You are given a string s consisting only of the characters '0' and '1'. In one operation, you can change any '0' to '1' or vice versa.
The string is called alternating if no two adjacent characters are equal. For example, the string "010 " is alternating, while the string "0100 " is not.
Return the minimum number of operations needed to make s alternating.
Example 1:
Input: s = "0100 " Output: 1 Explanation: If you change the last character to '1', s will be "0101 ", which is alternating.
Example 2:
Input: s = "10 " Output: 0 Explanation: s is already alternating.
Example 3:
Input: s = "1111 " Output: 2 Explanation: You need two operations to reach "0101 " or "1010 ".
Constraints:
1 <= s.length <= 104s[i] is either '0' or '1'.program main
implicit none
integer :: i, j, k, n, m
character(len=100) :: s
! Examples
write (*,*) "Example 1:"
s = "0100 "
call solve(s)
write (*,*) "Expected: 1"
write (*,*) "Actual:", s
write (*,*) "Example 2:"
s = "10 "
call solve(s)
write (*,*) "Expected: 0"
write (*,*) "Actual:", s
write (*,*) "Example 3:"
s = "1111 "
call solve(s)
write (*,*) "Expected: 2"
write (*,*) "Actual:", s
contains
subroutine solve(s)
implicit none
character(len=*), intent(inout) :: s
! Variables
integer :: i, j, k, n, m
! Initialize variables
n = len(s)
m = 0
! Check if string is already alternating
do i = 1, n - 1
if (s(i) == s(i+1)) then
exit
end if
end do
if (i == n) then
return
end if
! Find first non-alternating character
do i = 1, n
if (s(i) /= s(i+1)) then
exit
end if
end do
! Change character to opposite value
if (s(i) == '0') then
s(i) = '1'
else
s(i) = '0'
end if
! Update counter
m = m + 1
! Recursively call function until string is alternating
call solve(s)
end subroutine solve
end program maintemp.f95:28:22:
28 | subroutine solve(s)
| 1
Error: PROCEDURE attribute conflicts with INTENT attribute in ‘s’ at (1)
temp.f95:36:16:
36 | n = len(s)
| 1
Error: ‘string’ argument of ‘len’ intrinsic at (1) must be CHARACTER
temp.f95:59:12:
59 | s(i) = '1'
| 1
Error: The function result on the lhs of the assignment at (1) must have the pointer attribute.
temp.f95:61:12:
61 | s(i) = '0'
| 1
Error: The function result on the lhs of the assignment at (1) must have the pointer attribute.
temp.f95:68:21:
68 | call solve(s)
| 1
Error: SUBROUTINE ‘solve’ at (1) cannot be called recursively, as it is not RECURSIVE
temp.f95:10:15:
10 | call solve(s)
| 1
Error: Expected a procedure for argument ‘s’ at (1)
temp.f95:16:15:
16 | call solve(s)
| 1
Error: Expected a procedure for argument ‘s’ at (1)
temp.f95:22:15:
22 | call solve(s)
| 1
Error: Expected a procedure for argument ‘s’ at (1)
module alternating_string
implicit none
contains
integer function min_operations(s) result(res)
! Given a string s consisting only of the characters '0' and '1', return the minimum number of operations needed to make s alternating.
character(len=*), intent(in) :: s
integer :: i, n
! Initialize variables
n = len(s)
res = 0
! Check if the string is already alternating
if (mod(n, 2) == 0) then
! If the length of the string is even, check if all characters are equal
if (count(s == '0') == n/2) then
res = 1
else
! If the length of the string is even and not all characters are equal, check if the first and last characters are equal
if (s(1) == s(n)) then
res = 1
else
res = 2
endif
endif
else
! If the length of the string is odd, check if the first and last characters are equal
if (s(1) == s(n)) then
res = 1
else
res = 2
endif
endif
! If the string is not alternating, perform operations to make it alternating
if (res > 0) then
do i = 1, n-1
if (s(i) /= s(i+1)) then
res = res + 1
s(i) = s(i+1)
exit
endif
enddo
endif
end function min_operations
end module alternating_string
program test_alternating_string
use alternating_string
implicit none
! Test case 1:
call test("0100 ", 1)
! Test case 2:
call test("10 ", 0)
! Test case 3:
call test("1111 ", 2)
contains
subroutine test(s, expected)
character(len=*), intent(in) :: s
integer, intent(in) :: expected
integer :: res
res = min_operations(s)
if (res /= expected) then
write (*,*) "Error: min_operations(", s, ") returned ", res, " but expected ", expected
stop
endif
end subroutine test
end program test_alternating_string
temp.f95:34:6:
34 | if (s(1) == s(n)) then
| 1
Error: Syntax error in IF-expression at (1)
temp.f95:30:4:
30 | else
| 1
......
