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
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
}
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
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
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
}
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.
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
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
}
