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2363. Merge Similar Items

Description

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.

Solutions

  • 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;
        }
    }
    
  • 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;
        }
    };
    
  • 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())
    
    
  • 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()
        }
    }
    
    

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