Move Zeroes To End

Posted: 3 Dec, 2019
Difficulty: Moderate


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Given an unsorted array of integers, you have to move the array elements in a way such that all the zeroes are transferred to the end, and all the non-zero elements are moved to the front. The non-zero elements must be ordered in their order of appearance.

For example, if the input array is: [0, 1, -2, 3, 4, 0, 5, -27, 9, 0], then the output array must be:

[1, -2, 3, 4, 5, -27, 9, 0, 0, 0].

Expected Complexity: Try doing it in O(n) time complexity and O(1) space complexity. Here, ‘n’ is the size of the array.

Input format :

The first line of input contains a single integer ‘T’ representing the number of test cases.      

The first line of each test case contains a single integer ‘N’ representing the size of the array. The second line of each test case contains ‘N’ integers representing the elements of the array.

Output Format :

For each test case, modify the input array and print the output in a new line

Note :

You don’t need to print anything. It has already been taken care of. Just implement the given function.

Constraints :

1 <= T <= 50 
1 <= N <= 10^6
-10000 <= A[i] <= 10000

Where ‘T’ is the number of test cases, ‘N’ is the size of the array, A[i] is the value of the element present at the ith index.
Time Limit:1sec
Approach 1


We can create a new array of integers. As soon as we find a non-zero element, we can directly insert it into our new array. After that, we can insert all the left zero’s. 

We can easily calculate the number of left zeroes as :

Size of original array = size of new array + number of zero’s (since we only took non-zero elements.

Algorithm : 

  • Initialise a non-zero pointer with 0 value.
  • Start traversing the array until we reach the end,
    • If a non-zero element is found, insert it in new array. Else,
    • If a zero element is found, just continue.
  • After we reach the end, total number of zero’s will be original array size - new array size.
  • Insert these many zeros into new array.


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