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Sort Colors

Certification: Intermediate Level Accuracy: 100% Submissions: 1 Points: 10

Write a JavaScript program to sort an array containing only 0s, 1s, and 2s in-place. The colors red, white, and blue are represented by integers 0, 1, and 2 respectively. You must solve this problem without using the library's sort function and in a single pass.

Example 1
    
      
  • Input: nums = [2,0,2,1,1,0]
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  • Output: [0,0,1,1,2,2]
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  • Explanation: 
    • The array contains red (0), white (1), and blue (2) colors. 
    • We need to sort them so all reds come first, then whites, then blues. 
    • After sorting: [0,0,1,1,2,2] where 0s are grouped first, then 1s, then 2s. 
    • The array is now properly sorted with colors in order.
    •
Example 2
    
      
  • Input: nums = [2,0,1]
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  • Output: [0,1,2]
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  • Explanation: 
    • The array contains one of each color: red (0), white (1), blue (2). 
    • After sorting, red comes first, then white, then blue. 
    • The final sorted array is [0,1,2]. 
    • Each color is in its correct position.
    •
Constraints
    
      
  • n == nums.length
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  • 1 <= n <= 300
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  • nums[i] is either 0, 1, or 2
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  • Time Complexity: O(n)
    • 
      
  • Space Complexity: O(1)
    •
ArraysIBMPhillips
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Hint
Solution

Solution Hints

    
      
  • Use three pointers: low, mid, and high to partition the array
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  • low pointer tracks the boundary for 0s, high pointer tracks the boundary for 2s
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  • mid pointer is used to traverse the array and examine each element
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  • When nums[mid] is 0, swap with nums[low] and increment both low and mid
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  • When nums[mid] is 1, just increment mid pointer
    • 
      
  • When nums[mid] is 2, swap with nums[high] and decrement high (don't increment mid)
    •

Steps to solve by this approach:

 Step 1: Initialize three pointers - low at 0, mid at 0, and high at the last index.

 Step 2: The low pointer maintains the boundary for placing 0s, high pointer for 2s.

 Step 3: Traverse the array using mid pointer and examine each element.

 Step 4: If current element is 0, swap it with element at low pointer and increment both low and mid.

 Step 5: If current element is 1, simply increment mid pointer as 1s are in correct position.

 Step 6: If current element is 2, swap it with element at high pointer and decrement high only.

 Step 7: Continue until mid pointer crosses high pointer, ensuring all elements are properly sorted.

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