https://leetcode.com/problems/strong-password-checker/
https://discuss.leetcode.com/topic/63185/java-easy-solution-with-explanation
http://www.cnblogs.com/grandyang/p/5988792.html
https://discuss.leetcode.com/topic/63854/o-n-java-solution-by-analyzing-changes-allowed-to-fix-each-condition
https://discuss.leetcode.com/topic/63185/java-easy-solution-with-explanation
A password is considered strong if below conditions are all met:
- It has at least 6 characters and at most 20 characters.
- It must contain at least one lowercase letter, at least one uppercase letter, and at least one digit.
- It must NOT contain three repeating characters in a row ("...aaa..." is weak, but "...aa...a..." is strong, assuming other conditions are met).
Write a function strongPasswordChecker(s), that takes a string s as input, and return the MINIMUM change required to make s a strong password. If s is already strong, return 0.
Insertion, deletion and replace of any one character are all considered as one change.
https://discuss.leetcode.com/topic/63185/java-easy-solution-with-explanation
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));
}http://www.cnblogs.com/grandyang/p/5988792.html
https://discuss.leetcode.com/topic/63854/o-n-java-solution-by-analyzing-changes-allowed-to-fix-each-condition
The basic principle is straightforward: if we want to make MINIMUM change to turn s into a strong password, each change made should fix as many problems as possible.
So first let's figure out what changes are suitable for righting each condition. Just to clarify, each change here should be characterized by at least two factors: the type of operation it takes and the position in the string where the operation is applied.
(Note: Ideally we should also include the characters involved in the operation and the "power" of each operation for eliminating problems but they turn out to be partially relevant so I will mention them only when appropriate.)
- Length problem: if the total length is less than 6, the change that should be made is (insert, any position), which reads as "the operation is insertion and it can be applied to anywhere in the string". If the total length is greater than 20, then the change should be (delete, any position).
- Missing letter or digit: if any of the lowercase/uppercase letters or digits is missing, we can do either (insert, any position) or (replace, any position) to correct it. (Note here the characters for insertion or replacement can only be those missing.)
- Repeating characters: for repeating characters, all three operations are allowed but the positions where they can be applied are limited within the repeating characters. For example, to fix "aaaaa", we can do one replacement (replace the middle 'a') or two insertion (one after the second 'a' and one after the fourth 'a') or three deletion (delete any of the three 'a's). So the possible changes are (replace, repeating characters), (insert, repeating characters), (delete, repeating characters). (Note here the "power" of each operation for fixing the problem are different -- replacement is the strongest while deletion is the weakest.)
All right, what's next? If we want a change to eliminate as many problems as it can, it must be shared among the possible solutions to each problem it can fix. So our task is to find out possible overlappings among the changes for fixing each problem.
Since there are most (three) changes allowed for the third condition, we may start from combinations first condition & third conditionand second condition & third condition. It's not too hard to conclude that any change that can fix the first condition or second condition is also able to fix the third one (since the type of operation here is irrelevant, we are free to choose the position of the operation to match those of the repeating characters). For combination first condition & second condition, depending on the length of the string, there will be overlapping if length is less than 6 or no overlapping if length is greater than 20.
From the analyses above, it seems worthwhile to distinguish between the two cases: when the string length is too short or too long.
For the former case, it can be shown that the changes needed to fix the first and second conditions always outnumber those for the third one. Since whatever change used fixing the first two conditions can also correct the third one, we may concern ourselves with only the first two conditions. Also as there are overlappings between the changes for fixing these two conditions, we will prefer those overlapping ones, i.e. (insert, any position). Another point is that the characters involved in the operation matters now. To fix the first condition, only those missing characters can be inserted while for the second condition, it can be any character. Therefore correcting the first condition takes precedence over the second one.
For the latter case, there are overlappings between the first & third and second & third conditions, so those changes will be taken, i.e., first condition => (delete, any position), second condition => (replace, any position). The reason not to use (insert, any position) for the second condition is that it contradicts the changes made to the first condition (therefore has the tendency to cancel its effects). After fixing the first two conditions, what operations should we choose for the third one?
Now the "power" of each operation for eliminating problems comes into play. For the third condition, the "power" of each operation will be measured by the maximum number of repeating characters it is able to get rid of. For example, one replacement can eliminate at most 5 repeating characters while insertion and deletion can do at most 4 and 3, respectively. In this case, we say replacement has more "power" than insertion or deletion. Intuitively the more "powerful" the operation is, the less number of changes is needed for correcting the problem. Therefore (replace, repeating characters) triumphs in terms of fixing the third condition.
Further more, another very interesting point shows up when the "power" of operation is taken into consideration (And thank yicui for pointing it out). As I mentioned that there are overlappings between changes made for fixing the first two conditions and for the third one, which means the operations chosen above for the first two conditions will also be applied to the third one. For the second condition with change chosen as (replace, any position), we have no problem adapting it so that it coincides with the optimal change (replace, repeating characters) for the third condition. However, there is no way to do that for the first condition with change (delete, any position). We have a conflict now!
So how do we reconcile it? The trick is that for a sequence of repeating characters of length k (k >=3), instead of turning it all the way into a sequence of length 2 (so as to fix the repeating character problem) by the change (delete, any position), we will first reduce its length to (3m + 2), where (3m + 2) is the largest integer of the form yet no more than k. That is to say, if k is a multiple of 3, we apply once such change so its length will become (k - 1); else if k is a multiple of 3 plus 1, we apply twice such change to cut its length down to (k - 2), provided we have more such changes to spare (note here we need at least two changes but the remaining available changes may be less than that, so we should stick to the smaller one: 2 or the remaining available changes). The reason is that the optimal change (replace, repeating characters) for the third condition will be most powerful when the total length of the repeating characters is of this form. Of course, if we still have more changes (delete, any position) to do after that, then we are free to turn the repeating sequence all the way into a sequence of length 2.
public int strongPasswordChecker(String s) {
int res = 0, a = 1, A = 1, d = 1;
char[] carr = s.toCharArray();
int[] arr = new int[carr.length];
for (int i = 0; i < arr.length;) {
if (Character.isLowerCase(carr[i])) a = 0;
if (Character.isUpperCase(carr[i])) A = 0;
if (Character.isDigit(carr[i])) d = 0;
int j = i;
while (i < carr.length && carr[i] == carr[j]) i++;
arr[j] = i - j;
}
int total_missing = (a + A + d);
if (arr.length < 6) {
res += total_missing + Math.max(0, 6 - (arr.length + total_missing));
} else {
int over_len = Math.max(arr.length - 20, 0), left_over = 0;
res += over_len;
for (int k = 1; k < 3; k++) {
for (int i = 0; i < arr.length && over_len > 0; i++) {
if (arr[i] < 3 || arr[i] % 3 != (k - 1)) continue;
arr[i] -= Math.min(over_len, k);
over_len -= k;
}
}
for (int i = 0; i < arr.length; i++) {
if (arr[i] >= 3 && over_len > 0) {
int need = arr[i] - 2;
arr[i] -= over_len;
over_len -= need;
}
if (arr[i] >= 3) left_over += arr[i] / 3;
}
res += Math.max(total_missing, left_over);
}
return res;
}https://discuss.leetcode.com/topic/63185/java-easy-solution-with-explanation
The general idea is to record some states, and calculate the edit distance at the end. All detail are explained in the comments.
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));
}
}