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. 

// 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);
console.log("Array after pushing all zeros to end of array : ");
console.log(arr.join(" "));

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

Complexity Analysis:

• 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!

Move all zeros to end

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