Leetcode 420 Strong Password Checker Solution in Java | Hindi Coding Community

A password is considered strong if the below conditions are all met:

• It has at least 6 characters and at most 20 characters.
• It contains at least one lowercase letter, at least one uppercase letter, and at least one digit.
• It does not contain three repeating characters in a row (i.e., "...aaa..." is weak, but "...aa...a..." is strong, assuming other conditions are met).

Given a string password, return the minimum number of steps required to make password strong. if password is already strong, return 0.

In one step, you can:

• Insert one character to password,
• Delete one character from password, or
• Replace one character of password with another character.

 

Example 1:

Input: password = "a"
Output: 5

Example 2:

Input: password = "aA1"
Output: 3

Example 3:

Input: password = "1337C0d3"
Output: 0

 

Constraints:

• 1 <= password.length <= 50
• password consists of letters, digits, dot '.' or exclamation mark '!'.

Java Code :

public class Solution {

public int strongPasswordChecker(String s) {

if(s.length()<2) return 6-s.length();

//Initialize the states, including current ending character(end), existence of lowercase letter(lower), uppercase letter(upper), digit(digit) and number of replicates for ending character(end_rep)

char end = s.charAt(0);

boolean upper = end>='A'&&end<='Z', lower = end>='a'&&end<='z', digit = end>='0'&&end<='9';

//Also initialize the number of modification for repeated characters, total number needed for eliminate all consequnce 3 same character by replacement(change), and potential maximun operation of deleting characters(delete). Note delete[0] means maximum number of reduce 1 replacement operation by 1 deletion operation, delete[1] means maximun number of reduce 1 replacement by 2 deletion operation, delete[2] is no use here.

int end_rep = 1, change = 0;

int[] delete = new int[3];

for(int i = 1;i<s.length();++i){

if(s.charAt(i)==end) ++end_rep;

else{

change+=end_rep/3;

if(end_rep/3>0) ++delete[end_rep%3];

//updating the states

end = s.charAt(i);

upper = upper||end>='A'&&end<='Z';

lower = lower||end>='a'&&end<='z';

digit = digit||end>='0'&&end<='9';

end_rep = 1;

}

}

change+=end_rep/3;

if(end_rep/3>0) ++delete[end_rep%3];

//The number of replcement needed for missing of specific character(lower/upper/digit)

int check_req = (upper?0:1)+(lower?0:1)+(digit?0:1);

if(s.length()>20){

int del = s.length()-20;

//Reduce the number of replacement operation by deletion

if(del<=delete[0]) change-=del;

else if(del-delete[0]<=2*delete[1]) change-=delete[0]+(del-delete[0])/2;

else change-=delete[0]+delete[1]+(del-delete[0]-2*delete[1])/3;

return del+Math.max(check_req,change);

}

else return Math.max(6-s.length(), Math.max(check_req, change));

}

}