I need to move all 0's in an array to the end of the array.
Example: [1, 10, 0, 5, 7] should result in [1, 10, 5, 7, 0].
I am open to doing a reverse loop or a regular loop.
I cannot create a new array.
Here is what I have so far:
for (int i = arr.length; i <= 0; --i) {
if (arr[i] != 0) {
arr[i] = arr.length - 1;
}
}
Thanks!
Create an array that is the same size as the initial array you need to remove 0s from. Iterate over the original array and add each element to the new array provided it is not 0. When you encounter a 0, count it. Now, when you've reached the end of the first array, simply add the counted number of 0s to the end of the array. And, even simpler, since Java initializes arrays to 0, you can forget about adding the zeroes at the end.
Edit
Since you have added the additional constraint of not being able to create a new array, we need to take a slightly different approach than the one I've suggested above.
I assume the array needs to remain in the same order as it was before the 0s were moved to the end. If this is not the case there is another trivial solution as detailed in Brads answer: initialize a "last zero" index to the last element of the array and then iterate backwards swapping any zeros with the index of the last zero which is decremented each time you perform a swap or see a zero.
To move the 0s to the end without duplicating the array and keeping the elements in the proper order, you can do exactly as I've suggested without duplicating the array but keeping two indices over the same array.
Start with two indices over the array. Instead of copying the element to the new array if it is not zero, leave it where it is and increment both indices. When you reach a zero, increment only one index. Now, if the two indices are not the same, and you are not looking at a 0, swap current element the location of the index that has fallen behind (due to encountered 0s). In both cases, increment the other index provided the current element is not 0.
It will look something like this:
int max = arr.length;
for (int i = 0, int j = 0; j < max; j++) {
if (arr[j] != 0) {
if (i < j) {
swap(arr, i, j);
}
i++
}
}
Running this on:
{ 1, 2, 0, 0, 0, 3, 4, 0, 5, 0 }
yeilds:
{ 1, 2, 3, 4, 5, 0, 0, 0, 0, 0 }
I made a fully working version for anyone who's curious.
Two choices come to mind
Create a new array of the same size, then Iterate over your current array and only populate the new array with values. Then fill the remaining entries in the new array with "zeros"
Without creating a new array you can iterate over your current array backwards and when you encounter a "zero" swap it with the last element of your array. You'll need to keep a count of the number of "zero" elements swapped so that when you swap for a second time, you swap with the last-1 element, and so forth.
[Edit] following request for clarification
public class MyClass {
public static void main(String[] args) {
int[] elements = new int[] {1,2,3,4,0,5,6,0,7,8,9};
int swapCount = 0;
int lastIndex = elements.length-1;
for(int i = lastIndex-1; i >=0; i--) { // skip the very last element
if(elements[i] == 0) {
elements[i] = elements[lastIndex-swapCount];
elements[lastIndex-swapCount] = 0;
swapCount++;
}
}
for(int i : elements) {
System.out.print(i + ", ");
}
}
}
Gives you...
1, 2, 3, 4, 8, 5, 6, 9, 7, 0, 0,
Basic solution is to establish an inductive hypothesis that the subarray can be kept solved. Then extend the subarray by one element and maintain the hypothesis. In that case there are two branches - if next element is zero, do nothing. If next element is non-zero, swap it with the first zero in the row.
Anyway, the solution (in C# though) after this idea is optimized looks like this:
void MoveZeros(int[] a)
{
int i = 0;
for (int j = 0; j < a.Length; j++)
if (a[j] != 0)
a[i++] = a[j];
while (i < a.Length)
a[i++] = 0;
}
There is a bit of thinking that leads to this solution, starting from the inductive solution which can be formally proven correct. If you're interested, the whole analysis is here: Moving Zero Values to the End of the Array
var size = 10;
var elemnts = [0, 0, 1, 4, 5, 0,-1];
var pos = 0;
for (var i = 0; i < elemnts.length; i++) {
if (elemnts[i] != 0) {
elemnts[pos] = elemnts[i];
pos++;
console.log(elemnts[i]);
}
}
for (var i = pos; i < elemnts.length; i++) {
elemnts[pos++] = 0;
console.log(elemnts[pos]);
}
int arrNew[] = new int[arr.length];
int index = 0;
for(int i=0;i<arr.length;i++){
if(arr[i]!=0){
arrNew[index]=arr[i];
index++;
}
}
Since the array of int's is initialized to zero(according to the language spec). this will have the effect you want, and will move everything else up sequentially.
