These are simple questions asked in Facebook interview.
Given an array and a value, remove all instances of that value in place and return the new length.
The order of elements can be changed. It doesn’t matter what you leave beyond the new length.
The idea is to keep pointer to position index and store here if we found element that is not zero. The time complexity is O(n).
def remove(nums): """ :type nums: List[int] :type val: int :rtype: int """ if not nums: return 0 i = 0 for n in nums: if n != val: nums[i] = n i += 1 return i
Given an array nums, write a function to move all 0’s to the end of it while maintaining the relative order of the non-zero elements.
For example, given nums = [0, 1, 0, 3, 12], after calling your function, nums should be [1, 3, 12, 0, 0].
You must do this in-place without making a copy of the array.
Minimize the total number of operations.
The idea is similar to the first one. Keep a pointer to indicate the position of array to store to. The time complexity is O(n).
def move_zero(nums): i = 0 for n in nums: if n != 0: nums[i] = n i += 1 while i<len(nums): nums[i] = 0 i += 1