2570. Merge Two 2D Arrays by Summing Values

Description

You are given two 2D integer arrays nums1 and nums2.

• nums1[i] = [idi, vali] indicate that the number with the id idi has a value equal to vali.
• nums2[i] = [idi, vali] indicate that the number with the id idi has a value equal to vali.

Each array contains unique ids and is sorted in ascending order by id.

Merge the two arrays into one array that is sorted in ascending order by id, respecting the following conditions:

• Only ids that appear in at least one of the two arrays should be included in the resulting array.
• Each id should be included only once and its value should be the sum of the values of this id in the two arrays. If the id does not exist in one of the two arrays then its value in that array is considered to be 0.

Return the resulting array. The returned array must be sorted in ascending order by id.

 

Example 1:

Input: nums1 = [[1,2],[2,3],[4,5]], nums2 = [[1,4],[3,2],[4,1]]
Output: [[1,6],[2,3],[3,2],[4,6]]
Explanation: The resulting array contains the following:
- id = 1, the value of this id is 2 + 4 = 6.
- id = 2, the value of this id is 3.
- id = 3, the value of this id is 2.
- id = 4, the value of this id is 5 + 1 = 6.

Example 2:

Input: nums1 = [[2,4],[3,6],[5,5]], nums2 = [[1,3],[4,3]]
Output: [[1,3],[2,4],[3,6],[4,3],[5,5]]
Explanation: There are no common ids, so we just include each id with its value in the resulting list.

 

Constraints:

• 1 <= nums1.length, nums2.length <= 200
• nums1[i].length == nums2[j].length == 2
• 1 <= idi, vali <= 1000
• Both arrays contain unique ids.
• Both arrays are in strictly ascending order by id.

Solutions

Solution 1: Counting + Enumeration

We can use a hash table or an array cnt to count the frequency of each number in the two arrays.

Then we enumerate each number in cnt from small to large. If the frequency of a number is greater than $0$, we add it to the answer array.

The time complexity is $O(n + m)$, and the space complexity is $O(M)$. Where $n$ and $m$ are the lengths of the two arrays respectively; and $M$ is the maximum value in the two arrays, in this problem, $M = 1000$.

Python3JavaC++GoTypeScriptRust

class Solution:
    def mergeArrays(
        self, nums1: List[List[int]], nums2: List[List[int]]
    ) -> List[List[int]]:
        cnt = Counter()
        for i, v in nums1 + nums2:
            cnt[i] += v
        return sorted(cnt.items())

class Solution {
    public int[][] mergeArrays(int[][] nums1, int[][] nums2) {
        int[] cnt = new int[1001];
        for (var x : nums1) {
            cnt[x[0]] += x[1];
        }
        for (var x : nums2) {
            cnt[x[0]] += x[1];
        }
        int n = 0;
        for (int i = 0; i < 1001; ++i) {
            if (cnt[i] > 0) {
                ++n;
            }
        }
        int[][] ans = new int[n][2];
        for (int i = 0, j = 0; i < 1001; ++i) {
            if (cnt[i] > 0) {
                ans[j++] = new int[] {i, cnt[i]};
            }
        }
        return ans;
    }
}

class Solution {
public:
    vector<vector<int>> mergeArrays(vector<vector<int>>& nums1, vector<vector<int>>& nums2) {
        int cnt[1001]{};
        for (auto& x : nums1) {
            cnt[x[0]] += x[1];
        }
        for (auto& x : nums2) {
            cnt[x[0]] += x[1];
        }
        vector<vector<int>> ans;
        for (int i = 0; i < 1001; ++i) {
            if (cnt[i]) {
                ans.push_back({i, cnt[i]});
            }
        }
        return ans;
    }
};

func mergeArrays(nums1 [][]int, nums2 [][]int) (ans [][]int) {
    cnt := [1001]int{}
    for _, x := range nums1 {
        cnt[x[0]] += x[1]
    }
    for _, x := range nums2 {
        cnt[x[0]] += x[1]
    }
    for i, x := range cnt {
        if x > 0 {
            ans = append(ans, []int{i, x})
        }
    }
    return
}

function mergeArrays(nums1: number[][], nums2: number[][]): number[][] {
    const n = 1001;
    const cnt = new Array(n).fill(0);
    for (const [a, b] of nums1) {
        cnt[a] += b;
    }
    for (const [a, b] of nums2) {
        cnt[a] += b;
    }
    const ans: number[][] = [];
    for (let i = 0; i < n; ++i) {
        if (cnt[i] > 0) {
            ans.push([i, cnt[i]]);
        }
    }
    return ans;
}

impl Solution {
    pub fn merge_arrays(nums1: Vec<Vec<i32>>, nums2: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
        let mut cnt = vec![0; 1001];

        for x in &nums1 {
            cnt[x[0] as usize] += x[1];
        }

        for x in &nums2 {
            cnt[x[0] as usize] += x[1];
        }

        let mut ans = vec![];
        for i in 0..cnt.len() {
            if cnt[i] > 0 {
                ans.push(vec![i as i32, cnt[i] as i32]);
            }
        }

        ans
    }
}

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