How do you solve Sum of Imbalance Numbers of All Subarrays on LeetCode?

Sum of Imbalance Numbers of All Subarrays (LeetCode #2763) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n² log n) time complexity and O(n) space complexity. Key insight: The imbalance number only increases when there are gaps greater than 1 between sorted elements.

What is the brute force solution for Sum of Imbalance Numbers of All Subarrays?

The brute force approach for Sum of Imbalance Numbers of All Subarrays is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible subarrays and calculating their imbalance numbers by sorting each subarray. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n² log n).

What algorithm or data structure is used to solve Sum of Imbalance Numbers of All Subarrays?

Sum of Imbalance Numbers of All Subarrays uses the following concepts: Array, Hash Table, Enumeration. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Sum of Imbalance Numbers of All Subarrays?

Always clarify the problem statement and constraints before diving into coding. Think about edge cases, such as arrays with all identical elements or arrays with maximum length. Practice explaining your thought process as you code; this helps interviewers understand your approach.

What are common mistakes when solving Sum of Imbalance Numbers of All Subarrays?

Failing to account for duplicate elements when calculating the imbalance. Not properly managing the indices when iterating through subarrays.

#2763

Sum of Imbalance Numbers of All Subarrays

Hard

ArrayHash TableEnumerationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n² log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n² log n)Space O(n)

The optimal approach uses a HashMap to track the frequency of elements in the current subarray, allowing us to efficiently calculate the imbalance without sorting. This approach reduces the time complexity significantly.

⚙️

Algorithm

3 steps

1Step 1: Initialize a HashMap to count occurrences of elements in the current subarray.
2Step 2: For each starting index, expand the subarray to the right, updating the HashMap.
3Step 3: Calculate the imbalance based on the keys in the HashMap, counting gaps greater than 1.

solution.py11 lines

1def sum_of_imbalance(nums):
2    total_imbalance = 0
3    n = len(nums)
4    for left in range(n):
5        count = {}
6        for right in range(left, n):
7            count[nums[right]] = count.get(nums[right], 0) + 1
8            keys = sorted(count.keys())
9            imbalance = sum(1 for i in range(len(keys) - 1) if keys[i + 1] - keys[i] > 1)
10            total_imbalance += imbalance
11    return total_imbalance

ℹ

Complexity note: The time complexity is O(n² log n) due to the sorting of keys in the HashMap for each subarray, while the space complexity is O(n) for storing the frequency of elements.

1The imbalance number only increases when there are gaps greater than 1 between sorted elements.
2Using a HashMap allows for efficient counting of element frequencies, avoiding the need for repeated sorting.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.