backtracking and invariant during generating the parathese

righjt > left  (open bracket and cloase barckst)

class Solution {
//["((()))","(()())","(())()","()(())","()()()","())(()"] wrong case --> change right > left the numebr of bracket is the invariant
List<String> res = new ArrayList<>();
public List<String> generateParenthesis(int n) {
//back((new StringBuilder()).append('('),2*n, 1, n, n);
back((new StringBuilder()), 2*n, 0, n, n);
return res;
}
void back(StringBuilder temp, int n, int pos, int left, int right){//pos start from 1
if(pos >= n){
//temp.append(")"); // problem from here
System.out.println(pos);
res.add(temp.toString());
return;
}
if(left > 0 ){
temp.append("(");
back(temp,n, pos+1, left-1, right);
temp.setLength(temp.length()-1);
}
if(right > left ){
temp.append(")");
back(temp, n, pos+1, left, right-1);
temp.setLength(temp.length()-1);
} }
}

Restore IP Addresses

//insert element into the string

class Solution {
//invariant rule: each number are
// use the immuniateble of String
List<String> res = new ArrayList<String>();
public List<String> restoreIpAddresses(String s) {
back(0, s, new String(), 0);
return res;
} void back(int next, String s, String str , int num){ //num: there are only three dots.
if(num == 3){
//if(next==s.length()) return;
if(!valid(s.substring(next, s.length()))) return;
res.add(str+s.substring(next, s.length()));
return;
}
//for each step, move one digit or two or three
for(int i = 1; i <=3; i++ ){
//check string
if(next+i > s.length()) continue;
String sub = s.substring(next, next+i);//
if(valid(sub)){
back(next+i, s, str+sub+'.', num+1);
}
}
}
boolean valid(String sub){
if(sub.length() == 0 || sub.length()>=4) return false;
if(sub.charAt(0) == '0') {
//System.out.println(sub.equals("0"));
return sub.equals("0"); // not check '0' weired
}
int num = Integer.parseInt(sub);
if(num >255 || num <0) return false;
else return true;
}
}

//core idea: move one step or 2 step or three based on the question (0 - 255) also append . and substring

use string instead stringBuilder  (immuatable)

131. Palindrome Partitioning

class Solution {
//check palindrome, divide into small problems:
List<List<String>> res = new ArrayList<List<String>>();
public List<List<String>> partition(String s) {
back(s, new ArrayList<String>());
return res;
}
void back(String s, List<String> list){
if(s.length()==0){
List<String> temp = new ArrayList<>(list);
res.add(temp);
return ;
} for(int i = 0; i<s.length(); i++){//divide s into su and sub
String su = s.substring(0, i+1);
String sub = s.substring(i+1, s.length());
if(isPalindrome(su)){
list.add(su);
back(sub,list);
list.remove(list.size()-1);
} }
}
boolean isPalindrome(String su){
if(su.length()==0){
return true;
}else {
int i =0 , j = su.length()-1;
while(i<j){
if(su.charAt(i) != su.charAt(j)) return false;
i++; j--;
}
return true;
}
}
}

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