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SortColors.java
64 lines (61 loc) · 1.95 KB
/
SortColors.java
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package array;
/**
* Created by gouthamvidyapradhan on 06/08/2017. Given an array with n objects colored red, white or
* blue, sort them so that objects of the same color are adjacent, with the colors in the order red,
* white and blue.
*
* <p>Here, we will use the integers 0, 1, and 2 to represent the color red, white, and blue
* respectively.
*
* <p>Note: You are not suppose to use the library's sort function for this problem.
*
* <p>Follow up: A rather straight forward solution is a two-pass algorithm using counting sort.
* First, iterate the array counting number of 0's, 1's, and 2's, then overwrite array with total
* number of 0's, then 1's and followed by 2's.
*
* <p>Could you come up with an one-pass algorithm using only constant space?
*
* <p>Solution: The below solution works with one pass. The basic idea is to keep track of start and
* end index of contiguous 1s and push the 0s to left of 1s and 2 to right of 1s.
*/
public class SortColors {
/**
* Main method
*
* @param args
* @throws Exception
*/
public static void main(String[] args) throws Exception {
int[] nums = {2, 1, 0, 0, 1};
new SortColors().sortColors(nums);
for (int i : nums) System.out.println(i);
}
public void sortColors(int[] nums) {
int s = nums[0]; // save the first index value
nums[0] = 1; // overwrite with 1
int l = 0, r = 0; // left and right index indicating the start and end index of 1s
for (int i = 1; i < nums.length; i++) {
switch (nums[i]) {
case 0:
nums[l] = 0;
nums[r + 1] = 1;
if (r + 1 != i) {
nums[i] = 2;
}
l++;
r++;
break;
case 1:
nums[r + 1] = 1;
if (r + 1 != i) {
nums[i] = 2;
}
r++;
break;
}
}
// replace the initial overwritten value with the original value
if (s == 0) nums[l] = 0;
else if (s == 2) nums[r] = 2;
}
}