Two Out of Three

🏠 ⬅️ ➡️

Given three integer arrays nums1, nums2, and nums3, return a distinct array containing all the values that are present in at least two out of the three arrays. You may return the values in any order.

Example 1:

Input: nums1 = [1,1,3,2], nums2 = [2,3], nums3 = [3] Output: [3,2] Explanation: The values that are present in at least two arrays are: - 3, in all three arrays. - 2, in nums1 and nums2.

Example 2:

Input: nums1 = [3,1], nums2 = [2,3], nums3 = [1,2] Output: [2,3,1] Explanation: The values that are present in at least two arrays are: - 2, in nums2 and nums3. - 3, in nums1 and nums2. - 1, in nums1 and nums3.

Example 3:

Input: nums1 = [1,2,2], nums2 = [4,3,3], nums3 = [5] Output: [] Explanation: No value is present in at least two arrays.

Constraints:

  • 1 <= nums1.length, nums2.length, nums3.length <= 100
  • 1 <= nums1[i], nums2[j], nums3[k] <= 100
Note: This problem is from LeetCode.
❌ Compiled
❌ Executed
❌ Correct 
program main
    ! Solve the problem described in the task
    implicit none

    integer, parameter :: n = 3
    integer, dimension(n) :: nums1 = [1, 1, 3, 2]
    integer, dimension(n) :: nums2 = [2, 3]
    integer, dimension(n) :: nums3 = [3]
    integer, dimension(:), allocatable :: result

    call solve(nums1, nums2, nums3, result)

    write (*, '(*(g0, :, ", "))') result

contains

    subroutine solve(nums1, nums2, nums3, result)
        implicit none
        integer, dimension(:), intent(in) :: nums1
        integer, dimension(:), intent(in) :: nums2
        integer, dimension(:), intent(in) :: nums3
        integer, dimension(:), allocatable, intent(out) :: result
        logical, dimension(size(nums1)) :: mask1
        logical, dimension(size(nums2)) :: mask2
        logical, dimension(size(nums3)) :: mask3
        integer :: i, j

        ! Initialize the masks
        mask1 = .false.
        mask2 = .false.
        mask3 = .false.

        ! Set the masks for the elements that are present in at least two arrays
        do i = 1, size(nums1)
            if (any(nums1(i) == nums2)) then
                mask1(i) = .true.
            end if
            if (any(nums1(i) == nums3)) then
                mask1(i) = .true.
            end if
        end do
        do i = 1, size(nums2)
            if (any(nums2(i) == nums1)) then
                mask2(i) = .true.
            end if
            if (any(nums2(i) == nums3)) then
                mask2(i) = .true.
            end if
        end do
        do i = 1, size(nums3)
            if (any(nums3(i) == nums1)) then
                mask3(i) = .true.
            end if
            if (any(nums3(i) == nums2)) then
                mask3(i) = .true.
            end if
        end do

        ! Allocate the result array
        allocate(result(count(mask1 .or. mask2 .or. mask3)))

        ! Fill the result array
        j = 1
        do i = 1, size(nums1)
            if (mask1(i)) then
                result(j) = nums1(i)
                j = j + 1
            end if
        end do
        do i = 1, size(nums2)
            if (mask2(i)) then
                result(j) = nums2(i)
                j = j + 1
            end if
        end do
        do i = 1, size(nums3)
            if (mask3(i)) then
                result(j) = nums3(i)
                j = j + 1
            end if
        end do

    end subroutine solve
end program main
✅ Compiled
✅ Executed
❌ Correct 
module solve_problem

implicit none

contains

function solve(nums1, nums2, nums3) result(result)

integer, intent(in) :: nums1(:), nums2(:), nums3(:)
integer :: result(size(nums1))

integer :: i, j, k, count

! Initialize the result array
result = 0

! Loop through the arrays
do i = 1, size(nums1)
    ! Check if the current value is present in at least two arrays
    count = 0
    do j = 1, size(nums2)
        if (nums1(i) == nums2(j)) then
            count = count + 1
        end if
    end do
    do k = 1, size(nums3)
        if (nums1(i) == nums3(k)) then
            count = count + 1
        end if
    end do
    ! If the value is present in at least two arrays, add it to the result array
    if (count >= 2) then
        result(i) = nums1(i)
    end if
end do

end function solve

end module solve_problem

program test_solve_problem

use solve_problem

implicit none

integer, parameter :: nums1(4) = [1, 1, 3, 2]
integer, parameter :: nums2(2) = [2, 3]
integer, parameter :: nums3(1) = [3]
integer :: result(size(nums1))

result = solve(nums1, nums2, nums3)

write (*,*) "Result: ", result

end program test_solve_problem
🌐 Data from online sources 
def largest_odd(num: str) -> str:
    for i in range(len(num)):
        if (int(num[i]) % 2) == 1:
            return num[:i + 1]
    return ""

The algorithm used iterates through the input string num. For each character, it checks if it's an odd digit (by subtracting '0' and mod-ing 2 to check if it's equal to 1).

If an odd digit is found, it returns the substring from the beginning of the string to the odd digit (inclusive) as the largest odd integer substring.

If no odd digit is found, it returns an empty string.

🌐 Data from online sources 
std::string largestOdd(std::string num) {
    for (int i = 0; i < num.length(); ++i) {
        if ((num[i] - '0') % 2 == 1) {
            return num.substr(0, i + 1);
        }
    }
    return "";
}

The algorithm used iterates through the input string num. For each character, it checks if it's an odd digit (by subtracting '0' and mod-ing 2 to check if it's equal to 1).

If an odd digit is found, it returns the substring from the beginning of the string to the odd digit (inclusive) as the largest odd integer substring.

If no odd digit is found, it returns an empty string.