Python3 Program to Move all zeroes to end of array
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)
Please refer complete article on Move all zeroes to end of array for more details!
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