Minimum Window Substring - Two pointers #1 | Coding Interview

[@alperozdamar517]

This is a great explanation. I'm so surprised that nobody wrote any comment about your solution and explanation. Thank you for a clear and easy solution.

[@InvolveInInnovation]

Thanks for your kind words! Please share it with your friends if you find this video helpful.

[@Demigod23393]

The second while loop should be while(l

[@arunitbaidya6118]

The most simple and clear solution I found!

[@Deep-zk6dl]

This is the best explaination for this problem on youtube. Thank you so much 😊

[@purnimavijaya1231]

The best explanation I found for this problem!

[@shaikhirfan2799]

I am getting error in if condition it's asking that cannot find variable (invalue()) this is showing me cannot find

[@amanr11314]

Thanks man you saved me lots of time

[@anilkishorereddy9246]

Great explaination

[@GAURAVGUPTA-zu2bu]

What is the need of if statement res[0]==-1 || res[0]>r-l+1 in 2nd while loop.I didn't get?

[@InvolveInInnovation]

That is check and replace the result only if result is not already found (res[0]==-1, this is what we initialized), second is to check if the newly found one is minimum then the existing one(res[0]>r-l+1).

[@iark1]

Very nice explanation! Thanks a lot!

[@anushacheedella2291]

Very great explanation 👌

[@vishnukhanna9246]

Best explanation

[@rodrickedwards2759]

Great explanation. Thanks.

[@animeshsaxena3626]

Simplified approach ! :)

[@shaikhirfan2799]

I am doing in normal editor with main method

[@matpoliquin]

Well explained and concise! Thank you!

[@mayanksikarwal8431]

nice explanation bro.

[@kaushalrajsinghbhati5076]

Good explanation. Pronunciation and tone needs a bit of improvement btw.

[@premchandru05]

s = "aa" and t= "aa", this test case is not working

[@premchandru05]

Please check my working solution and please correct me if I am wrong.
class Solution {
public String minWindow(String s, String t) {
if(s ==null || t==null || s.length()==0 || t.length() == 0 || s.length() < t.length())
return "";
Map window = new HashMap();
for(char c:t.toCharArray())
window.put(c,window.getOrDefault(c,0)+1);
Map minWindow = new HashMap();
int l =0;
int r =0;
int expe = 0;
int real = t.length();
int[] output = {-1,0,0};
while(r < s.length()){
char c = s.charAt(r);
minWindow.put(c,minWindow.getOrDefault(c, 0)+1);
if(window.containsKey(c) && minWindow.get(c).intValue() (r-l)+1){
output[0] = (r-l)+1;
output[1] = l;
output[2] = r;
}
char cc = s.charAt(l);
minWindow.put(cc,minWindow.getOrDefault(cc, 0)-1);
if(window.containsKey(cc) && minWindow.get(cc).intValue() < window.get(cc).intValue()){
expe--;
}
l++;
}
r++;
}
return (output[0]==-1?"":s.substring(output[1],output[2]+1));
}
}

[@maidul13]

Because this solution only works if t only contains unique chars.

[@RiteshSingh-ru1sk]

The simplest explaination u will find on internet..