Edit: Based on your edit that you cannot use a new array this answer doesnt cover your requirements. You would instead need to check for a zero(starting at the end of the array and working to the start) and swap with the last element of the array and then decrease the index of your last-nonzero element that you would then swap with next. Ex:
int lastZero = arr.length - 1;
if(arr[i] == 0){
//perform swap and decrement lastZero by 1 I will leave this part up to you
}
Reimplemented this in Python:
Pythonic way:
lst = [ 1, 2, 0, 0, 0, 3, 4, 0, 5, 0 ]
for i, val in enumerate(lst):
if lst[i] == 0:
lst.pop(i)
lst.append(0)
print("{}".format(lst))
@dcow's implementation in python:
lst = [ 1, 2, 0, 0, 0, 3, 4, 0, 5, 0 ]
i = 0 # init the index value
for j in range(len(lst)): # using the length of list as the range
if lst[j] != 0:
if i < j:
lst[i], lst[j] = lst[j], lst[i] # swap the 2 elems.
i += 1
print("{}".format(lst))
/// <summary>
/// From a given array send al zeros to the end (C# solution)
/// </summary>
/// <param name="numbersArray">The array of numbers</param>
/// <returns>Array with zeros at the end</returns>
public static int[] SendZerosToEnd(int[] numbersArray)
{
// Edge case if the array is null or is not long enough then we do
// something in this case we return the same array with no changes
// You can always decide to return an exception as well
if (numbersArray == null ||
numbersArray.Length < 2)
{
return numbersArray;
}
// We keep track of our second pointer and the last element index
int secondIndex = numbersArray.Length - 2,
lastIndex = secondIndex;
for (int i = secondIndex; i >= 0; i--)
{
secondIndex = i;
if (numbersArray[i] == 0)
{
// While our pointer reaches the end or the next element
// is a zero we swap elements
while (secondIndex != lastIndex ||
numbersArray[secondIndex + 1] != 0)
{
numbersArray[secondIndex] = numbersArray[secondIndex + 1];
numbersArray[secondIndex + 1] = 0;
++secondIndex;
}
// This solution is like having two pointers
// Also if we look at the solution you do not pass more than
// 2 times actual array will be resume as O(N) complexity
}
}
// We return the same array with no new one created
return numbersArray;
}
In case if question adds following condition.
Then following implementation will work fine.
Steps to be followed :
public static void main(String args[]) { int[] array = { 1, 0, 3, 0, 0, 4, 0, 6, 0, 9 }; // Maintaining count of non zero elements int count = -1; // Iterating through array and copying non zero elements in front of array. for (int i = 0; i < array.length; i++) { if (array[i] != 0) array[++count] = array[i]; } // Replacing end elements with zero while (count < array.length - 1) array[++count] = 0; for (int i = 0; i < array.length; i++) { System.out.print(array[i] + " "); } }
Time complexity = O(n), Space Complexity = O(1)
import java.util.Scanner;
public class ShiftZeroesToBack {
int[] array;
void shiftZeroes(int[] array) {
int previousK = 0;
int firstTime = 0;
for(int k = 0; k < array.length - 1; k++) {
if(array[k] == 0 && array[k + 1] != 0 && firstTime != 0) {
int temp = array[previousK];
array[previousK] = array[k + 1];
array[k + 1] = temp;
previousK = previousK + 1;
continue;
}
if(array[k] == 0 && array[k + 1] != 0) {
int temp = array[k];
array[k] = array[k + 1];
array[k + 1] = temp;
continue;
}
if(array[k] == 0 && array[k + 1] == 0) {
if(firstTime == 0) {
previousK = k;
firstTime = 1;
}
}
}
}
int[] input(Scanner scanner, int size) {
array = new int[size];
for(int i = 0; i < size; i++) {
array[i] = scanner.nextInt();
}
return array;
}
void print() {
System.out.println();
for(int i = 0; i < array.length; i++) {
System.out.print(array[i] + " ");
}
System.out.println();
}
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
ShiftZeroesToBack sztb = new ShiftZeroesToBack();
System.out.print("Enter Size of Array\t");
int size = scanner.nextInt();
int[] input = sztb.input(scanner, size);
sztb.shiftZeroes(input);
sztb.print();
}
}
let's say we have an array
[5,4,0,0,6,7,0,8,9]
Let's assume we have array elements between 0-100, now our goal is to move all 0's at the end of the array. now hold 0 from the array and check it with non zero element if any non zero element found swap with that element and so on, at the end of the loop we will find the solution.
here is the code
for (int i = 0; i < outputArr.length; i++)
{
for (int j = i+1; j < outputArr.length; j++)
{
if(outputArr[i]==0 && outputArr[j]!=0){
int temp = outputArr[i];
outputArr[i] = outputArr[j];
outputArr[j] = temp;
}
}
print outputArr[i]....
}
output:
[5,4,6,7,8,9,0,0,0]
Here is how i have implemented this. Time complexity: O(n) and Space Complexity: O(1) Below is the code snippet:
int arr[] = {0, 1, 0, 0, 2, 3, 0, 4, 0};
int i=0, k = 0;
int n = sizeof(arr)/sizeof(arr[0]);
for(i = 0; i<n; i++)
{
if(arr[i]==0)
continue;
arr[k++]=arr[i];
}
for(i=k;i<n;i++)
{
arr[i]=0;
}
output: {1, 2, 3, 4, 0, 0, 0, 0, 0}
Explanation: using another variable k to hold index location, non-zero elements are shifted to the front while maintaining the order. After traversing the array, non-zero elements are shifted to the front of array and another loop starting from k is used to override the remaining positions with zeros.
