Given an array of n numbers. The problem is to move all the 0’s to the end of the array while maintaining the order of the other elements. Only single traversal of the array is required.
Examples:
Input : arr[] = {1, 2, 0, 0, 0, 3, 6}
Output : 1 2 3 6 0 0 0
Input: arr[] = {0, 1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0, 9}
Output: 1 9 8 4 2 7 6 9 0 0 0 0 0
Algorithm:
moveZerosToEnd(arr, n)
Initialize count = 0
for i = 0 to n-1
if (arr[i] != 0) then
swap(arr[count++], arr[i])
Javascript
<script> // JavaScript implementation to move all zeroes at // the end of array // function to move all zeroes at // the end of array function moveZerosToEnd(arr, n) { // Count of non-zero elements let count = 0; // Traverse the array. If arr[i] is non-zero, then // swap the element at index 'count' with the // element at index 'i' for (let i = 0; i < n; i++) if (arr[i] != 0) { temp = arr[count]; arr[count] = arr[i]; arr[i] = temp; count = count + 1; } } // function to print the array elements function printArray(arr, n) { for (let i = 0; i < n; i++) document.write(arr[i] + " "); } // Driver program to test above let arr = [ 0, 1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0, 9 ]; let n = arr.length; document.write("Original array: "); printArray(arr, n); moveZerosToEnd(arr, n); document.write("<br>" + "Modified array: "); printArray(arr, n); //This code is contributed by Mayank Tyagi </script> |
Output:
Original array: 0 1 9 8 4 0 0 2 7 0 6 0 9 Modified array: 1 9 8 4 2 7 6 9 0 0 0 0 0
Time Complexity: O(n).
Auxiliary Space: O(1).
Please refer complete article on Move all zeroes to end of array | Set-2 (Using single traversal) for more details!
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