The core approach: Sort Colors is solved with the Dutch National Flag algorithm in O(n) time and O(1) space. Maintain three pointers — low, mid, and high. Swap elements to ensure 0s accumulate before low, 1s between low and mid, and 2s after high. Each swap advances at least one pointer, guaranteeing a single-pass sort without extra space.

You’re in a coding interview. The interviewer says: “Sort an array of 0s, 1s, and 2s in place without using a library sort.” The two-pass counting sort is obvious — but the one everyone remembers is the Dutch National Flag algorithm: a single-pass, O(1) space solution using three pointers.

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The Problem

Given an array nums with objects colored red (0), white (1), or blue (2), sort them in place so same-colored objects are adjacent, in order 0, 1, 2.

Example

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

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What Interviewers Are Actually Testing

Skill	What Interviewers Observe
Dutch National Flag	Do you know the three-pointer algorithm?
Pointer invariants	Can you state what each pointer represents?
Loop termination	Do you use mid <= high (not < high)?
One-pass vs. two-pass	Can you optimize from counting sort to DNF?

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Step 1: The Key Insight

Three pointers maintain three regions: low marks the 0-boundary, high marks the 2-boundary, mid scans. When nums[mid] is 0, swap to low and advance both. When 2, swap to high and retreat high only (don’t advance mid — the swapped value is unseen). When 1, advance mid.

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Python Implementation

def sortColors(nums: list[int]) -> None:
    low, mid, high = 0, 0, len(nums) - 1

    while mid <= high:
        if nums[mid] == 0:
            nums[low], nums[mid] = nums[mid], nums[low]
            low += 1
            mid += 1
        elif nums[mid] == 1:
            mid += 1
        else:
            nums[mid], nums[high] = nums[high], nums[mid]
            high -= 1
            # Don't increment mid — swapped value needs examination

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Time & Space Complexity

Aspect	Complexity
Time	O(n) — single pass
Space	O(1) — in place

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Common Interview Mistakes

1. Incrementing mid after swapping with high. The value from high is unseen — don’t skip it.
2. Using mid < high instead of mid <= high. When equal, nums[mid] is still unsorted.
3. Proposing a general sort. Use the constrained value set — that’s the point.

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What a Strong Candidate Sounds Like

“I’ll use the Dutch National Flag algorithm. Three pointers: low, mid, high. Everything left of low is 0, between low and mid is 1, right of high is 2. When I see a 0, swap to left and advance both. When 2, swap to right and retreat high only. One pass, O(1) space.”

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Related Problems to Practice

• Two Sum — Array manipulation fundamentals.
• Container With Most Water — Two-pointer pattern.
• 3Sum — Multi-pointer array traversal.

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Frequently Asked Questions

How do you sort an array seamlessly in a single iteration pass?

You employ the Dutch National Flag algorithm utilizing three distinct pointers. A low pointer acts as a gatekeeper for zeroes pushing them forward, a high pointer safely sequesters two’s pushing them backwards, and a mid pointer acts as the scanning agent navigating unclassified elements.

What remains the primary advantage of the Dutch National Flag approach?

Normal sorting functions inherently demand O(N log N) runtime overhead. But by actively swapping exactly three known numerical values cleanly via opposing pointer bounds, this algorithm achieves perfectly sorted array states dynamically in exactly O(N) linear time and O(1) in-place space.

Practice This in a Mock Interview

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Further reading:

• How to Prepare for a Coding Interview in 2026, the complete roadmap
• Why LeetCode alone isn’t enough, and what to practice instead
• Practice any LeetCode problem as a live mock interview
• The Grind 75 Pathway, a structured study plan with AI practice links
• Browse all walkthroughs on GitHub, condensed solutions with code