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Move Zeros to the Left

editor-page-cover

Problem Statement

Given an integer array, move all elements that are 0 to the left while maintaining the order of other elements in the array. The array has to be modified in-place.

Let’s look at the following integer array:

widget

After moving all zeros to the left, the array should like this:

widget

Remember to maintain the order of non-zero elements


Hint

  • Use the concept of reader/writer indexes

Try it yourself

void move_zeros_to_left(int A[], int n) {
//TODO: Write - Your - Code
}

Solution

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.

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