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Move Zeroes

time O(n)space O(1)

Use a slow pointer to mark the next open slot for a non-zero value, and a fast pointer to scan ahead. Whenever fast finds a non-zero, swapping it into slow's slot never disturbs anything already placed — the elements before slow are already confirmed non-zero and in order — so one pass is enough to push every non-zero to the front while zeroes naturally settle at the back.

Details

Animated on: Given nums = [0, 1, 0, 3, 12], move all zeroes to the end while maintaining the relative order of the non-zero elements. Answer: [1, 3, 12, 0, 0].

Problem:Move ZeroesEasy

Two Pointers

step 1 / 7

0
[0]
1
[1]
0
[2]
3
[3]
12
[4]

slow = 0, fast = 0: slow marks the next slot for a non-zero; fast will scan every element.

  • slow=0· arr[0] = 0
  • fast=0· arr[0] = 0
Pseudocode
1FUNCTION moveZeroes(nums):
2  slow = 0
3  FOR fast from 0 to n-1:
4    IF nums[fast] != 0:
5      swap nums[slow] and nums[fast]
6      slow = slow + 1

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Every step in words

Show all 7 steps
  1. slow = 0, fast = 0: slow marks the next slot for a non-zero; fast will scan every element.
  2. fast=0: nums[0]=0 is zero, skip. fast=1: nums[1]=1 is non-zero — swap nums[slow=0] and nums[fast=1]: [1,0,0,3,12]. slow advances to 1.
  3. fast=2: nums[2]=0 is zero. Do nothing; just advance fast. slow stays at 1. Array unchanged: [1,0,0,3,12].
  4. fast=3: nums[3]=3 is non-zero — swap nums[slow=1] and nums[fast=3]: [1,3,0,0,12]. slow advances to 2.
  5. fast=4: nums[4]=12 is non-zero — swap nums[slow=2] and nums[fast=4]: [1,3,12,0,0]. slow advances to 3.
  6. fast reaches n=5, the loop ends. All non-zero elements [1,3,12] are at the front in their original order; the two zeroes settled at the back.
  7. Final: [1, 3, 12, 0, 0].

Now solve it yourself

Watching is the first half. The second is writing it without the animation.

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