This is my code with 2 for loops:
int [] arr = {0, 4, 2, 0, 0, 1, 0, 1, 5, 0, 9,};
int temp;
for (int i = 0; i < arr.length; i++)
{
if (arr[i] == 0)
{
for (int j = i + 1; j < arr.length; j++)
{
if (arr[j] != 0)
{
temp = arr[j];
arr[j] = arr[i];
arr[i] = temp;
break;
}
}
}
}
System.out.println(Arrays.toString(arr));
output: [4, 2, 1, 1, 5, 9, 0, 0, 0, 0, 0]
public static void main(String[] args) {
Integer a[] = {10,0,0,0,6,0,0,0,70,6,7,8,0,4,0};
int flag = 0;int count =0;
for(int i=0;i<a.length;i++) {
if(a[i]==0 ) {
flag=1;
count++;
}else if(a[i]!=0) {
flag=0;
}
if(flag==0 && count>0 && a[i]!=0) {
a[i-count] = a[i];
a[i]=0;
}
}
}
This is One method of Moving the zeroes to the end of the array.
public class SeparateZeroes1 {
public static void main(String[] args) {
int[] a = {0,1,0,3,0,0,345,12,0,13};
movezeroes(a);
}
static void movezeroes(int[] a) {
int lastNonZeroIndex = 0;
// If the current element is not 0, then we need to
// append it just in front of last non 0 element we found.
for (int i = 0; i < a.length; i++) {
if (a[i] != 0 ) {
a[lastNonZeroIndex++] = a[i];
}
}//for
// We just need to fill remaining array with 0's.
for (int i = lastNonZeroIndex; i < a.length; i++) {
a[i] = 0;
}
System.out.println( lastNonZeroIndex );
System.out.println(Arrays.toString(a));
}
}
This is very simple in Python. We will do it with list comprehension
a =[0,1,0,3,0,0,345,12,0,13]
def fix(a):
return ([x for x in a if x != 0] + [x for x in a if x ==0])
print(fix(a))
int[] nums = { 3, 1, 2, 5, 4, 6, 3, 2, 1, 6, 7, 9, 3, 8, 0, 4, 2, 4, 6, 4 };
List<Integer> list1 = new ArrayList<>();
List<Integer> list2 = new ArrayList<>();
for (int i = 0; i < nums.length; i++) {
if (nums[i] == 3) {
list1.add(nums[i]);
} else if (nums[i] != 3) {
list2.add(nums[i]);
}
}
List<Integer> finalList = new ArrayList<>(list2);
finalList.addAll(list1);
}
}
int[] nums = {3,1,2,5,4,6,3,2,1,6,7,9,3,8,0,4,2,4,6,4};
int i = 0;
for(int j = 0, s = nums.length; j < s;) {
if(nums[j] == 3)
j++;
else {
int temp = nums[i];
nums[i] = nums[j];``
nums[j] = temp;
i ++;
j ++;
}
}
For Integer array it can be as simple as
Integer[] numbers = { 1, 10, 0, 5, 7 };
Arrays.sort(numbers, Comparator.comparing(n -> n == 0));
For int array :
int[] numbers = { 1, 10, 0, 5, 7 };
numbers = IntStream.of(numbers).boxed()
.sorted(Comparator.comparing(n -> n == 0))
.mapToInt(i->i).toArray();
Output of both:
[1, 10, 5, 7, 0]
Lets say, you have an array like
int a[] = {0,3,0,4,5,6,0};
Then you can sort it like,
for(int i=0;i<a.length;i++){
for(int j=i+1;j<a.length;j++){
if(a[i]==0 && a[j]!=0){
a[i]==a[j];
a[j]=0;
}
}
}
public void test1(){
int [] arr = {0,1,0,4,0,3,2};
System.out.println("Lenght of array is " + arr.length);
for(int i=0; i < arr.length; i++){
for(int j=0; j < arr.length - i - 1; j++){
if(arr[j] == 0){
int temp = arr[j];
arr[j] = arr[arr.length - i - 1];
arr[arr.length - i - 1] = temp;
}
}
}
for(int x = 0; x < arr.length; x++){
System.out.println(arr[x]);
}
}
a = [ 1, 2, 0, 0, 0, 3, 4, 0, 5, 0 ]
count = 0
for i in range(len(a)):
if a[i] != 0:
a[count], a[i] = a[i], a[count]
count += 1
print(a)
#op [1, 2, 3, 4, 5, 0, 0, 0, 0, 0]
