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:                                                          Example 2: 

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

Output: [1,3,12,0,0]                                           Output: [0]   

 First, i used this     (Two-Pointer Approach)

class Solution {

    public void moveZeroes(int[] nums) {

      int n= nums.length;int j=0;int k=0;

     

      for(int i=0;i<n;i++){

        if(nums[i]==0){

        k++;

        }

        else{

            nums[j]=nums[i];

            j++;

        }

       

      }

      while(k>0){

        nums[n-1]=0;

       n--;

       k--;

      }

    }

}

But this is not optimised.

T.C-o(n^2) S.C-o(1)

Most optimised (Swap Approach) -

class Solution {

    public void moveZeroes(int[] nums) {

      int n= nums.length;int j=0;

     

      for(int i=0;i<n;i++){

        if(nums[i]!=0){

           if(i!=j){

            int temp=nums[i];

            nums[i]=nums[j];

            nums[j]=temp;

           }

           j++;

        }

       

      }

     

    }

}

Let’s take the input:
[0, 1, 0, 3, 12]

1️⃣ i = 0, j = 0 → nums[i] == 0, so do nothing.
2️⃣ i = 1, j = 0 → nums[i] == 1 (non-zero), swap nums[0] ↔ nums[1] → [1, 0, 0, 3, 12]

• Move j to 1
  3️⃣ i = 2, j = 1 → nums[i] == 0, so do nothing.
  4️⃣ i = 3, j = 1 → nums[i] == 3 (non-zero), swap nums[1] ↔ nums[3] → [1, 3, 0, 0, 12]
• Move j to 2
  5️⃣ i = 4, j = 2 → nums[i] == 12 (non-zero), swap nums[2] ↔ nums[4] → [1, 3, 12, 0, 0]
• Move j to 3

💡 Final Output: [1, 3, 12, 0, 0]