1940. Longest Common Subsequence Between Sorted Arrays

Description

Given an array of integer arrays arrays where each arrays[i] is sorted in strictly increasing order, return an integer array representing the longest common subsequence between all the arrays.

A subsequence is a sequence that can be derived from another sequence by deleting some elements (possibly none) without changing the order of the remaining elements.

 

Example 1:

Input: arrays = [[1,3,4],
                 [1,4,7,9]]
Output: [1,4]
Explanation: The longest common subsequence in the two arrays is [1,4].

Example 2:

Input: arrays = [[2,3,6,8],
                 [1,2,3,5,6,7,10],
                 [2,3,4,6,9]]
Output: [2,3,6]
Explanation: The longest common subsequence in all three arrays is [2,3,6].

Example 3:

Input: arrays = [[1,2,3,4,5],
                 [6,7,8]]
Output: []
Explanation: There is no common subsequence between the two arrays.

 

Constraints:

  • 2 <= arrays.length <= 100
  • 1 <= arrays[i].length <= 100
  • 1 <= arrays[i][j] <= 100
  • arrays[i] is sorted in strictly increasing order.

Solutions

Solution 1

Python Code
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class Solution:
    def longestCommomSubsequence(self, arrays: List[List[int]]) -> List[int]:
        n = len(arrays)
        counter = defaultdict(int)
        for array in arrays:
            for e in array:
                counter[e] += 1
        return [e for e, count in counter.items() if count == n]

Java Code
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class Solution {
    public List<Integer> longestCommomSubsequence(int[][] arrays) {
        Map<Integer, Integer> counter = new HashMap<>();
        for (int[] array : arrays) {
            for (int e : array) {
                counter.put(e, counter.getOrDefault(e, 0) + 1);
            }
        }
        int n = arrays.length;
        List<Integer> res = new ArrayList<>();
        for (Map.Entry<Integer, Integer> entry : counter.entrySet()) {
            if (entry.getValue() == n) {
                res.add(entry.getKey());
            }
        }
        return res;
    }
}

C++ Code
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class Solution {
public:
    vector<int> longestCommomSubsequence(vector<vector<int>>& arrays) {
        unordered_map<int, int> counter;
        vector<int> res;
        int n = arrays.size();
        for (auto array : arrays) {
            for (auto e : array) {
                counter[e] += 1;
                if (counter[e] == n) {
                    res.push_back(e);
                }
            }
        }
        return res;
    }
};

Go Code
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func longestCommomSubsequence(arrays [][]int) []int {
	counter := make(map[int]int)
	n := len(arrays)
	var res []int
	for _, array := range arrays {
		for _, e := range array {
			counter[e]++
			if counter[e] == n {
				res = append(res, e)
			}
		}
	}
	return res
}

JavaScript Code
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/**
 * @param {number[][]} arrays
 * @return {number[]}
 */
var longestCommonSubsequence = function (arrays) {
    const m = new Map();
    const rs = [];
    const len = arrays.length;
    for (let i = 0; i < len; i++) {
        for (let j = 0; j < arrays[i].length; j++) {
            m.set(arrays[i][j], (m.get(arrays[i][j]) || 0) + 1);
            if (m.get(arrays[i][j]) === len) rs.push(arrays[i][j]);
        }
    }
    return rs;
};

Solution 2

Python Code
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class Solution:
    def longestCommomSubsequence(self, arrays: List[List[int]]) -> List[int]:
        def common(l1, l2):
            i, j, n1, n2 = 0, 0, len(l1), len(l2)
            res = []
            while i < n1 and j < n2:
                if l1[i] == l2[j]:
                    res.append(l1[i])
                    i += 1
                    j += 1
                elif l1[i] > l2[j]:
                    j += 1
                else:
                    i += 1
            return res

        n = len(arrays)
        for i in range(1, n):
            arrays[i] = common(arrays[i - 1], arrays[i])
        return arrays[n - 1]