LeetCode 290. Word Pattern Solution Explained - Java

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[NoWarForever]

map.containsValue has O(N) time complexity if instead of 1 Map you use 2 Maps you can improve the time complexity while sacrificing the space

[wolfofthelight]

One thing I had an issue with here was that the pattern length must equal the string length. I passed the leetcode problem before coming here, but it was the last issue I went into. The question is ambiguous. If it is a pattern, there is no reason the string cannot be longer. For example "abba" and "dog cat cat dog dog cat cat dog". In my original solution to the problem, I accounted for this.... the sentence still "fits" the pattern. The problem does not accept this, however, and I removed it and passed the leetcode problem easily. The reason I comment is, if you are asked this in an interview or something, be prepared to ask whether the pattern can recur or if its length is an actual integral part (in which case, it really isn't a pattern so much as simply a one-to-one mapping).

[pushpendrasingh]

Tried this same algorithm using the c++ was not able to get it but now it seems easy...

[samandarboymurodov]

Thank you. it is clean and easy to understand

[ramsharma]

doest looks a "easy " problem to me

[sriramsatvikatukuri]

class Solution {
public boolean wordPattern(String pattern, String str) {
String []sa=str.split(" ");
if(pattern == null || sa.length!=pattern.length())
return false;

int n=pattern.length();
Map<Character, Integer> m=new HashMap<>();
Map<String, Integer> m1=new HashMap<>();

for(int i=0;i<n;i++)
{
char ch=pattern.charAt(i);
String s=sa[i];

if(m.containsKey(ch) && m1.containsKey(s))
{
if(m.get(ch)!=m1.get(s))
return false;
}
else if(m.containsKey(ch) || m1.containsKey(s))
return false;

m.put(ch, i+1);
m1.put(s, i+1);
}
return true;
}
}

what's the mistake in this code can u please tell

[aayushpagare]

It's a request please don't use Java ....use c++ in your videos🙏🙏🙏🙏🙏🙏