## sorting 0 and 1 in Java

given an array of 0 and 1, find minimum no. of swaps to bring all 1s together

sort array of 0, 1, 2 leetcode

segregate 0s and 1s in an array in javascript

sort by color java

linear array sort java

binary sort array java

sort array in one loop

**This question already has answers here**:

These lines

while(arr[right]==1 && left<right){ right++; }

should be

while(arr[right]==1 && left<right){ right--; }

**Segregate 0s and 1s in an array,** array only once. Input array = [0, 1, 0, 1, 0, 0, 1, 1, 1, 0] Output array = [0, 0, 0, 0, 0, 1, 1, 1, 1, 1] Java code to Segregate 0s and 1s in an array. class GFG {. Unable to figure out the problem in the code It says Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 6 I have gone through it a lot of times,still unable to figure whats wrong

If it's just 0-1 array, you should count zeros and ones in it and print all the ones after all the zeros:

int[] arr = new int[]{1,0,0,1,1,0}; int zerosCount = 0; for (int e: arr) { if (e == 0) { ++zerosCount; } } for (int i = 0; i < zerosCount; ++i) { System.out.print(0); } for (int i = zerosCount; i < arr.length; ++i) { System.out.print(1); }

This approach has linear complexity.

**Sort an array of 0s, 1s and 2s,** Java program to sort an array of 0, 1 and 2. import java.io.*;. class countzot {. // Sort the input array, the array is assumed to. // have values in {0, 1, 2}. static void Actually standard off-the-shelf sorting algorithms will typically work on O(n*log(n)). You could just run through the array once adding all the values (i.e. the number of 1). Let's say you put this in count1. Then go once more over the array setting the first count1 positions to 1, and the rest to 0.

A more efficient approach (only valid for arrays which contain 0's and 1's):

int[] arr = new int[]{1, 0, 0, 1, 1, 0}; int current = 0; int total = 0; while (current < arr.legth) { total += arr[current]; current++; } current = 0; int[] sortedArr = new int[arr.length]; while (current < sortedArr.legth) { if (current < sortedArr.legth - total) { sortedArr[current] = 0; } else { sortedArr[current] = 1; } } for (int i = 0; i < sortedArr.length; i++) { System.out.print(sortedArr[i]) }

**Sorting in java for array only containing 0 and 1,** use any sorting algorithm to do it. For beginner use bubble sort (easy to understand) Refer Wiki public static void bubble_srt( int a[], int n ){ int i, j,t=0; for(i = 0; In this core java programming tutorial we will Write a Program to Sort Binary array(Array of 0 and 1 only) in java.

**What is the best way to sort an array if it contains only 0 and 1?,** [code]import java.*; import java.util.*; class testQ { public static void main(String args[]) throws Exception{ int[] arr = { 1,1,0,1,0,0,0,1,0,1,0,1,0,1,0,1,0,1}; String str Solution 2: Algorithm: Lets say array is arr [] Initialise two index variable , left=0 and right=arr.length-1. Increment left variable until you get 1’s. Decrement right variable until you get 0’s. If left < right, swap arr [left] and arr [right] In the end, you will see that you have 0’s on left side and 1’s on right side. Java code:

**Sort an array containing 0's, 1's and 2's (Dutch national flag problem ,** Given an array containing only 0's, 1's and 2's, sort the array in linear time and is similar to three-way Partitioning for Dutch national flag problem. C; Java For sorting objects in an array, we need only assume that we can compare two elements to see whether the first is bigger than, smaller than, or equal to the second. Java provides the Comparable interface for this purpose. Insertion.java implements insertion sort so that it sorts arrays of Comparable objects. Empirical analysis.

**Sort Binary Array in Linear Time,** Alternatively, we can also count number of 1's present in the array (say k) and fill last k indices in the array by 1 and all remaining indices by 0. C; Java; Python. C. At first, the full array is unknown. There are three indices – low, mid and high. Initially, the value of low = mid = 1 and high = N. If the ith element is 0 then swap the element to the low range, thus shrinking the unknown range. Similarly, if the element is 1 then keep it as it is but shrink the unknown range.

##### Comments

- The highest element should be
`5`

, as it’s 0-based. You’re trying to access`6`

, hence the error. - you should always check that
`left`

and`right`

are valid indices before using them to access an index in your array. - limit the for loop to array length-1: for(int i=0; i<arr.length - 1;i++)