Python3 Program to Move all zeroes to end of array (original) (raw)

Given an array of random numbers, Push all the zero's of a given array to the end of the array. For example, if the given arrays is {1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0}, it should be changed to {1, 9, 8, 4, 2, 7, 6, 0, 0, 0, 0}. The order of all other elements should be same. Expected time complexity is O(n) and extra space is O(1).
**Example:

Input : arr[] = {1, 2, 0, 4, 3, 0, 5, 0};
Output : arr[] = {1, 2, 4, 3, 5, 0, 0};

Input : arr[] = {1, 2, 0, 0, 0, 3, 6};
Output : arr[] = {1, 2, 3, 6, 0, 0, 0};

There can be many ways to solve this problem. Following is a simple and interesting way to solve this problem.
Traverse the given array 'arr' from left to right. While traversing, maintain count of non-zero elements in array. Let the count be 'count'. For every non-zero element arr[i], put the element at 'arr[count]' and increment 'count'. After complete traversal, all non-zero elements have already been shifted to front end and 'count' is set as index of first 0. Now all we need to do is that run a loop which makes all elements zero from 'count' till end of the array.
Below is the implementation of the above approach.

Python3 `

Python3 code to move all zeroes

at the end of array

Function which pushes all

zeros to end of an array.

def pushZerosToEnd(arr, n): count = 0 # Count of non-zero elements

# Traverse the array. If element 
# encountered is non-zero, then
# replace the element at index
# 'count' with this element
for i in range(n):
    if arr[i] != 0:
        
        # here count is incremented
        arr[count] = arr[i]
        count+=1

# Now all non-zero elements have been
# shifted to front and 'count' is set
# as index of first 0. Make all 
# elements 0 from count to end.
while count < n:
    arr[count] = 0
    count += 1
    

Driver code

arr = [1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0, 9] n = len(arr) pushZerosToEnd(arr, n) print("Array after pushing all zeros to end of array:") print(arr)

This code is contributed by "Abhishek Sharma 44"

`

Output

Array after pushing all zeros to end of array: [1, 9, 8, 4, 2, 7, 6, 9, 0, 0, 0, 0]

**Time Complexity: O(n) where n is number of elements in input array.
**Auxiliary Space: O(1)

**Method 2: Using two pointer

• Initialize a pointer variable 'left' to 0 and another pointer variable 'right' to n-1, where n is the length of the array.
• Repeat the following steps until the value of 'left' is less than or equal to 'right':
  a. If the value at index 'left' is non-zero, increment the value of 'left'.
  b. If the value at index 'right' is zero, decrement the value of 'right'.
  c. If the value at index 'left' is zero and the value at index 'right' is non-zero, swap the values at these indices and then increment the value of 'left' and decrement the value of 'right'.
• The above steps will move all the zeros to the end of the array and all the non-zero values to the front of the array. Python `

def pushZerosToEnd(arr, n): # Initialize two pointers - 'left' and 'right' - pointing to the first and last elements of the array respectively left = 0 right = n - 1

# Loop until 'left' pointer crosses 'right' pointer
while left <= right:
    # If the element at the 'left' pointer is non-zero, increment 'left'
    if arr[left] != 0:
        left += 1
    # If the element at the 'right' pointer is zero, decrement 'right'
    elif arr[right] == 0:
        right -= 1
    # If the element at the 'left' pointer is zero and the element at the 'right' pointer is non-zero, swap the two elements and increment 'left' and decrement 'right'
    else:
        # Move all non-zero elements to the left of the array
        for i in range(left, right):
            if arr[i] == 0:
                arr[i], arr[i+1] = arr[i+1], arr[i]
        # Decrement the 'right' pointer after each swap
        right -= 1
      

Driver code

arr = [1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0, 9] n = len(arr) pushZerosToEnd(arr, n) print("Array after pushing all zeros to end of array:") print(arr)

`

Output

Array after pushing all zeros to end of array: [1, 9, 8, 4, 2, 7, 6, 9, 0, 0, 0, 0]

**Time complexity: The time complexity of this approach is O(n), where n is the length of the array.
**Auxiliary space: The space complexity of this approach is O(1), as it only uses two pointers and a few variables for swapping.

Please refer complete article on Move all zeroes to end of array for more details!