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Given a string s and a non-empty string p, find all the start indices of p's anagrams in s.
Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.
The order of output does not matter.
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".
The text was updated successfully, but these errors were encountered:
varfindAnagrams=function(s,p){constoutput=[];for(leti=0;i<s.length-p.length+1;i++){if(p.indexOf(s[i])>-1){letinitial=i;//console.log('initial', i)letclone=p.slice();console.log(clone)//console.∆log('original clone', clone)for(letj=i;j<p.length+i;j++){letexists=clone.indexOf(s[j]);if(exists>-1){clone=clone.slice(0,exists)+clone.slice(exists+1);}}//console.log('after', clone)if(clone.length===0){output.push(initial);}}}returnoutput;};//Alternative://Create a freq object for target string//Build up freq object for each substring//Compare and output.push(initial) if equal//{a: 1, b: 2, c:1}
Given a string s and a non-empty string p, find all the start indices of p's anagrams in s.
Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.
The order of output does not matter.
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".
The text was updated successfully, but these errors were encountered: