Permutation in String || NEETCODE 150

[@soumyopandey7974]

Good Approach...didi...keep it up.😊.

[@ankitraj2283]

Day 18 Done ✅✅
i have doubt after looking this Question directly Sidling Window Concept come in mind not Brute Force Approach
Should i first go for brute Force

[@prathamgoel3386]

class Solution {
public boolean checkInclusion(String s1, String s2) {
if (s1.length() > s2.length()) {
return false;
}
int[] c1 = new int[26];
int[] c2 = new int[26];
for (int i = 0; i < s1.length(); i++) {
c1[s1.charAt(i) - 'a']++;
c2[s2.charAt(i) - 'a']++;
}
for (int i = s1.length(); i < s2.length(); i++) {
if (isEqual(c1, c2)) {
return true;
}
c2[s2.charAt(i) - 'a']++;
c2[s2.charAt(i - s1.length()) - 'a']--;
}
if (isEqual(c1, c2)) {
return true;
}
return false;
}
private boolean isEqual(int[] c1, int[] c2) {
for (int i = 0; i < c1.length; i++) {
if (c1[i] != c2[i]) {
return false;
}
}
return true;
}
}

[@ankitraj2283]

My Approach
class Solution {
public boolean checkInclusion(String s1, String s2) {
HashMap map = new HashMap();
int i = 0;
int j = 0;
int k = s1.length();
int n = s2.length();
while (j < n) {
map.put(s2.charAt(j), map.getOrDefault(s2.charAt(j), 0) + 1);
while (j - i + 1 == k) {
if (check(map, s1)) {
return true;
}
map.put(s2.charAt(i), map.getOrDefault(s2.charAt(i), 0) - 1);
if (map.get(s2.charAt(i)) == 0) {
map.remove(s2.charAt(i));
}
i++;
}
j++;
}
return false;
}
static boolean check(HashMap map, String s1) {
HashMap targetMap = new HashMap();
for (char c : s1.toCharArray()) {
targetMap.put(c, targetMap.getOrDefault(c, 0) + 1);
}
return map.equals(targetMap);
}
}

[@mayappapujari9259]

Done ✌

[@prathamgoel3386]

My Brute force approach
class Solution {
private void generatePermutations(String prefix, String remain, Set pers) {
if (remain.length() == 0) {
pers.add(prefix);
} else {
for (int i = 0; i < remain.length(); i++) {
char curr = remain.charAt(i);
String newPrefix = prefix + curr;
String newRemain = remain.substring(0, i) + remain.substring(i + 1);
generatePermutations(newPrefix, newRemain, pers); // Corrected variable name
}
}
}
public boolean checkInclusion(String s1, String s2) {
if (s1.length() == 0) {
return false;
}
Set permutations = new HashSet(); // Corrected variable name
generatePermutations("", s1, permutations);
for (String permu : permutations) {
if (s2.contains(permu)) {
return true;
}
}
return false;
}
}

[@prathamgoel3386]

Time Limit Exceeded 😥