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.
Javascript
<script> // A JavaScript program to move all zeroes at the end of array // Function which pushes all zeros to end of an array. function pushZerosToEnd(arr, n) { let 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 (let i = 0; i < n; i++) if (arr[i] != 0) arr[count++] = arr[i]; // here count is // incremented // 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; } // Driver code let arr = [1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0, 9]; let n = arr.length; pushZerosToEnd(arr, n); document.write("Array after pushing all zeros to end of array :<br>"); for (let i = 0; i < n; i++) document.write(arr[i] + " "); // This code is contributed by Surbhi Tyagi. </script> |
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!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!