Sort Colors - Problem

🎨 Sort Colors (Dutch National Flag Problem)

You're organizing a collection of colored objects that need to be arranged in a specific order! Given an array nums with n objects colored red, white, or blue, your task is to sort them in-place so that objects of the same color are grouped together.

The Challenge:

Objects are represented by integers: 0 = red, 1 = white, 2 = blue
Final arrangement must be: red → white → blue
Sort in-place (modify the original array)
No built-in sort functions allowed!

Example: [2,0,2,1,1,0] becomes [0,0,1,1,2,2]

This classic problem is also known as the Dutch National Flag Problem, named after the three colors of the Netherlands flag! 🇳🇱

Input & Output

example_1.py — Basic Case

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

› Output: [0,0,1,1,2,2]

💡 Note: The array contains 2 red objects (0), 2 white objects (1), and 2 blue objects (2). After sorting in-place, all reds come first, then whites, then blues.

example_2.py — Already Sorted

$ Input: [0,1,2]

› Output: [0,1,2]

💡 Note: The array is already sorted with red, white, and blue in correct order. No changes needed.

example_3.py — Single Color

$ Input: [1,1,1]

› Output: [1,1,1]

💡 Note: All objects are the same color (white), so the array remains unchanged after sorting.

Visualization

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Understanding the Visualization

Setup Zones

Create three zones with boundaries: red zone, processing zone, blue zone

Process Each Ball

Examine current ball: if red, send to red zone; if white, keep in middle; if blue, send to blue zone

Adjust Boundaries

After each placement, update zone boundaries to reflect new organization

Complete Sorting

When processing zone is empty, all balls are perfectly organized!

Key Takeaway

🎯 Key Insight: The Dutch National Flag algorithm maintains three distinct regions simultaneously, allowing us to sort in a single pass by intelligently moving boundaries as we process each element. This transforms a potentially complex sorting problem into an elegant three-way partitioning solution!

Time & Space Complexity

Time Complexity

⏱️

O(n²)

In worst case, we need O(n) passes and each pass takes O(n) time to scan the array

⚠ Quadratic Growth

Space Complexity

O(1)

Only uses a constant amount of extra space for loop variables and swapping

✓ Linear Space

Constraints

n == nums.length
1 ≤ n ≤ 300
nums[i] is either 0, 1, or 2
You must solve this problem without using the library's sort function

Asked in

f Facebook 85 ⊞ Microsoft 72 a Amazon 68 Apple 45 G Google 38

The optimal solution uses the Dutch National Flag algorithm with three pointers to partition the array in a single pass. This O(n) time, O(1) space approach maintains three regions: processed 0s (left), unprocessed elements (middle), and processed 2s (right). By carefully managing pointer movements during swaps, we achieve perfect in-place sorting without using built-in sort functions.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Bubble Sort Style)	O(n²)	O(1)	Repeatedly scan and swap elements to move them to correct positions
Three-Way Partitioning (Dutch National Flag)	O(n)	O(1)	Use three pointers to partition array into three sections in one pass
Two-Pass Counting	O(n)	O(1)	Count each color first, then overwrite array with correct counts

Brute Force (Bubble Sort Style) — Algorithm Steps

Make multiple passes through the array
In each pass, compare adjacent elements
If left element > right element, swap them
Continue until no more swaps are needed
This naturally groups 0s, 1s, and 2s together

Visualization

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Step-by-Step Walkthrough

Initial Array

Start with mixed colors [2,0,2,1,1,0]

First Pass

Compare adjacent elements, swap if left > right

Continue Passes

Repeat until no more swaps needed

Sorted Result

All colors grouped: [0,0,1,1,2,2]

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>

void sortColors(int* nums, int numsSize) {
    // Bubble sort approach
    for (int i = 0; i < numsSize; i++) {
        int swapped = 0;
        for (int j = 0; j < numsSize - i - 1; j++) {
            if (nums[j] > nums[j + 1]) {
                int temp = nums[j];
                nums[j] = nums[j + 1];
                nums[j + 1] = temp;
                swapped = 1;
            }
        }
        if (!swapped) break;
    }
}

int main() {
    int nums[1000];
    int size = 0;
    char ch;
    int num = 0;
    int reading = 0;
    
    while ((ch = getchar()) != '\n' && ch != EOF) {
        if (ch >= '0' && ch <= '9') {
            num = num * 10 + (ch - '0');
            reading = 1;
        } else if (reading) {
            nums[size++] = num;
            num = 0;
            reading = 0;
        }
    }
    if (reading) nums[size++] = num;
    
    sortColors(nums, size);
    
    printf("[");
    for (int i = 0; i < size; i++) {
        printf("%d", nums[i]);
        if (i < size - 1) printf(",");
    }
    printf("]\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

In worst case, we need O(n) passes and each pass takes O(n) time to scan the array

⚠ Quadratic Growth

Space Complexity

O(1)

Only uses a constant amount of extra space for loop variables and swapping

✓ Linear Space

Constraints

n == nums.length
1 ≤ n ≤ 300
nums[i] is either 0, 1, or 2
You must solve this problem without using the library's sort function

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🎨 Sort Colors (Dutch National Flag Problem)

Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force (Bubble Sort Style) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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