Is Count the Number of K-Big Indices easy or hard?

The problem is rated Hard on LeetCode because it requires recognizing the pattern of counting smaller elements on both sides efficiently. While the logic is straightforward once identified, implementing Fenwick trees or segment trees correctly requires strong data structure knowledge.

Count the Number of K-Big Indices Python or Java solution

Most implementations use a Fenwick Tree with coordinate compression. Python, Java, C++, Go, and TypeScript solutions all follow the same steps: compress values, compute left smaller counts, compute right smaller counts, and count indices where both are at least k. Each update and query runs in O(log n).

How to solve Count the Number of K-Big Indices in O(n log n)?

Use coordinate compression followed by a Binary Indexed Tree. During a left-to-right pass, query how many numbers smaller than the current value already appeared. During a right-to-left pass, repeat the process to count smaller elements on the right. Count indices where both results are at least k.

What is the best approach for Count the Number of K-Big Indices?

The most efficient approach uses a Binary Indexed Tree (Fenwick Tree). Scan the array left-to-right to count how many smaller elements appear before each index, then scan right-to-left to count smaller elements after it. Both operations use prefix frequency queries in O(log n), producing an overall time complexity of O(n log n).

Is Count the Number of K-Big Indices asked at Google Amazon Meta?

Problems involving counting smaller elements and Fenwick trees frequently appear in interviews at companies like Google, Amazon, and Meta. Variants such as 'count of smaller numbers after self' or range frequency queries are common interview patterns that test data structures like Binary Indexed Trees and Segment Trees.

What data structure is used in Count the Number of K-Big Indices?

The key data structure is a Binary Indexed Tree (Fenwick Tree). It efficiently maintains prefix frequencies and supports updates and queries in O(log n). Segment trees or ordered sets can also solve the problem, but Fenwick trees are typically simpler and faster for this pattern.

What is the time complexity of Count the Number of K-Big Indices?

The optimal solution runs in O(n log n) time using a Binary Indexed Tree or Segment Tree. Each element performs a logarithmic query and update while counting smaller elements on the left and right. The brute force method takes O(n²) time because it scans both sides for every index.

Description Hints 3 Solution Video 10:44

2519. Count the Number of K-Big Indices

Hard53.2% acceptance1 companyPremium elsewhere, free here

You are given a 0-indexed integer array nums and a positive integer k.

We call an index i k-big if the following conditions are satisfied:

There exist at least k different indices idx1 such that idx1 < i and nums[idx1] < nums[i].
There exist at least k different indices idx2 such that idx2 > i and nums[idx2] < nums[i].

Return the number of k-big indices.

Example 1:

Input: nums = [2,3,6,5,2,3], k = 2
Output: 2
Explanation: There are only two 2-big indices in nums:
- i = 2 --> There are two valid idx1: 0 and 1. There are three valid idx2: 2, 3, and 4.
- i = 3 --> There are two valid idx1: 0 and 1. There are two valid idx2: 3 and 4.

Example 2:

Input: nums = [1,1,1], k = 3
Output: 0
Explanation: There are no 3-big indices in nums.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i], k <= nums.length

Asked at 1 company

Amazon

Topics

Hints open one at a time

Hint 1

The intended solution uses Fenwick Tree.

Hint 2

Let’s describe the solution for counting the number of elements smaller than nums[i] in the range [0, i - 1], and counting in the range [i + 1, nums.length - 1] can be done similarly by simply reversing the array.

Hint 3

Iterate from left to right and maintain the Fenwick Tree. Save the value of the query(nums[i]) and update nums[i] + 1 by 1.

10:44

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