README.md

567. Permutation in String

Given two strings s1 and s2, return true if s2 contains the permutation of s1.

In other words, one of s1's permutations is the substring of s2.

Example 1:

Input: s1 = "ab", s2 = "eidbaooo" Output: true Explanation: s2 contains one permutation of s1 ("ba"). 

Example 2:

Input: s1 = "ab", s2 = "eidboaoo" Output: false 

Constraints:

• 1 <= s1.length, s2.length <= 104
• s1 and s2 consist of lowercase English letters.

Solutions (Ruby)

1. Sliding Window

# @param {String} s1# @param {String} s2# @return {Boolean}defcheck_inclusion(s1,s2)returnfalseifs1.size > s2.sizecount1=[0] * 26count2=[0] * 26(0...s1.size).eachdo |i| count1[s1[i].ord - 97] += 1count2[s2[i].ord - 97] += 1end(0..s2.size - s1.size).eachdo |i| returntrueifcount1 == count2ifi + s1.size < s2.sizecount2[s2[i].ord - 97] -= 1count2[s2[i + s1.size].ord - 97] += 1endendfalseend

Solutions (Rust)

1. Sliding Window

implSolution{pubfncheck_inclusion(s1:String,s2:String) -> bool{if s1.len() > s2.len(){returnfalse;}let s1 = s1.as_bytes();let s2 = s2.as_bytes();letmut count1 = [0;26];letmut count2 = [0;26];for i in0..s1.len(){ count1[(s1[i] - b'a')asusize] += 1; count2[(s2[i] - b'a')asusize] += 1;}for i in0..=s2.len() - s1.len(){if count1 == count2 {returntrue;}if i + s1.len() < s2.len(){ count2[(s2[i] - b'a')asusize] -= 1; count2[(s2[i + s1.len()] - b'a')asusize] += 1;}}false}}