LeetCode #2363 — EASY

Merge Similar Items

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

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] = [value_i, weight_i] where value_i represents the value and weight_i represents the weight of the i^th item.
The value of each item in items is unique.

Return a 2D integer array ret where ret[i] = [value_i, weight_i], with weight_i being the sum of weights of all items with value value_i.

Note: ret should be returned in ascending order by value.

Example 1:

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]].

Example 2:

Input: items1 = [[1,1],[3,2],[2,3]], items2 = [[2,1],[3,2],[1,3]]
Output: [[1,4],[2,4],[3,4]]
Explanation: 
The item with value = 1 occurs in items1 with weight = 1 and in items2 with weight = 3, total weight = 1 + 3 = 4.
The item with value = 2 occurs in items1 with weight = 3 and in items2 with weight = 1, total weight = 3 + 1 = 4.
The item with value = 3 occurs in items1 with weight = 2 and in items2 with weight = 2, total weight = 2 + 2 = 4.
Therefore, we return [[1,4],[2,4],[3,4]].

Example 3:

Input: items1 = [[1,3],[2,2]], items2 = [[7,1],[2,2],[1,4]]
Output: [[1,7],[2,4],[7,1]]
Explanation:
The item with value = 1 occurs in items1 with weight = 3 and in items2 with weight = 4, total weight = 3 + 4 = 7. 
The item with value = 2 occurs in items1 with weight = 2 and in items2 with weight = 2, total weight = 2 + 2 = 4. 
The item with value = 7 occurs in items2 with weight = 1, total weight = 1.
Therefore, we return [[1,7],[2,4],[7,1]].

Constraints:

1 <= items1.length, items2.length <= 1000
items1[i].length == items2[i].length == 2
1 <= value_i, weight_i <= 1000
Each value_i in items1 is unique.
Each value_i in items2 is unique.

Step 01

Brute Force Baseline

Problem summary: 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. The value of each item in items is unique. Return a 2D integer array ret where ret[i] = [valuei, weighti], with weighti being the sum of weights of all items with value valuei. Note: ret should be returned in ascending order by value.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Segment Tree

Example 1

[[1,1],[4,5],[3,8]]
[[3,1],[1,5]]

Example 2

[[1,1],[3,2],[2,3]]
[[2,1],[3,2],[1,3]]

Example 3

[[1,3],[2,2]]
[[7,1],[2,2],[1,4]]

Core Insight

What unlocks the optimal approach

Map the weights using the corresponding values as keys.
Make sure your output is sorted in ascending order by value.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2363: Merge Similar Items
class Solution {
    public List<List<Integer>> mergeSimilarItems(int[][] items1, int[][] items2) {
        int[] cnt = new int[1010];
        for (var x : items1) {
            cnt[x[0]] += x[1];
        }
        for (var x : items2) {
            cnt[x[0]] += x[1];
        }
        List<List<Integer>> ans = new ArrayList<>();
        for (int i = 0; i < cnt.length; ++i) {
            if (cnt[i] > 0) {
                ans.add(List.of(i, cnt[i]));
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2363: Merge Similar Items
func mergeSimilarItems(items1 [][]int, items2 [][]int) (ans [][]int) {
	cnt := [1010]int{}
	for _, x := range items1 {
		cnt[x[0]] += x[1]
	}
	for _, x := range items2 {
		cnt[x[0]] += x[1]
	}
	for i, x := range cnt {
		if x > 0 {
			ans = append(ans, []int{i, x})
		}
	}
	return
}

# Accepted solution for LeetCode #2363: Merge Similar Items
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())

// Accepted solution for LeetCode #2363: Merge Similar Items
impl Solution {
    pub fn merge_similar_items(items1: Vec<Vec<i32>>, items2: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
        let mut count = [0; 1001];
        for item in items1.iter() {
            count[item[0] as usize] += item[1];
        }
        for item in items2.iter() {
            count[item[0] as usize] += item[1];
        }
        count
            .iter()
            .enumerate()
            .filter_map(|(i, &v)| {
                if v == 0 {
                    return None;
                }
                Some(vec![i as i32, v])
            })
            .collect()
    }
}

// Accepted solution for LeetCode #2363: Merge Similar Items
function mergeSimilarItems(items1: number[][], items2: number[][]): number[][] {
    const count = new Array(1001).fill(0);
    for (const [v, w] of items1) {
        count[v] += w;
    }
    for (const [v, w] of items2) {
        count[v] += w;
    }
    return [...count.entries()].filter(v => v[1] !== 0);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n + m)

Space

O(n + m)

Approach Breakdown

BRUTE FORCE

O(n × q) time

O(1) space

For each of q queries, scan the entire range to compute the aggregate (sum, min, max). Each range scan takes up to O(n) for a full-array query. With q queries: O(n × q) total. Point updates are O(1) but queries dominate.

SEGMENT TREE

O(n + q log n) time

O(n) space

Building the tree is O(n). Each query or update traverses O(log n) nodes (tree height). For q queries: O(n + q log n) total. Space is O(4n) ≈ O(n) for the tree array. Lazy propagation adds O(1) per node for deferred updates.

Shortcut: Build O(n), query/update O(log n) each. When you need both range queries AND point updates.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.