2570. Merge Two 2D Arrays by Summing Values - Detailed Explanation

Problem Statement

You are given two 2D integer arrays, nums1 and nums2, where each element is in the form [id, value]. Each array represents a collection of pairs with unique ids within that array. Your task is to merge these two arrays by summing the values for pairs that have the same id. The result should be a 2D array where each element is [id, totalValue], and the array is sorted in ascending order by id.

For example, if:

• nums1 = [[1, 2], [2, 3], [4, 5]]
• nums2 = [[1, 4], [3, 2], [4, 1]]

Then the merged result should be:

• [[1, 6], [2, 3], [3, 2], [4, 6]]

Explanation:

• For id = 1: 2 (from nums1) + 4 (from nums2) = 6
• For id = 2: Only present in nums1 with value 3
• For id = 3: Only present in nums2 with value 2
• For id = 4: 5 (from nums1) + 1 (from nums2) = 6

Hints

1. Using a Hash Map / Dictionary:
   You can use a hash map to record the sum of values for each id. Iterate through both arrays, update the sum for each id, and then convert the map back into a sorted 2D array.

2. Using Two Pointers (Merge Technique):
   Since each input array is sorted by id, you can use a two-pointer approach to merge them in one pass. Compare the current id from both arrays:

   • If they are equal, sum the values.
   • If one id is smaller, append that pair as is and move the corresponding pointer.
   • Continue until all pairs from both arrays are processed.

Approaches

Approach 1: Hash Map / Dictionary

• Steps:
  1. Create an empty dictionary.
  2. Iterate over each pair in nums1 and add its value to the dictionary using the id as the key.
  3. Iterate over each pair in nums2 and update the dictionary by adding to the existing value (or inserting a new key if not present).
  4. Extract the dictionary items, sort by id, and construct the final 2D array.
• Complexity:
  • Time: O(n₁ + n₂ + k log k) where n₁ and n₂ are the lengths of nums1 and nums2 respectively and k is the number of unique ids.
  • Space: O(k)

Approach 2: Two Pointers (Merge Technique)

• Steps:
  1. Initialize two pointers for nums1 and nums2.
  2. Compare the ids at the pointers:
     • If equal, sum the values, add the pair to the result, and move both pointers.
     • If one id is smaller, add that pair to the result and move that pointer.
  3. Append the remaining pairs from the array that is not yet fully processed.
• Complexity:
  • Time: O(n₁ + n₂) because each array is traversed once.
  • Space: O(1) extra space (excluding the output array).

Complexity Analysis

• Time Complexity:
  The two approaches yield efficient solutions. The hash map approach has an extra sorting step but is still efficient if the number of unique ids is small. The two pointers approach leverages the sorted order for linear time merging.

• Space Complexity:
  The hash map approach uses extra space proportional to the number of unique ids. The two pointers approach uses minimal extra space.

Python Code

Java Code