Sort Colors

Sort Colors - Problem

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.

Input & Output

Example 1 — Basic Case

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

› Output: [0,0,1,1,2,2]

💡 Note: Sort the colors in-place: all 0s first, then 1s, then 2s

Example 2 — Already Sorted

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

› Output: [0,1,2]

💡 Note: Array is already sorted with one of each color

Example 3 — All Same Color

$ Input: nums = [1,1,1]

› Output: [1,1,1]

💡 Note: All elements are the same color, no changes needed

Constraints

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

Visualization

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Asked in

M Microsoft 45 a Amazon 38 Apple 32 f Facebook 28

The key insight is using the Dutch National Flag algorithm with three pointers to partition the array in a single pass. The optimal approach maintains regions for each color and swaps elements to their correct positions. Time: O(n), Space: O(1)

Common Approaches

✓ Counting Sort

⏱️ Time: O(n) Space: O(1)

First pass counts how many 0s, 1s, and 2s exist. Second pass overwrites the array with the correct number of each color in order.

Dutch National Flag (Optimal)

⏱️ Time: O(n) Space: O(1)

Maintains three pointers: left boundary for 0s, right boundary for 2s, and current pointer. Swaps elements to correct regions based on current value.

Counting Sort — Algorithm Steps

Count occurrences of 0, 1, and 2
Overwrite array with counted values in order

Visualization

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Step-by-Step Walkthrough

Count Phase

Count occurrences of each color

Fill Phase

Overwrite array with correct counts

Result

Array is sorted in-place

Code -

solution.c — C

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

void solution(int* nums, int n) {
    int count0 = 0, count1 = 0, count2 = 0;
    
    // Count occurrences
    for (int i = 0; i < n; i++) {
        if (nums[i] == 0) count0++;
        else if (nums[i] == 1) count1++;
        else count2++;
    }
    
    // Fill array
    int idx = 0;
    for (int i = 0; i < count0; i++) nums[idx++] = 0;
    for (int i = 0; i < count1; i++) nums[idx++] = 1;
    for (int i = 0; i < count2; i++) nums[idx++] = 2;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    int nums[100];
    int n = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL && n < 100) {
        // Skip whitespace
        while (isspace(*token)) token++;
        if (*token != '\0') {
            nums[n++] = atoi(token);
        }
        token = strtok(NULL, ",]");
    }
    
    solution(nums, n);
    
    printf("[");
    for (int i = 0; i < n; i++) {
        printf("%d", nums[i]);
        if (i < n - 1) printf(",");
    }
    printf("]\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Two passes through the array: one to count, one to fill

✓ Linear Growth

Space Complexity

O(1)

Only uses three variables to store counts

✓ Linear Space

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Ln 1, Col 1

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Counting Sort — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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