38 | else
| 2
Error: Duplicate ELSE statements at (1) and (2)
temp.f95:46:4:
46 | else
| 1
Error: Unexpected ELSE statement at (1)
temp.f95:50:6:
50 | if (s(1) == s(n)) then
| 1
Error: Syntax error in IF-expression at (1)
temp.f95:54:4:
54 | else
| 1
Error: Unexpected ELSE statement at (1)
temp.f95:58:3:
58 | endif
| 1
Error: Expecting END FUNCTION statement at (1)
temp.f95:60:3:
60 | endif
| 1
Error: Expecting END FUNCTION statement at (1)
temp.f95:68:6:
68 | if (s(i) /= s(i+1)) then
| 1
Error: Syntax error in IF-expression at (1)
temp.f95:76:3:
76 | endif
| 1
Error: Expecting END DO statement at (1)
temp.f95:26:10:
26 | if (count(s == '0') == n/2) then
| 1
Error: ‘mask’ argument of ‘count’ intrinsic at (1) must be a logical array
temp.f95:88:5:
88 | use alternating_string
| 1
Fatal Error: Cannot open module file ‘alternating_string.mod’ for reading at (1): No such file or directory
compilation terminated.
from typing import List
def canDistribute(nums: List[int], quantity: List[int]) -> bool:
counts = [0] * 51
for num in nums:
counts[num] += 1
quantity.sort(reverse=True)
def dfs(idx: int) -> bool:
if idx == len(quantity):
return True
for i in range(1, 51):
if counts[i] >= quantity[idx]:
counts[i] -= quantity[idx]
if dfs(idx + 1):
return True
counts[i] += quantity[idx]
return False
return dfs(0)
1. Count the occurrences of each unique value in the `nums` list and store them in an array `counts`.
quantity array in descending order.dfs that takes the index idx of the current customer.
a. If idx reaches the total number of customers, all orders have been fulfilled so return true.
b. For each unique value, if its count in counts is greater than or equal to the current customer's order quantity, subtract the order quantity from its count and call dfs for the next customer (idx + 1).
c. If the quantity can be distributed successfully for the next customers, return true. Otherwise, restore the count in counts and continue to check the next unique value.dfs(0) for the first customer's order fulfillment.#include <vector>
#include <algorithm>
using namespace std;
bool canDistribute(vector<int>& nums, vector<int>& quantity) {
vector<int> counts(51, 0);
for (int num : nums) {
counts[num]++;
}
sort(quantity.rbegin(), quantity.rend());
function<bool(int)> dfs = [&](int idx) {
if (idx == quantity.size()) {
return true;
}
for (int i = 1; i <= 50; ++i) {
if (counts[i] >= quantity[idx]) {
counts[i] -= quantity[idx];
if (dfs(idx + 1)) {
return true;
}
counts[i] += quantity[idx];
}
}
return false;
};
return dfs(0);
}
1. Count the occurrences of each unique value in the `nums` list and store them in an array `counts`.
quantity array in descending order.dfs that takes the index idx of the current customer.
a. If idx reaches the total number of customers, all orders have been fulfilled so return true.
b. For each unique value, if its count in counts is greater than or equal to the current customer's order quantity, subtract the order quantity from its count and call dfs for the next customer (idx + 1).
c. If the quantity can be distributed successfully for the next customers, return true. Otherwise, restore the count in counts and continue to check the next unique value.dfs(0) for the first customer's order fulfillment.