Leetcode #2653: Sliding Subarray Beauty
In this guide, we solve Leetcode #2653 Sliding Subarray Beauty 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
Given an integer array nums containing n integers, find the beauty of each subarray of size k. The beauty of a subarray is the xth smallest integer in the subarray if it is negative, or 0 if there are fewer than x negative integers.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, Hash Table, Sliding Window
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: nums = [1,-1,-3,-2,3], k = 3, x = 2
Output: [-1,-2,-2]
Explanation: There are 3 subarrays with size k = 3.
The first subarray is [1, -1, -3] and the 2nd smallest negative integer is -1.
The second subarray is [-1, -3, -2] and the 2nd smallest negative integer is -2.
The third subarray is [-3, -2, 3] and the 2nd smallest negative integer is -2.
Python Solution
class Solution:
def getSubarrayBeauty(self, nums: List[int], k: int, x: int) -> List[int]:
def f(x: int) -> int:
s = 0
for i in range(50):
s += cnt[i]
if s >= x:
return i - 50
return 0
cnt = [0] * 101
for v in nums[:k]:
cnt[v + 50] += 1
ans = [f(x)]
for i in range(k, len(nums)):
cnt[nums[i] + 50] += 1
cnt[nums[i - k] + 50] -= 1
ans.append(f(x))
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
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.