Find All Anagrams in a String-(Amazon, Microsoft, Flipkart):Explanation ➕ Live Coding 🧑🏻‍💻👩🏻‍💻

This is the 2nd video on Sliding Window Playlist .
We will be going through the Sliding Window in the easiest way possible and will make it one of the easiest topics.

Today, we will do our very first Qn 'Find All Anagrams in a String'

Problem Name : Find All Anagrams in a String

Company Tags 😱🤯 : Amazon, Intuit, Microsoft, Flipkart

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#coding #helpajobseeker #easyrecipes
#interviewpreparation #interview_ds_algo #hinglish

[sakshiarora]

you gave the best explanation man.
reach++

[ankitshaw_]

Same sliding window approach using unordered_map:-
class Solution {
public:
vector<int> findAnagrams(string s, string p) {
unordered_map<char, int> mp;
for(int i=0;i<p.size();i++){
mp[p[i]]++;
}
int i=0, j=0;
vector<int> ans;
unordered_map<char, int> mp2;
while(j<s.size()){
mp2[s[j]]++;
if(j-i+1<p.size()) j++;
else{
if(mp==mp2) ans.push_back(i);
/*if there is only a single frequency of a character then remove it completely from the map otherwise it will store char->0
*/
if(mp2[s[i]]==1){
mp2.erase(s[i]);
}else{
mp2[s[i]]--;
}
i++;
j++;
}
}
return ans;
}
};

[ayusharyan]

Hi Bhaiya, bhot hi clear solution tha kya ek video manacher's algorithm pe bnaoge kya. YT pe tutorials pade h but aapse achha koi ni smjha skta so please wo ek request thi

[ankitshaw_]

Java Code:-
class Solution {
boolean allzero(int[] counter){
for(int i=0;i<26;i++){
if(counter[i]!=0) return false;
}
return true;
}
public List<Integer> findAnagrams(String s, String p) {
int[] counter=new int[26];
for(int i=0;i<p.length();i++){
counter[p.charAt(i)-'a']++;
}
int i=0, j=0;
List<Integer> ans=new ArrayList<>();
while(j<s.length()){
counter[s.charAt(j)-'a']--;
if(j-i+1==p.length()){
if(allzero(counter)) ans.add(i);
counter[s.charAt(i)-'a']++;
i++;
}
j++;
}
return ans;
}
}

[jagadeeshp]

def check(self, counter):
for i in counter:
if i!=0:
return 0
return 1
def findAnagrams(self, s: str, p: str) -> List[int]:
k=len(p)
ans=[]
counter=[0 for i in range(26)]
for i in p:
counter[ord(i)-ord('a')]+=1
i=0
for j in range(len(s)):

while(j-i+1==k):
if self.check(counter):
ans.append(i)

i+=1
return ans