76. Minimum Window Substring || NEETCODE 150
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- Опубликовано: 17 сен 2024
- 76. Minimum Window Substring || NEETCODE 150
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NEETCODE 150 QUESTIONS:- • Contains Duplicate - L...
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In if(mp.get(curr) == 0), it should have been a >= condition instead of ==. I have verified that == case fails because it doesn't account for same character repeating in t.
you can check with s = "aa" and t = "aa"
Done ✅✅
Optimized Approach
class Solution {
public String minWindow(String s, String t) {
Map mp = new HashMap();
int start = 0;
int end = 0;
int i = 0;
int j = 0;
int minSize = Integer.MAX_VALUE;
for (char ch : t.toCharArray()) {
mp.put(ch, mp.getOrDefault(ch, 0) + 1);
}
int n = mp.size();
while (j < s.length()) {
char curr = s.charAt(j);
mp.put(curr, mp.getOrDefault(curr, 0) - 1);
if (mp.get(curr) == 0) {
n--;
}
while (n == 0) {
if (minSize > j - i + 1) {
minSize = j - i + 1;
start = i;
}
mp.put(s.charAt(i), mp.getOrDefault(s.charAt(i), 0) + 1);
if (mp.get(s.charAt(i)) > 0) {
n++;
}
i++;
}
j++;
}
return minSize == Integer.MAX_VALUE ? "" : s.substring(start, start + minSize);
}
}
No worry 😊😊
Pata nahi mere jeevan me aap jaisi ladki h ki nahi. At the age of 26 bohot haara haara mehsus karta hu.😔
Brute force approach
class Solution {
public String minWindow(String s, String t) {
if (t.length() > s.length()) {
return "";
}
int minLength = Integer.MAX_VALUE;
String minWindow = "";
for (int i = 0; i < s.length(); i++) {
for (int j = i + t.length(); j 0) {
return false;
}
}
return true;
}
}
Me Bsss yaha inko dekhne aaata hu , padhai se to mera vese bhi dur dur tak koi lena dena nhi h...
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mujhe software eng banna hai, walmart main naukri karni hai, 27 lakh ka package h, video bna do codehers pe