> 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/2022/06/2022-06-01.md). # 2022-06-01 ## [438. Find All Anagrams in a String](https://leetcode.com/problems/find-all-anagrams-in-a-string/) ### Description Given two strings `s` and `p`, return *an array of all the start indices of* `p`*'s anagrams in* `s`. You may return the answer in **any order**. An **Anagram** is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once. **Example 1:** ``` Input: s = "cbaebabacd", p = "abc" Output: [0,6] Explanation: The substring with start index = 0 is "cba", which is an anagram of "abc". The substring with start index = 6 is "bac", which is an anagram of "abc". ``` **Example 2:** ``` Input: s = "abab", p = "ab" Output: [0,1,2] Explanation: The substring with start index = 0 is "ab", which is an anagram of "ab". The substring with start index = 1 is "ba", which is an anagram of "ab". The substring with start index = 2 is "ab", which is an anagram of "ab". ``` **Constraints:** * `1 <= s.length, p.length <= 3 * 104` * `s` and `p` consist of lowercase English letters. ### Solution #### Approach #0 ```go func findAnagrams(s string, p string) (ans []int) { target, current := make([]int, 26), make([]int, 26) count := 0 for _, ch := range p { target[ch-'a']++ } for _, t := range target { if t != 0 { count++ } } i, j, sig := 0, 0, 0 for j < len(s) { in := s[j] - 'a' j++ if target[in] > 0 { current[in]++ if target[in] == current[in] { sig++ } } for j-i >= len(p) { if sig == count { ans = append(ans, i) } out := s[i] - 'a' i++ if target[out] > 0 { if current[out] == target[out] { sig-- } current[out]-- } } } return ans } ``` ## [713. Subarray Product Less Than K](https://leetcode.com/problems/subarray-product-less-than-k/) ### Description Given an array of integers `nums` and an integer `k`, return *the number of contiguous subarrays where the product of all the elements in the subarray is strictly less than* `k`. **Example 1:** ``` Input: nums = [10,5,2,6], k = 100 Output: 8 Explanation: The 8 subarrays that have product less than 100 are: [10], [5], [2], [6], [10, 5], [5, 2], [2, 6], [5, 2, 6] Note that [10, 5, 2] is not included as the product of 100 is not strictly less than k. ``` **Example 2:** ``` Input: nums = [1,2,3], k = 0 Output: 0 ``` **Constraints:** * `1 <= nums.length <= 3 * 10^4` * `1 <= nums[i] <= 1000` * `0 <= k <= 10^6` ### Solution #### Approach #0 ```go func numSubarrayProductLessThanK(nums []int, k int) (ans int) { cur := 1 i, j := 0, 0 for i < len(nums) { if j >= len(nums) { cur = 1 i++ j = i continue } in := nums[j] if cur*in < k { ans++ cur = cur * in j++ } else { cur = 1 i++ j = i } } return ans } ``` #### Approach #1 ```go func numSubarrayProductLessThanK(nums []int, k int) (ans int) { i, cur := 0, 1 for j, num := range nums { cur *= num for ; i <= j && cur >= k; i++ { cur /= nums[i] } ans += j - i + 1 } return } ``` ## [209. Minimum Size Subarray Sum](https://leetcode.com/problems/minimum-size-subarray-sum/) ### Description Given an array of positive integers `nums` and a positive integer `target`, return the minimal length of a **contiguous subarray** `[numsl, numsl+1, ..., numsr-1, numsr]` of which the sum is greater than or equal to `target`. If there is no such subarray, return `0` instead. **Example 1:** ``` Input: target = 7, nums = [2,3,1,2,4,3] Output: 2 Explanation: The subarray [4,3] has the minimal length under the problem constraint. ``` **Example 2:** ``` Input: target = 4, nums = [1,4,4] Output: 1 ``` **Example 3:** ``` Input: target = 11, nums = [1,1,1,1,1,1,1,1] Output: 0 ``` **Constraints:** * `1 <= target <= 10^9` * `1 <= nums.length <= 10^5` * `1 <= nums[i] <= 10^5` **Follow up:** If you have figured out the `O(n)` solution, try coding another solution of which the time complexity is `O(n log(n))`. ### Solution #### Approach #0 ```go func minSubArrayLen(target int, nums []int) (ans int) { ans = len(nums) + 1 i, cur := 0, 0 for j, num := range nums { cur += num for ; i <= j && cur >= target; i++ { if cur >= target { ans = min(ans, j-i+1) } cur -= nums[i] } } if ans == len(nums)+1 { ans = 0 } return } func min(a, b int) int { if a < b { return a } return b } ``` --- # Agent Instructions This documentation is published with GitBook. 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