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Leetcode 3069. Distribute Elements Into Two Arrays I

Problem Statement

Given an array nums and an integer k, your task is to distribute the nums array’s elements into two arrays a and b such that:

  1. Each array contains exactly k elements.
  2. Maximize the sum of absolute differences between the elements of a and the elements of b.

Clarifying Questions

  1. Constraints on the elements of nums
    • Are the elements of nums negative, non-negative, or can they be both?
    • If nums array length is less than 2k, how should we handle it? (assuming input constraint)
  2. Output Format
    • Should the function return the two arrays a and b, or just the maximum sum of absolute differences?
  3. Input Constraints
    • Maximum possible size of nums?
    • Range of values for nums and k?

Assuming that elements can be positive or negative, the size of nums is guaranteed to be at least 2k, and we need to return exactly two arrays a and b.

Strategy

  1. Sort the nums array in descending order to simplify the selection of the largest differences.
  2. Select elements for arrays a and b:
    • One way to potentially maximize the sum of absolute differences is to place the k largest elements in one array (a) and the next k largest elements in another array (b). This will maximize the differences because we are pairing large numbers against the next set of smaller numbers.
  3. Compute the sum of absolute differences:
    • Having selected the elements for a and b, iterate through both arrays and sum the absolute differences.

Code

Here’s a C++ implementation based on the above strategy:

#include <iostream>
#include <vector>
#include <algorithm>

std::pair<std::vector<int>, std::vector<int>> distributeElements(std::vector<int>& nums, int k) {
    // Sort the array in descending order
    std::sort(nums.begin(), nums.end(), std::greater<int>());

    std::vector<int> a(nums.begin(), nums.begin() + k);
    std::vector<int> b(nums.begin() + k, nums.begin() + 2 * k);

    return {a, b};
}

int main() {
    std::vector<int> nums = {1, 3, 2, 2, 5, 2, 7, 8, 1, 9, 10, 6};
    int k = 3;

    auto result = distributeElements(nums, k);
    std::vector<int> a = result.first;
    std::vector<int> b = result.second;

    std::cout << "Array a: ";
    for (const auto& num : a) {
        std::cout << num << " ";
    }
    std::cout << "\nArray b: ";
    for (const auto& num : b) {
        std::cout << num << " ";
    }

    return 0;
}

Time Complexity

The time complexity for sorting the array is (O(n \log n)), where (n) is the size of the nums array. The selection of elements takes linear time, i.e., (O(n)). Hence, the overall time complexity is dominated by the sorting step, making it (O(n \log n)).

Space Complexity

The space complexity is (O(1)) additional space if not counting the storage for the arrays a and b, making it essentially (O(n)) accounting for the inputs and outputs together.

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