How do you solve Distribute Elements Into Two Arrays II on LeetCode?

Distribute Elements Into Two Arrays II (LeetCode #3072) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Using efficient data structures can significantly reduce time complexity.

What is the time complexity of Distribute Elements Into Two Arrays II?

The optimal solution for Distribute Elements Into Two Arrays II runs in O(n log n) time and uses O(n) space. The BIT allows us to update and query counts in logarithmic time, leading to an overall complexity of O(n log n) for processing all elements.

What is the brute force solution for Distribute Elements Into Two Arrays II?

The brute force approach for Distribute Elements Into Two Arrays II is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we will iterate through the array and for each element, count how many elements in both arr1 and arr2 are greater than the current element. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n log n).

What are the interview tips for Distribute Elements Into Two Arrays II?

Clarify the requirements and constraints of the problem. Discuss your thought process and approach before coding. Test your solution with edge cases to ensure correctness.

What are common mistakes when solving Distribute Elements Into Two Arrays II?

Not updating the data structure correctly after each insertion. Misunderstanding how to calculate greater counts efficiently.

#3072

Distribute Elements Into Two Arrays II

Hard

ArrayBinary Indexed TreeSegment TreeSimulationBinary Indexed TreeSegment Tree

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log n)Space O(n)

Using a data structure like a Binary Indexed Tree (BIT) allows us to efficiently count the number of elements greater than a given value while supporting quick updates as we build arr1 and arr2.

⚙️

Algorithm

3 steps

1Step 1: Normalize the values in nums to fit into a range [1, n] for BIT usage.
2Step 2: Initialize a BIT to keep track of counts of elements in arr1 and arr2.
3Step 3: For each element in nums, use the BIT to get greaterCount for both arrays and append to the appropriate array.

solution.py42 lines

1class BIT:
2    def __init__(self, size):
3        self.size = size
4        self.tree = [0] * (size + 1)
5
6    def update(self, index, delta):
7        while index <= self.size:
8            self.tree[index] += delta
9            index += index & -index
10
11    def query(self, index):
12        total = 0
13        while index > 0:
14            total += self.tree[index]
15            index -= index & -index
16        return total
17
18def distributeElements(nums):
19    n = len(nums)
20    max_val = max(nums)
21    bit = BIT(max_val)
22    arr1 = [nums[0]]
23    arr2 = [nums[1]]
24    bit.update(nums[0], 1)
25    bit.update(nums[1], 1)
26    for i in range(2, n):
27        count1 = bit.query(max_val) - bit.query(nums[i])
28        count2 = bit.query(max_val) - bit.query(nums[i])
29        if count1 > count2:
30            arr1.append(nums[i])
31            bit.update(nums[i], 1)
32        elif count1 < count2:
33            arr2.append(nums[i])
34            bit.update(nums[i], 1)
35        else:
36            if len(arr1) < len(arr2):
37                arr1.append(nums[i])
38                bit.update(nums[i], 1)
39            else:
40                arr2.append(nums[i])
41                bit.update(nums[i], 1)
42    return arr1 + arr2

ℹ

Complexity note: The BIT allows us to update and query counts in logarithmic time, leading to an overall complexity of O(n log n) for processing all elements.

1Using efficient data structures can significantly reduce time complexity.
2Understanding how to normalize data for specific algorithms is crucial.

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