Sort Colors

Sure! This problem is known as the Dutch national flag problem, and it can be solved using a three-pointer approach. The idea is to maintain three pointers to divide the array into four sections: red, white, unknown, and blue.

Here’s how you can implement this in the Solution class:

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class Solution:
    def sortColors(self, nums):
        red, white, blue = 0, 0, len(nums) - 1

        while white <= blue:
            if nums[white] == 0:
                nums[red], nums[white] = nums[white], nums[red]
                red += 1
                white += 1
            elif nums[white] == 1:
                white += 1
            else:
                nums[white], nums[blue] = nums[blue], nums[white]
                blue -= 1

Here’s a breakdown of the logic:

  • If the current element nums[white] is 0 (red), we swap it with the element at the red pointer and move both red and white pointers forward.
  • If the current element is 1 (white), we don’t need to swap, and we just move the white pointer forward.
  • If the current element is 2 (blue), we swap it with the element at the blue pointer and move the blue pointer backward.

This approach ensures that the colors are sorted in-place, and the time complexity is (O(n)), where (n) is the length of the array nums.

Identifying Problem Isomorphism

“Sort Colors” involves sorting an array of integers where each element is either 0, 1, or 2. It’s an application of the Dutch national flag problem and requires understanding of in-place array manipulation.

An easier problem is “Move Zeroes”, where you need to move all the 0’s to the end of an array while maintaining the relative order of the non-zero elements. It’s simpler than “Sort Colors” as it only involves two types of elements, zeroes and non-zeroes, rather than three distinct values. However, both problems require a similar approach of swapping elements in the array to achieve the required order.

A more complex problem is “Sort an Array”, where you have to sort an array of integers. This problem does not limit the array’s values to just 0, 1, and 2, meaning a more general sorting algorithm is required. This elevates the complexity as it requires a broad understanding of sorting algorithms and their efficiency.

In increasing complexity:

  • “Move Zeroes”
  • “Sort Colors”
  • “Sort an Array”.

These problems all involve reordering elements in an array to fulfill certain conditions.

10 Prerequisite LeetCode Problems

Understand concepts like array manipulation, two pointers, and sorting. Here are 10 problems to prepare for “75. Sort Colors”.

  1. 283. Move Zeroes: This problem is about re-arranging an array, which is a simpler version of the “Sort Colors” problem.

  2. 27. Remove Element: This problem is also about array manipulation and will help you practice using two pointers to maintain the original order of an array.

  3. 88. Merge Sorted Array: This problem will help you learn how to merge two sorted arrays into one.

  4. 26. Remove Duplicates from Sorted Array: It’s a problem to help understand the use of pointers in array manipulation.

  5. 167. Two Sum II - Input array is sorted: This problem will help you understand the use of two pointers.

  6. 344. Reverse String: This problem is about in-place reversal using two pointers.

  7. 349. Intersection of Two Arrays: This problem helps understand basic set operations, a foundational concept.

  8. 268. Missing Number: This problem is about finding missing elements in an array, a similar concept to arranging or sorting.

  9. 414. Third Maximum Number: This problem involves finding the third maximum number in a non-sorted array, which can provide good practice with sorting concepts.

  10. 283. Move Zeroes: This problem involves moving all zeros in an array to the end while maintaining the relative order of the non-zero elements, which is a similar concept to sorting colors.

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.

Here, we will use the integers 0, 1, and 2 to represent the color red, white, and blue respectively.

Note: You are not suppose to use the library’s sort function for this problem.

Here is Python code that solves the problem:

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def sortColors(nums):
    red, white, blue = 0, 0, len(nums)-1

    while white <= blue:
        if nums[white] == 0:
            nums[red], nums[white] = nums[white], nums[red]
            white += 1
            red += 1
        elif nums[white] == 1:
            white += 1
        else:
            nums[white], nums[blue] = nums[blue], nums[white]
            blue -= 1
    return nums

This code uses a variation of the Dutch national flag problem’s solution by Edsger Dijkstra. The idea is to move all 0s to the beginning, and all 2s to the end. We maintain three pointers: one for the current element (white), one for the next position where the 0 should be put (red), and one for the next position where the 2 should be put (blue). The white pointer is used to scan the array from the start to the end.

