Obsidian Notes

Sort Colors

2 min read

Description

Sort Colors
Given an array nums 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.

We will use the integers 0, 1, and 2 to represent the color red, white, and blue, respectively.

You must solve this problem without using the library’s sort function.

Example 1:
Input: nums = [2,0,2,1,1,0]
Output: [0,0,1,1,2,2]

Example 2:
Input: nums = [2,0,1]
Output: [0,1,2]

Constraints:

  • n == nums.length
  • 1 <= n <= 300
  • nums[i] is either 0, 1, or 2.

Follow up: Could you come up with a one-pass algorithm using only constant extra space?

Approach - Dutch flag (3 pointers)

  • We have 3 options if it is 0 then swap with front if 1 then let it be there if 2 then swap with last
  • It is basically a question asking us to sort the array but simple sorts will at max give us nlogn and not n which we get here
  • Time: O(n) Space: O(1)
class Solution {
    public void sortColors(int[] nums) {
        int low = 0, mid = 0, high = nums.length - 1;
        while (mid <= high) {
            if (nums[mid] == 0)
                swap(nums, low++, mid++);
            else if (nums[mid] == 1)
                mid++;
            else
                swap(nums, mid, high--);
        }
    }
 
    void swap(int[] nums, int a, int b) {
        int t = nums[b];
        nums[b] = nums[a];
        nums[a] = t;
    }
}

Here is the ultra-simple rule to remember:

  • Swapping with low (for 0): You swap with an area you’ve already processed. You know a 1 is coming back to mid, so it’s safe to move mid++.
  • Swapping with high (for 2): You swap with unexplored territory. You don’t know what value just landed at mid, so mid must stay put to inspect it.

Think of it like inspecting mail:

  • Found a 0? Throw it to the left stack (low). You get a known, checked item (1) in return. Move to the next item (mid++).
  • Found a 2? Throw it to the right stack (high). You get an unopened package in return. Inspect it right now—don’t move forward (mid stays)!

The reason we use m <= r (instead of m < r) comes down to one simple rule:

The r pointer points to an element that hasn’t been checked yet.

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