Can someone tell me the time complexity of this code?

[sscrock1234]

class Solution {
    public void moveZeroes(int[] nums) {
        int left=0;
        int right=0;

        for(int i=0;i<nums.length;i++){
            if(nums[i]==0){
                left=i;
                right=i+1;
                break;
            }
        }

        while(right<nums.length){
            if(nums[right]!=0 && nums[left]==0){
                int temp=nums[right];
                nums[right]=nums[left];
                nums[left]=temp;
                left++;
                right++;
            }
            else{
                right++;   
            }
        }
    }
}

 

[FResher]

public void moveZeroes(int[] nums) {

  int temp[]; // use a temp array
  int i = 0; // iterator index
  int j=0; // iterator index for temp array

  while(i<nums.length){ // fetching the Array with values

    if(nums[i] == 0){ // if the value is 0 : skip the indexes by incrementing
        i++;
        continue; // jump to the next lap of the loop, without executing the lines after
    }

        temp[j] = nums[i]; // allocating the value in nums at index i.
        i++; // incrementing i for the next value
        j++;
  }

  nums[] = temp[] ; // old array is replace by new array without 0.

}

It's a solution too, maybe the syntax and statement are to be fix.

 

[WhiteCube]

class Solution {
    public void moveZeroes(int[] nums) {
        int left=0;
        int right=0;

        for(int i=0;i<nums.length;i++){
            if(nums[i]==0){
                left=i;
                right=i+1;
                break;
            }
        }

        while(right<nums.length){
            if(nums[right]!=0 && nums[left]==0){
                int temp=nums[right];
                nums[right]=nums[left];
                nums[left]=temp;
                left++;
                right++;
            }
            else{
                right++; 
            }
        }
    }
}

The number of iterations is equal to the length of the array, so that would be linear time. Double the array, double the time, triple the array, triple the time etc.
I don't know the official "big O" notation for this case, I do not study computer science.

 

[raymbs]

class Solution {
    public void moveZeroes(int[] nums) {
        int left=0;
        int right=0;

        for(int i=0;i<nums.length;i++){
            if(nums[i]==0){
                left=i;
                right=i+1;
                break;
            }
        }

        while(right<nums.length){
            if(nums[right]!=0 && nums[left]==0){
                int temp=nums[right];
                nums[right]=nums[left];
                nums[left]=temp;
                left++;
                right++;
            }
            else{
                right++;  
            }
        }
    }
}

O(n) linear

 

[raymbs]

The number of iterations is equal to the length of the array, so that would be linear time. Double the array, double the time, triple the array, triple the time etc.
I don't know the official "big O" notation for this case, I do not study computer science.i

 

[raymbs]

it's Big O notation will be O(n) linear as you mentioned

 

[alexcray]

class Solution {
    public void moveZeroes(int[] nums) {
        int left=0;
        int right=0;

        for(int i=0;i<nums.length;i++){
            if(nums[i]==0){
                left=i;
                right=i+1;
                break;
            }
        }

        while(right<nums.length){
            if(nums[right]!=0 && nums[left]==0){
                int temp=nums[right];
                nums[right]=nums[left];
                nums[left]=temp;
                left++;
                right++;
            }
            else{
                right++;  
            }
        }
    }
}

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