Reverse Pairs

We can use a merge-sort-based approach to count reverse pairs in the given array. This will give us an efficient way to solve the problem.

Here’s the Python code that performs this task:

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class Solution:
    def reversePairs(self, nums: List[int]) -> int:
        def mergeSort(start, end):
            if start >= end:
                return 0

            mid = start + (end - start) // 2
            count = mergeSort(start, mid) + mergeSort(mid + 1, end)

            j = mid + 1
            for i in range(start, mid + 1):
                while j <= end and nums[i] > 2 * nums[j]:
                    j += 1
                count += j - (mid + 1)

            temp = []
            left, right = start, mid + 1
            while left <= mid and right <= end:
                if nums[left] <= nums[right]:
                    temp.append(nums[left])
                    left += 1
                else:
                    temp.append(nums[right])
                    right += 1
            while left <= mid:
                temp.append(nums[left])
                left += 1
            while right <= end:
                temp.append(nums[right])
                right += 1
            for i in range(len(temp)):
                nums[start + i] = temp[i]

            return count

        return mergeSort(0, len(nums) - 1)

This code first divides the array into smaller parts and then merges them back, counting reverse pairs in the process. The merging step ensures that the original array remains sorted, which is essential for the correct calculation of reverse pairs.

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class Solution:
    def reversePairs(self, nums: List[int]) -> int:
        def recursiveFunction(lowerIndex = 0, upperIndex = len(nums) - 1):

            if lowerIndex >= upperIndex:
                return 0

            midIndex = (lowerIndex + upperIndex) // 2
            count = recursiveFunction(lowerIndex, midIndex) + recursiveFunction(midIndex + 1, upperIndex)

            index_i = lowerIndex
            for rightNumber in nums[midIndex + 1: upperIndex + 1]:
                while index_i <= midIndex and nums[index_i] <= rightNumber * 2:
                    index_i += 1
                count += midIndex + 1 - index_i
                if index_i > midIndex:
                    break

            nums[lowerIndex: upperIndex + 1] = sorted(nums[lowerIndex: upperIndex + 1])

            return count

        return recursiveFunction()

Problem Classification

The task is to count pairs in an array where one element is larger than the other’s double. This is a Pair Counting Problem.

Language Agnostic Coding Drills

The code is a solution to count the important reverse pairs in an array where 2*i > j holds true. It uses recursion and sorting, which are both important concepts in many modern programming languages. Here are the drills, sorted in order of increasing difficulty:

Understanding and Using Variables and Basic Data Structures: Practice creating and manipulating simple variables like integers, and complex ones like arrays (list in Python).
Conditionals: Understand the use of conditionals (if, else, and else if statements). These allow for decision making in the code based on the evaluation of certain conditions.
Loops: Familiarize yourself with loops, specifically for loops. These are used to repeatedly execute a block of code until a certain condition is met.
Recursion: Understand the principle of recursion, a method where the solution to a problem depends on solutions to smaller instances of the same problem.
Sorting: Learn about different sorting algorithms (like quicksort, mergesort, etc.). Sorting is an essential tool and is used heavily in the given code.
Divide and Conquer: Learn the divide and conquer strategy used in algorithms. This involves dividing the problem into smaller subproblems and combining their results to solve the original problem.

To arrive at the final solution, the approach would be:

Define the main problem which is to find all the important reverse pairs in the list.
Identify that the problem can be broken down into smaller subproblems using a divide and conquer strategy.
Implement a recursive function that will count the number of important reverse pairs in a given range of indices in the list.
Use a sort function to maintain a sorted list at every recursive function call.
Traverse through the right half of the list for every element and count the number of important reverse pairs.
Finally, return the total count of the important reverse pairs.

This thought process and steps are fundamental to many algorithmic problems and learning them will be beneficial in solving a multitude of problems in computer science and programming.

Targeted Drills in Python

Let’s look at the Python coding drills corresponding to each concept. The following drills will help you grasp the underlying concepts used in the provided solution.

Drill 1: Understanding and Using Variables and Basic Data Structures

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# Drill: Create an integer variable and a list, manipulate them
integer_variable = 5
print(integer_variable)  # output: 5

list_variable = [1, 2, 3, 4, 5]
list_variable.append(6)
print(list_variable)  # output: [1, 2, 3, 4, 5, 6]

Drill 2: Conditionals

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# Drill: Write an if-else statement
x = 10
if x > 5:
    print("x is greater than 5")
else:
    print("x is not greater than 5")

Drill 3: Loops

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# Drill: Write a for loop to iterate over a list
for i in list_variable:
    print(i)

Drill 4: Recursion

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# Drill: Write a simple recursive function for factorial calculation
def factorial(n):
    if n == 0 or n == 1:
        return 1
    else:
        return n * factorial(n - 1)
print(factorial(5))  # output: 120

Drill 5: Sorting

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# Drill: Sort a list of integers
unsorted_list = [5, 3, 2, 4, 1]
sorted_list = sorted(unsorted_list)
print(sorted_list)  # output: [1, 2, 3, 4, 5]

Drill 6: Divide and Conquer

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# Drill: Write a function that uses the divide and conquer approach to find the maximum element in a list
def find_max(nums, left, right):
    if left == right:
        return nums[left]
    mid = (left + right) // 2
    max1 = find_max(nums, left, mid)
    max2 = find_max(nums, mid+1, right)
    return max(max1, max2)

nums = [1, 5, 3, 2, 6, 7, 4]
print(find_max(nums, 0, len(nums)-1))  # output: 7

Problem-Specific Drill:

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# Drill: Write a function to count the number of pairs in a list where i*2 > j
def count_pairs(nums):
    count = 0
    for i in range(len(nums)):
        for j in range(i+1, len(nums)):
            if nums[i] * 2 > nums[j]:
                count += 1
    return count

nums = [1, 3, 2]
print(count_pairs(nums))  # output: 2

These drills should help you get comfortable with the individual concepts used in the provided solution. The final solution involves combining these concepts in a particular way to solve the problem.