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Two Pointers

Sort Colors

medium

max (21)341224132;341225102;21365487109
Count: 10
abcValid triangle requires:a + b > c AND a + c > b AND b + c > a(every pair must sum to more than the third side)3511SOURCE23211SOURCE23UNREACHABLE$100$100$100$5000SRC123DST$100$100$1000SRC123DST01233141Threshold: 4Answer: 32 reachable01234231118Threshold: 2Answer: 01 reachable1102233321432263321
DESCRIPTION (inspired by Leetcode.com)

Write a function to sort a given integer array nums in-place (and without the built-in sort function), where the array contains n integers that are either 0, 1, and 2 and represent the colors red, white, and blue. Arrange the objects so that same-colored ones are adjacent, in the order of red, white, and blue (0, 1, 2).

Input:

nums = [2,1,2,0,1,0,1,0,1]

Output:

[0,0,0,1,1,1,1,2,2]

Explanation

We can understand this algorithm by looking at the invariants which hold true after each iteration:
  1. All elements to the left of the left are 0s.
  2. All elements between left and i - 1 are 1s.
  3. All elements between i and right are unsorted.
  4. All elements to the right of right are 2s.
00110120222
Let's now see how we maintain these invariants as we iterate through the array.

Sorting 0s

When nums[i] is equal to 0, invariant 2 tells us there are two possible values for left: 0 or 1.
Let's consider the case when left == 1 first. We swap i with left. This allows us to increment the left pointer to maintain invariant 1. Since we know that the new item at i is a 1, we can also increment i to maintain variant 2.
00110120222ileftrightzeroestwosunsortedones
Now let's consider what happens when left == 0, which happens when we haven't encountered any 1s yet and i == left. We swap i with left (which is itself) and increment left to maintain invariant 1. Since we still haven't encountered any 1s, we can increment i to maintain invariant 2. In other words, the "ones" region remains empty.
0020011022ileftrightzeroestwosunsorted

Sorting 1s

When nums[i] == 1, we can simply increment i to maintain invariant 2.
00011120222ileftrightzeroestwosunsortedones

Sorting 2s

When nums[i] == 2, we swap nums[i] with nums[right]. This allows us to decrement right to maintain invariant 4. But since the new item at i came from the unsorted region, the new item at i is still unsorted, so we have to go through another iteration to correctly sort it.
00011120222ileftrightzeroestwosunsortedones

Termination

When i surpasses right the unsorted region is empty and the entire array is sorted.
00011102222ileftrightzeroestwosunsortedones

Solution

Visualization
Python
Language
Try these examples:
def sortColors(nums):
left, right = 0, len(nums) - 1
i = 0
while i <= right:
if nums[i] == 0:
nums[i], nums[left] = nums[left], nums[i]
left += 1
i += 1
elif nums[i] == 2:
nums[i], nums[right] = nums[right], nums[i]
right -= 1
else:
i += 1
212010101

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Explanation

Sorting 0s

Sorting 1s

Sorting 2s

Termination

Solution

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