from the second room (i=1), we loop through all rooms and update each dp[i] value. To calculate dp[i], we need to perform a cumulative summation of the days it has taken to visit the previous room (dp[i - 1]) plus 1 extra day (day i) and subtract the value at the index stored in nextVisit for the previous room (dp[nextVisit[i - 1]]).
Finally, we add the calculated value modulo 10^9 + 7 to the answer ans. Once we exit the loop, we return ans as the result.