#3072

Hard

auto_awesomeArray plus Binary Indexed Tree

LeetCode Problem Workspace

Distribute Elements Into Two Arrays II

Distribute elements into two arrays based on conditions, utilizing a Binary Indexed Tree for efficient counting and simulation.

array binary indexed tree segment tree simulation

Problem Statement

You are given a 1-indexed array nums of length n. Your task is to distribute the elements of nums into two arrays, arr1 and arr2, using n operations. The first operation appends nums[1] to arr1, and the second appends nums[2] to arr2. After the second operation, for every subsequent operation, the decision on where to insert the current element is based on the number of elements greater than the current value in each array and their relative sizes.

The function greaterCount(arr, val) returns the count of elements in arr that are strictly greater than val. The element from nums must be added to the array where the count of greater elements is lesser, with a tie-breaker based on the array sizes. The process continues until all elements are distributed. The goal is to concatenate arr1 and arr2 and return the result.

Examples

Example 1

Input: nums = [2,1,3,3]

Output: [2,3,1,3]

After the first 2 operations, arr1 = [2] and arr2 = [1]. In the 3rd operation, the number of elements greater than 3 is zero in both arrays. Also, the lengths are equal, hence, append nums[3] to arr1. In the 4th operation, the number of elements greater than 3 is zero in both arrays. As the length of arr2 is lesser, hence, append nums[4] to arr2. After 4 operations, arr1 = [2,3] and arr2 = [1,3]. Hence, the array result formed by concatenation is [2,3,1,3].

Example 2

Input: nums = [5,14,3,1,2]

Output: [5,3,1,2,14]

After the first 2 operations, arr1 = [5] and arr2 = [14]. In the 3rd operation, the number of elements greater than 3 is one in both arrays. Also, the lengths are equal, hence, append nums[3] to arr1. In the 4th operation, the number of elements greater than 1 is greater in arr1 than arr2 (2 > 1). Hence, append nums[4] to arr1. In the 5th operation, the number of elements greater than 2 is greater in arr1 than arr2 (2 > 1). Hence, append nums[5] to arr1. After 5 operations, arr1 = [5,3,1,2] and arr2 = [14]. Hence, the array result formed by concatenation is [5,3,1,2,14].

Example 3

Input: nums = [3,3,3,3]

Output: [3,3,3,3]

At the end of 4 operations, arr1 = [3,3] and arr2 = [3,3]. Hence, the array result formed by concatenation is [3,3,3,3].

Constraints

3 <= n <= 105
1 <= nums[i] <= 109

Solution Approach

Binary Indexed Tree

Using a Binary Indexed Tree (BIT), we can efficiently count the number of elements greater than a given value in arr1 and arr2. This data structure supports both dynamic insertion and efficient range queries, which helps manage the counts as elements are inserted.

Simulation of Distribution

The problem requires simulating the process of inserting elements into two arrays based on conditions. For each element, we calculate the number of elements greater than it in both arrays using BIT and append it to the appropriate array based on the described rules.

Efficient Insertion and Querying

Efficient querying and insertion are critical for solving this problem, as the array length n can go up to 105. Using BIT allows for O(log n) time complexity for both operations, making the solution feasible within the problem's constraints.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is O(n log n) due to the use of Binary Indexed Tree for each insertion and querying step. The space complexity is O(n) as we store the elements of both arrays and the BIT structure.

What Interviewers Usually Probe

Look for an understanding of Binary Indexed Tree (BIT) and its application to dynamic counting.
Evaluate the candidate’s ability to simulate the problem conditions with efficient data structures.
Check for a clear explanation of handling the array insertion logic while maintaining efficiency.

Common Pitfalls or Variants

Common pitfalls

Failing to efficiently track counts of greater elements using an appropriate data structure.
Incorrectly handling tie-breakers when both arrays have the same count of greater elements.
Not managing the balance between the two arrays properly, leading to incorrect results.

Follow-up variants

Change the distribution rule, for example, append to the array with fewer elements instead of the one with fewer greater elements.
Instead of using a Binary Indexed Tree, try using a Segment Tree for the counting operation.
Increase the number of arrays from two to three or more, with different insertion conditions.

FAQ

What is the main technique used to solve Distribute Elements Into Two Arrays II?

The main technique used in this problem is the Binary Indexed Tree (BIT), which allows efficient tracking of counts of elements greater than a given value.

How does the Binary Indexed Tree help with the distribution process?

The Binary Indexed Tree allows for efficient querying and updating of counts of elements greater than a specific value, enabling quick decision-making for each insertion into the arrays.

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

The time complexity of the optimal solution is O(n log n) due to the use of Binary Indexed Tree for insertion and querying operations.

Can we use a Segment Tree instead of a Binary Indexed Tree?

Yes, a Segment Tree can be used instead of a Binary Indexed Tree, though it may be less efficient for this specific problem.

How do we handle tie-breakers in this problem?

When both arrays have the same count of elements greater than the current value, the decision is based on the array sizes, where the element is appended to the smaller array.

terminal

Solution

Solution 1: Discretization + Binary Indexed Tree

We can use two binary indexed trees `tree1` and `tree2` to maintain the number of elements in `arr1` and `arr2` that are less than or equal to a certain number. Each time, we query the number of elements that are less than or equal to the current number in the binary indexed tree, then the number of elements that are greater than the current number is the length of the current array minus the query result. Then we can decide which array to add the current number to based on this difference.

class BinaryIndexedTree:
    __slots__ = "n", "c"

    def __init__(self, n: int):
        self.n = n
        self.c = [0] * (n + 1)

    def update(self, x: int, delta: int) -> None:
        while x <= self.n:
            self.c[x] += delta
            x += x & -x

    def query(self, x: int) -> int:
        s = 0
        while x:
            s += self.c[x]
            x -= x & -x
        return s


class Solution:
    def resultArray(self, nums: List[int]) -> List[int]:
        st = sorted(set(nums))
        m = len(st)
        tree1 = BinaryIndexedTree(m + 1)
        tree2 = BinaryIndexedTree(m + 1)
        tree1.update(bisect_left(st, nums[0]) + 1, 1)
        tree2.update(bisect_left(st, nums[1]) + 1, 1)
        arr1 = [nums[0]]
        arr2 = [nums[1]]
        for x in nums[2:]:
            i = bisect_left(st, x) + 1
            a = len(arr1) - tree1.query(i)
            b = len(arr2) - tree2.query(i)
            if a > b:
                arr1.append(x)
                tree1.update(i, 1)
            elif a < b:
                arr2.append(x)
                tree2.update(i, 1)
            elif len(arr1) <= len(arr2):
                arr1.append(x)
                tree1.update(i, 1)
            else:
                arr2.append(x)
                tree2.update(i, 1)
        return arr1 + arr2

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