Leetcode #2763: Sum of Imbalance Numbers of All Subarrays
In this guide, we solve Leetcode #2763 Sum of Imbalance Numbers of All Subarrays 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
The imbalance number of a 0-indexed integer array arr of length n is defined as the number of indices in sarr = sorted(arr) such that: 0 <= i < n - 1, and sarr[i+1] - sarr[i] > 1 Here, sorted(arr) is the function that returns the sorted version of arr. Given a 0-indexed integer array nums, return the sum of imbalance numbers of all its subarrays.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Array, Hash Table, Enumeration
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 = [2,3,1,4]
Output: 3
Explanation: There are 3 subarrays with non-zero imbalance numbers:
- Subarray [3, 1] with an imbalance number of 1.
- Subarray [3, 1, 4] with an imbalance number of 1.
- Subarray [1, 4] with an imbalance number of 1.
The imbalance number of all other subarrays is 0. Hence, the sum of imbalance numbers of all the subarrays of nums is 3.
Python Solution
class Solution:
def sumImbalanceNumbers(self, nums: List[int]) -> int:
n = len(nums)
ans = 0
for i in range(n):
sl = SortedList()
cnt = 0
for j in range(i, n):
k = sl.bisect_left(nums[j])
h = k - 1
if h >= 0 and nums[j] - sl[h] > 1:
cnt += 1
if k < len(sl) and sl[k] - nums[j] > 1:
cnt += 1
if h >= 0 and k < len(sl) and sl[k] - sl[h] > 1:
cnt -= 1
sl.add(nums[j])
ans += cnt
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.