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