Leetcode #2363: Merge Similar Items
In this guide, we solve Leetcode #2363 Merge Similar Items in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given two 2D integer arrays, items1 and items2, representing two sets of items. Each array items has the following properties: items[i] = [valuei, weighti] where valuei represents the value and weighti represents the weight of the ith item.
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: Array, Hash Table, Ordered Set, Sorting
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: items1 = [[1,1],[4,5],[3,8]], items2 = [[3,1],[1,5]]
Output: [[1,6],[3,9],[4,5]]
Explanation:
The item with value = 1 occurs in items1 with weight = 1 and in items2 with weight = 5, total weight = 1 + 5 = 6.
The item with value = 3 occurs in items1 with weight = 8 and in items2 with weight = 1, total weight = 8 + 1 = 9.
The item with value = 4 occurs in items1 with weight = 5, total weight = 5.
Therefore, we return [[1,6],[3,9],[4,5]].
Python Solution
class Solution:
def mergeSimilarItems(
self, items1: List[List[int]], items2: List[List[int]]
) -> List[List[int]]:
cnt = Counter()
for v, w in chain(items1, items2):
cnt[v] += w
return sorted(cnt.items())
Complexity
The time complexity is and the space complexity is , where and are the lengths of items1 and items2 respectively. The space complexity is , where and are the lengths of items1 and items2 respectively.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.