Move Zeros to the Left

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Problem Statement Hint Try it yourself Solution Solution Explanation Runtime complexity Memory complexity Breakdown of solution

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Oct 15, 2025

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void move_zeros_to_left(int A[], int n) {
   
  if (n < 1) return;
  
  int write_index = n - 1;
  int read_index = n - 1;
  while(read_index >= 0) {
    if(A[read_index] != 0) {
      A[write_index] = A[read_index];
      write_index--;
    }
    read_index--;
  }
  while(write_index >= 0) {
    A[write_index] = 0;
    write_index--;
  }
}
int main() {
  int v[] = {1, 10, 20, 0, 59, 63, 0, 88, 0};
  int n = sizeof(v) / sizeof(v[0]);
  cout << "Original Array" <<endl;
  
  for(int x=0 ; x<n; x++) {
    cout << v[x];
    cout << ", ";
  }  
  
  move_zeros_to_left(v, n);
  
  cout << endl<< "After Moving Zeroes to Left"<< endl;
  for(int i=0 ; i<n; i++) {
    cout << v[i];
    cout << ", ";
  }  
}

Solution Explanation#

Runtime complexity#

The runtime complexity if this solution is linear, O(n).

Memory complexity#

The memory complexity of this solution is constant, O(1).

Breakdown of solution#

Keep two markers: read_index and write_index and point them to the end of the array. Let’s take a look at an overview of the algorithm:

While moving read_index towards the start of the array:

If read_index points to 0, skip.
If read_index points to a non-zero value, write the value at read_index to write_index and decrement write_index.
Assign zeros to all the values before the write_index and to the current position of write_index as well.