Move Zeroes

🏠 ⬅️ ➡️

Given an integer array nums, move all 0's to the end of it while maintaining the relative order of the non-zero elements.

Note that you must do this in-place without making a copy of the array.

Example 1:

Input: nums = [0,1,0,3,12] Output: [1,3,12,0,0]

Example 2:

Input: nums = [0] Output: [0]

Constraints:

  • 1 <= nums.length <= 104
  • -231 <= nums[i] <= 231 - 1

Follow up: Could you minimize the total number of operations done?

Note: This problem is from LeetCode.
✅ Compiled
✅ Executed
✅ Correct 
program main
    implicit none

    integer, parameter :: n = 5
    integer :: nums(n) = [0, 1, 0, 3, 12]
    integer :: i, j

    call move_zeros(nums)

    write (*, '(A)') 'Expected:'
    write (*, '(A)') '[1, 3, 12, 0, 0]'
    write (*, '(A)') 'Actual:'
    write (*, '(A)') nums

contains

subroutine move_zeros(nums)
    implicit none
    integer, intent(inout) :: nums(:)
    integer :: i, j

    i = 1
    j = 1
    do while (j <= size(nums))
        if (nums(j) /= 0) then
            nums(i) = nums(j)
            i = i + 1
        end if
        j = j + 1
    end do
end subroutine move_zeros

end program main
❌ Compiled
❌ Executed
❌ Correct 
module move_zeros
contains
    subroutine move_zeros_solution(nums)
        implicit none
        integer, dimension(:), intent(inout) :: nums
        integer :: i, j, n

        n = size(nums)
        i = 1
        j = 1
        do while (i <= n)
            if (nums(i) /= 0) then
                nums(j) = nums(i)
                j = j + 1
            end if
            i = i + 1
        end do

        do while (j <= n)
            nums(j) = 0
            j = j + 1
        end do
    end subroutine move_zeros_solution
end module move_zeros

program test_move_zeros
    use move_zeros
    implicit none
    integer, dimension(5) :: nums

    nums = [0, 1, 0, 3, 12]
    call move_zeros_solution(nums)
    write (*,*) nums

    nums = [0]
    call move_zeros_solution(nums)
    write (*,*) nums
end program test_move_zeros
🌐 Data from online sources 
def moveZeroes(nums):
    index = 0
    for i in range(len(nums)):
        if nums[i] != 0:
            nums[index] = nums[i]
            index += 1
    for i in range(index, len(nums)):
        nums[i] = 0

In each language, the algorithm starts by initializing an index variable with the value of 0. Next, it loops through the input array nums. If the current element is not zero, it assigns the value of the current element to the position indicated by the index and then increments the index. After the first loop, all non-zero elements are placed before the zero elements in the array.

Finally, in the second loop, it assigns the value of 0 to all remaining positions in the array from the current index to the end of the array.

This algorithm maintains the relative order of non-zero elements as they are moved to the left at the beginning of the array, and ensures that zeros are moved to the end of the array. The algorithm also modifies the given array in-place as required.

🌐 Data from online sources 
void moveZeroes(vector<int>& nums) {
    int index = 0;
    for (int i = 0; i < nums.size(); i++) {
        if (nums[i] != 0) {
            nums[index++] = nums[i];
        }
    }
    for (int i = index; i < nums.size(); i++) {
        nums[i] = 0;
    }
}

In each language, the algorithm starts by initializing an index variable with the value of 0. Next, it loops through the input array nums. If the current element is not zero, it assigns the value of the current element to the position indicated by the index and then increments the index. After the first loop, all non-zero elements are placed before the zero elements in the array.

Finally, in the second loop, it assigns the value of 0 to all remaining positions in the array from the current index to the end of the array.

This algorithm maintains the relative order of non-zero elements as they are moved to the left at the beginning of the array, and ensures that zeros are moved to the end of the array. The algorithm also modifies the given array in-place as required.