Merge Similar Items LeetCode Solution

Problem – Merge Similar Items LeetCode Solution

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.

Merge Similar Items LeetCode Solution in C++

vector<vector<int>> mergeSimilarItems(vector<vector<int>>& items1, vector<vector<int>>& items2) 
    {
        map<int,int> m;
        for(int i=0;i<items1.size();i++)
        {
            m[items1[i][0]]=items1[i][1];
        }
        for(int i=0;i<items2.size();i++)
        {
            if(m.find(items2[i][0])!=m.end())
            {
                m[items2[i][0]]+=items2[i][1];
            }
            else
            {
                m[items2[i][0]]=items2[i][1];
            }
        }
        vector<vector<int>> ans;
        for(auto it : m)
        {
            ans.push_back({it.first,it.second});
        }
        return ans;
        
    }
	```

Merge Similar Items LeetCode Solution in Java

    public List<List<Integer>> mergeSimilarItems(int[][] items1, int[][] items2) {
        TreeMap<Integer, Integer> cnt = new TreeMap<>();
        for (int[] it : items1) {
            cnt.merge(it[0], it[1], Integer::sum);
        }
        for (int[] it : items2) {
            cnt.merge(it[0], it[1], Integer::sum);
        }
        List<List<Integer>> ans = new ArrayList<>();
        for (var e : cnt.entrySet()) {
            ans.add(Arrays.asList(e.getKey(), e.getValue()));
        }
        return ans;
    }

Merge Similar Items LeetCode Solution in Python

class Solution:
    from collections import defaultdict
    def mergeSimilarItems(self, items1: List[List[int]], items2: List[List[int]]) -> List[List[int]]:
		
		# The idea here was to create a hashmap of value to weight mapping
        # iterate through items1 and populate the hashmap
		value_weight = defaultdict(int)
        for i, j in items1:
            value_weight[i] = j
		
        # compare values with hashmap and update the weights
		for i, j in items2:
            value_weight[i] += j
			
		# return the hashmap in list format
        ans = []
        for value in sorted(value_weight.keys()):
            ans.append([value, value_weight[value]])
        return ans

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