### Problem Statement

Given an array with

*n*objects colored red, white or blue, sort them**in-place**so that objects of the same color are adjacent, with the colors in the order red, white and blue.
Here, we will use the integers 0, 1, and 2 to represent the color red, white, and blue respectively.

**Note:**You are not suppose to use the library's sort function for this problem.

**Example:**

Input:[2,0,2,1,1,0]Output:[0,0,1,1,2,2]

**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. - Could you come up with a one-pass algorithm using only constant space?

### Video Tutorial

You can find the detailed video tutorial here### Thought Process

This is the classic Dutch National Flag Problem.Two pass with constant space is easy with count sort. (Reminds me of Radix Sort somehow) An extension would be what happens if there are K elements (K >=3)

If do it in one pass, then the idea must involve pointers and swap. Start from left to right, if 1, pass; if 0, move to the left; if 2, move to the right

### Solutions

#### Two pass

Time Complexity: O(N), where N is the array size, N + 3*N = 4N = O(N)Space Complexity: O(1) since only need 3 elements in this case, still considered constant

#### One Pass

Time Complexity: O(N), where N is the array sizeSpace Complexity: O(1)

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