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Formatted question description: https://leetcode.ca/all/2363.html

# 2363. Merge Similar Items

• Difficulty: Easy.
• Related Topics: Array, Hash Table, Sorting, Ordered Set.
• Similar Questions: .

## Problem

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.

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 <= valuei, weighti <= 1000

• Each valuei in items1 is unique.

• Each valuei in items2 is unique.

## Solution (Java, C++, Python)

• class Solution {
public List<List<Integer>> mergeSimilarItems(int[][] arr1, int[][] arr2) {
int[] cache = new int[1001];
for (int[] num : arr1) {
cache[num[0]] += num[1];
}
for (int[] num : arr2) {
cache[num[0]] += num[1];
}
List<List<Integer>> result = new ArrayList<>();
for (int i = 0; i < cache.length; i++) {
int weight = cache[i];
if (weight > 0) {
int value = i;
}
}
return result;
}
}

############

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) {
}
}
return ans;
}
}

• class Solution:
def mergeSimilarItems(
self, items1: List[List[int]], items2: List[List[int]]
) -> List[List[int]]:
cnt = [0] * 1010
for v, w in chain(items1, items2):
cnt[v] += w
return [[i, v] for i, v in enumerate(cnt) if v]

############

# 2363. Merge Similar Items
# https://leetcode.com/problems/merge-similar-items/

class Solution:
def mergeSimilarItems(self, items1: List[List[int]], items2: List[List[int]]) -> List[List[int]]:
mp = defaultdict(int)

for value, weight in items1 + items2:
mp[value] += weight

A = list(mp.items())
A.sort(key = lambda x : x[0])

return [[v, w] for v, w in A]


• class Solution {
public:
vector<vector<int>> mergeSimilarItems(vector<vector<int>>& items1, vector<vector<int>>& items2) {
int cnt[1010]{};
for (auto& x : items1) {
cnt[x[0]] += x[1];
}
for (auto& x : items2) {
cnt[x[0]] += x[1];
}
vector<vector<int>> ans;
for (int i = 0; i < 1010; ++i) {
if (cnt[i]) {
ans.push_back({i, cnt[i]});
}
}
return ans;
}
};

• 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
}

• 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);
}


• 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()
}
}



Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).