algoadvance

You are given a 0-indexed integer array nums. You have to distribute all the integers in the array into two arrays array1 and array2 while respecting the following conditions:

  1. Condition 1: All elements of nums must be in either array1 or array2.
  2. Condition 2: The sum of elements in array1 must be strictly greater than the sum of elements in array2.

Return the two arrays array1 and array2 such that the mentioned conditions are satisfied. If there are multiple answers, return any of them.

Clarifying Questions

  1. Are there any constraints on the size of nums and its elements?
    • Typically, there are constraints in competitive programming problems, this helps in understanding if the given solution will perform efficiently within limits.
  2. Is there any preferred ordering for elements in the resultant arrays, or any other constraints?
    • This helps us decide if we need to maintain some order while distributing the elements.

For now, we will assume general competitive programming constraints and no specific ordering required for results.

Strategy

To tackle this problem I will:

  1. Sort the array nums in descending order.
  2. Iterate through the sorted array, and greedily distribute elements to array1 until the sum of array1 is greater than the remaining elements.
  3. Place the rest of the elements into array2.

Steps:

  1. Sort the array in descending order.
  2. Initialize two lists: array1 and array2, and their respective sums.
  3. Traverse the sorted list and add elements to array1 until its sum exceeds the remaining sum of elements.
  4. Add remaining elements to array2.

This ensures we achieve a solution where sum(array1) is likely to be strictly greater than sum(array2).

Code

def distribute_elements(nums):
    # Sort the array in descending order
    nums.sort(reverse=True)
    
    array1, array2 = [], []
    sum1, sum2 = 0, 0
    
    # Iterate over the array and distribute the elements
    for num in nums:
        if sum1 <= sum2:
            array1.append(num)
            sum1 += num
        else:
            array2.append(num)
            sum2 += num
            
    return array1, array2

Time Complexity

Thus, the overall time complexity is ( O(n \log n) ).

This approach ensures that we greedily achieve the required distribution while keeping the algorithm efficient in terms of both time and space complexity.

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