feat: add solutions to lc problem: No.1940

No.1940.Longest Common Subsequence Between Sorted Arrays

Author: yanglbme

solution/1900-1999/1940.Longest Common Subsequence Between Sorted Arrays/README.md

@@ -67,42 +67,50 @@ tags:
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：计数
+
+我们注意到，元素的范围是 $[1, 100]$，我们可以用一个长度为 $101$ 的数组 $\textit{cnt}$ 来记录每个元素出现的次数。
+
+由于数组 $\textit{arrays}$ 中的每个数组都是严格递增排序的，因此，公共子序列的元素必然是单调递增，并且这些元素的出现次数都等于 $\textit{arrays}$ 的长度。
+
+因此，我们可以遍历 $\textit{arrays}$ 中的每个数组，统计每个元素的出现次数，最后，从小到大遍历 $\textit{cnt}$ 的每个元素，如果出现次数等于 $\textit{arrays}$ 的长度，那么这个元素就是公共子序列的元素之一，我们将其加入答案数组中。
+
+遍历结束后，返回答案数组即可。
+
+时间复杂度 $O(M + N)$，空间复杂度 $O(M)$。其中 $M$ 为元素的范围，本题中 $M = 101$，而 $N$ 为数组所有元素的个数。
 
 <!-- tabs:start -->
 
 #### Python3
 
 ```python
 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]
+    def longestCommonSubsequence(self, arrays: List[List[int]]) -> List[int]:
+        cnt = [0] * 101
+        for row in arrays:
+            for x in row:
+                cnt[x] += 1
+        return [x for x, v in enumerate(cnt) if v == len(arrays)]
 ```
 
 #### Java
 
 ```java
 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);
+    public List<Integer> longestCommonSubsequence(int[][] arrays) {
+        int[] cnt = new int[101];
+        for (var row : arrays) {
+            for (int x : row) {
+                ++cnt[x];
             }
         }
-        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());
+        List<Integer> ans = new ArrayList<>();
+        for (int i = 0; i < 101; ++i) {
+            if (cnt[i] == arrays.length) {
+                ans.add(i);
             }
         }
-        return res;
+        return ans;
     }
 }
 ```
@@ -112,39 +120,60 @@ class Solution {
 ```cpp
 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);
-                }
+    vector<int> longestCommonSubsequence(vector<vector<int>>& arrays) {
+        int cnt[101]{};
+        for (const auto& row : arrays) {
+            for (int x : row) {
+                ++cnt[x];
             }
         }
-        return res;
+        vector<int> ans;
+        for (int i = 0; i < 101; ++i) {
+            if (cnt[i] == arrays.size()) {
+                ans.push_back(i);
+            }
+        }
+        return ans;
     }
 };
 ```
 
 #### Go
 
 ```go
-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)
-			}
+func longestCommonSubsequence(arrays [][]int) (ans []int) {
+	cnt := [101]int{}
+	for _, row := range arrays {
+		for _, x := range row {
+			cnt[x]++
+		}
+	}
+	for x, v := range cnt {
+		if v == len(arrays) {
+			ans = append(ans, x)
 		}
 	}
-	return res
+	return
+}
+```
+
+#### TypeScript
+
+```ts
+function longestCommonSubsequence(arrays: number[][]): number[] {
+    const cnt: number[] = Array(101).fill(0);
+    for (const row of arrays) {
+        for (const x of row) {
+            ++cnt[x];
+        }
+    }
+    const ans: number[] = [];
+    for (let i = 0; i < 101; ++i) {
+        if (cnt[i] === arrays.length) {
+            ans.push(i);
+        }
+    }
+    return ans;
 }
 ```
 
@@ -156,56 +185,24 @@ func longestCommomSubsequence(arrays [][]int) []int {
  * @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]);
+    const cnt = Array(101).fill(0);
+    for (const row of arrays) {
+        for (const x of row) {
+            ++cnt[x];
         }
     }
-    return rs;
+    const ans = [];
+    for (let i = 0; i < 101; ++i) {
+        if (cnt[i] === arrays.length) {
+            ans.push(i);
+        }
+    }
+    return ans;
 };
 ```
 
 <!-- tabs:end -->
 
 <!-- solution:end -->
 
-<!-- solution:start -->
-
-### 方法二
-
-<!-- tabs:start -->
-
-#### Python3
-
-```python
-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]
-```
-
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
 <!-- problem:end -->