Introduction

In this blog post, we will explore how to solve the LeetCode problem "315" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given an integer array nums, return an integer array counts where counts[i] is the number of smaller elements to the right of nums[i]. Example 1: Input: nums = [5,2,6,1] Output: [2,1,1,0] Explanation: To the right of 5 there are 2 smaller elements (2 and 1). To the right of 2 there is only 1 smaller element (1). To the right of 6 there is 1 smaller element (1). To the right of 1 there is 0 smaller element. Example 2: Input: nums = [-1] Output: [0] Example 3: Input: nums = [-1,-1] Output: [0,0] Constraints: 1 <= nums.length <= 105 -104 <= nums[i] <= 104

Explanation

• Binary Indexed Tree (BIT): The core idea is to iterate through the input array nums from right to left. For each element, we query a Binary Indexed Tree (BIT) to find the number of elements smaller than the current element that we've already processed (i.e., elements to its right). We then update the BIT with the current element's value so that subsequent queries can account for it.

• Value Mapping: Since the input numbers can be negative, we need to map them to positive indices for the BIT. We achieve this by first sorting the unique elements of nums and then using binary search to find the index (rank) of each element within the sorted unique elements. This rank serves as the index into the BIT.

• BIT Operations: The BIT supports two main operations: update(i, val) which increments the value at index i by val, and query(i) which returns the sum of all values from index 1 up to index i.

• Time Complexity: O(n log n), Space Complexity: O(n)

Code

    class Solution:
    def countSmaller(self, nums: list[int]) -> list[int]:
        """
        Counts the number of smaller elements to the right of each element in the input array.

        Args:
            nums: The input integer array.

        Returns:
            A list of integers, where counts[i] is the number of smaller elements to the right of nums[i].
        """

        n = len(nums)
        counts = [0] * n
        sorted_nums = sorted(list(set(nums)))  # Sort unique numbers
        rank = {num: i + 1 for i, num in enumerate(sorted_nums)}  # Rank each number

        bit = [0] * (len(sorted_nums) + 1)  # Initialize Binary Indexed Tree

        def update(i, val):
            while i < len(bit):
                bit[i] += val
                i += i & -i

        def query(i):
            res = 0
            while i > 0:
                res += bit[i]
                i -= i & -i
            return res

        for i in range(n - 1, -1, -1):
            counts[i] = query(rank[nums[i]] - 1)  # Query smaller numbers
            update(rank[nums[i]], 1)  # Update with current number

        return counts