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.
Input: s1 = "ab", s2 = "eidbaooo" Output: true Explanation: s2 contains one permutation of s1 ("ba"). Input: s1 = "ab", s2 = "eidboaoo" Output: false
1 <= s1.length, s2.length <= 104s1ands2consist of lowercase English letters.
# @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] += 1endendfalseendimplSolution{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}}