> For the complete documentation index, see [llms.txt](https://a.b.cr/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://a.b.cr/dictionary/algorithm/diary/2022/06/2022-06-12.md). # 2022-06-12 ## [300. Longest Increasing Subsequence](https://leetcode.com/problems/longest-increasing-subsequence/) ### Description Given an integer array `nums`, return the length of the longest strictly increasing subsequence. A **subsequence** is a sequence that can be derived from an array by deleting some or no elements without changing the order of the remaining elements. For example, `[3,6,2,7]` is a subsequence of the array `[0,3,1,6,2,2,7]`. **Example 1:** ``` Input: nums = [10,9,2,5,3,7,101,18] Output: 4 Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4. ``` **Example 2:** ``` Input: nums = [0,1,0,3,2,3] Output: 4 ``` **Example 3:** ``` Input: nums = [7,7,7,7,7,7,7] Output: 1 ``` **Constraints:** * `1 <= nums.length <= 2500` * `-10^4 <= nums[i] <= 10^4` **Follow up:** Can you come up with an algorithm that runs in `O(n log(n))` time complexity? ### Solution #### Approach #0 ```go func lengthOfLIS(nums []int) (ans int) { n := len(nums) dp := make([]int, n) dp[0] = 1 ans = 1 for i := 1; i < n; i++ { dp[i] = 1 for j := 0; j < i; j++ { if nums[i] > nums[j] { dp[i] = max(dp[i], dp[j]+1) } } ans = max(ans, dp[i]) } return } func max(a, b int) int { if a > b { return a } return b } ``` ## [673. Number of Longest Increasing Subsequence](https://leetcode.com/problems/number-of-longest-increasing-subsequence/) ### Description Given an integer array `nums`, return *the number of longest increasing subsequences.* **Notice** that the sequence has to be **strictly** increasing. **Example 1:** ``` Input: nums = [1,3,5,4,7] Output: 2 Explanation: The two longest increasing subsequences are [1, 3, 4, 7] and [1, 3, 5, 7]. ``` **Example 2:** ``` Input: nums = [2,2,2,2,2] Output: 5 Explanation: The length of longest continuous increasing subsequence is 1, and there are 5 subsequences' length is 1, so output 5. ``` **Constraints:** * `1 <= nums.length <= 2000` * `-10^6 <= nums[i] <= 10^6` ### Solution #### Approach #0 ```go func findNumberOfLIS(nums []int) (ans int) { n := len(nums) dp := make([]int, n) cnt := make([]int, n) maxLen := 0 for i := 0; i < n; i++ { dp[i] = 1 cnt[i] = 1 for j := 0; j < i; j++ { if nums[i] > nums[j] { if dp[j]+1 == dp[i] { cnt[i] += cnt[j] } if dp[j]+1 > dp[i] { dp[i] = dp[j] + 1 cnt[i] = cnt[j] } } } if dp[i] > maxLen { maxLen = dp[i] ans = cnt[i] } else if dp[i] == maxLen { ans += cnt[i] } } return } ``` ## [890. Find and Replace Pattern](https://leetcode.com/problems/find-and-replace-pattern/) ### Description Given a list of strings `words` and a string `pattern`, return *a list of* `words[i]` *that match* `pattern`. You may return the answer in **any order**. A word matches the pattern if there exists a permutation of letters `p` so that after replacing every letter `x` in the pattern with `p(x)`, we get the desired word. Recall that a permutation of letters is a bijection from letters to letters: every letter maps to another letter, and no two letters map to the same letter. **Example 1:** ``` Input: words = ["abc","deq","mee","aqq","dkd","ccc"], pattern = "abb" Output: ["mee","aqq"] Explanation: "mee" matches the pattern because there is a permutation {a -> m, b -> e, ...}. "ccc" does not match the pattern because {a -> c, b -> c, ...} is not a permutation, since a and b map to the same letter. ``` **Example 2:** ``` Input: words = ["a","b","c"], pattern = "a" Output: ["a","b","c"] ``` **Constraints:** * `1 <= pattern.length <= 20` * `1 <= words.length <= 50` * `words[i].length == pattern.length` * `pattern` and `words[i]` are lowercase English letters. ### Solution #### Approach #0 ```go func findAndReplacePattern(words []string, pattern string) (ans []string) { if len(pattern) == 1 { return words } for _, word := range words { if match(word, pattern) && match(pattern, word) { ans = append(ans, word) } } return } func match(word, pattern string) bool { m := make(map[rune]byte) for i, a := range word { b := pattern[i] if m[a] == 0 { m[a] = b } else { if m[a] != b { return false } } } return true } ``` --- # Agent Instructions This documentation is published with GitBook. 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