Leetcode #2519: Count the Number of K-Big Indices
In this guide, we solve Leetcode #2519 Count the Number of K-Big Indices in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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].
Quick Facts
- Difficulty: Hard
- Premium: Yes
- Tags: Binary Indexed Tree, Segment Tree, Array, Binary Search, Divide and Conquer, Ordered Set, Merge Sort
Intuition
The problem structure suggests a monotonic decision, which makes binary search a natural fit.
By halving the search space each step, we reach the answer efficiently.
Approach
Search either directly on a sorted array or on the answer space using a check function.
Each check is fast, and the logarithmic search keeps the overall runtime low.
Steps:
- Define the search bounds.
- Check the mid point condition.
- Narrow the bounds until convergence.
Example
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.
Python Solution
class BinaryIndexedTree:
def __init__(self, n):
self.n = n
self.c = [0] * (n + 1)
def update(self, x, delta):
while x <= self.n:
self.c[x] += delta
x += x & -x
def query(self, x):
s = 0
while x:
s += self.c[x]
x -= x & -x
return s
class Solution:
def kBigIndices(self, nums: List[int], k: int) -> int:
n = len(nums)
tree1 = BinaryIndexedTree(n)
tree2 = BinaryIndexedTree(n)
for v in nums:
tree2.update(v, 1)
ans = 0
for v in nums:
tree2.update(v, -1)
ans += tree1.query(v - 1) >= k and tree2.query(v - 1) >= k
tree1.update(v, 1)
return ans
Complexity
The time complexity is , and the space complexity is , where is the length of the array. The space complexity is , where is the length of the array.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.