Problem Classification

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.

Example 1:

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

Example 2:

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

Constraints:

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

Follow up: Could you come up with a one-pass algorithm using only constant extra space?

Language Agnostic Coding Drills

This is an implementation of the Dutch national flag problem, which involves sorting an array of 0s, 1s, and 2s (which represent the three colors in the Dutch national flag). This is solved using a two-pointer approach.

  1. Variable Declarations and Initialization: Understand the purpose of each variable and how they’re initialized (red, white, blue).

  2. Control Structures (While Loop): Understand how to use a while loop and when to use it. The loop here continues until the white pointer has passed the blue pointer.

  3. Control Structures (If-elif-else statements): Understand how to use if-elif-else statements to control program flow. In this code, different actions are taken based on the value at the current index pointed to by white.

  4. Working with Lists: Understand how to manipulate lists, in this case, swapping elements in a list.

  5. Incrementing and decrementing pointers: Know how and when to increment or decrement pointers, in this case, red, white and blue.

  6. Combining the above: The ultimate problem-solving drill would be to understand the two-pointer method and how it is applied in sorting this array. You start with red and white at the beginning of the array and blue at the end. You go through each element (white), and if it’s 0, swap it with red and move both red and white forward. If it’s 1, just move white forward. If it’s 2, swap it with blue and move blue backward.

This algorithm allows you to sort the array in a single pass with constant space complexity. The general problem-solving approach here is understanding the behavior of the elements you are sorting and setting up pointers that help you track the position of these elements as you go through the array.

Targeted Drills in Python

Drill 1: Variable Declarations and Initialization

Create three variables, red, white, blue, and initialize them with the values 0, 0, and 9 respectively. Print the variables.

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red, white, blue = 0, 0, 9
print(red, white, blue)

Drill 2: Control Structures (While Loop)

Create a while loop that continues running until the variable white is greater than blue. In each iteration, increment white by one. Print white in each iteration.

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while white <= blue:
    print(white)
    white += 1

Drill 3: Control Structures (If-elif-else statements)

In a for loop from 0 to 10, check each number. If the number is divisible by 3, print “Divisible by 3”. If it’s divisible by 2, print “Divisible by 2”. Else, print the number.

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for i in range(10):
    if i % 3 == 0:
        print("Divisible by 3")
    elif i % 2 == 0:
        print("Divisible by 2")
    else:
        print(i)

Drill 4: Working with Lists

Given a list nums = [1, 2, 3, 4, 5], swap the first and last elements.

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nums = [1, 2, 3, 4, 5]
nums[0], nums[-1] = nums[-1], nums[0]
print(nums)

Drill 5: Incrementing and decrementing pointers

Start with variables left and right at values 0 and 10 respectively. In a while loop, continue until left is equal to or greater than right. In each iteration, increment left and decrement right.

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left, right = 0, 10
while left < right:
    print(left, right)
    left += 1
    right -= 1

Drill 6: Combining the above

For the final drill, implement the sortColors function on a sample list and print the result.

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class Solution:
    def sortColors(self, nums: List[int]) -> None:
        red, white, blue = 0, 0, len(nums) - 1
        while white <= blue:
            if nums[white] == 0:
                nums[white], nums[red] = nums[red], nums[white]
                red += 1
                white += 1
            elif nums[white] == 1:
                white += 1
            else:
                nums[white], nums[blue] = nums[blue], nums[white]
                blue -= 1
        print(nums)

sol = Solution()
sol.sortColors([2,0,2,1,1,0])

These drills should help in understanding each concept separately. By combining them, one can grasp the entire algorithm.