What is the minimum substring length of a string that contains all the characters of another string?

Question:
Given A string and A string set A, evaluate the shortest substring length of the string containing all characters in.
 
Solution 1:
The most direct method is to directly start traversing: Find whether any two substrings contain str2. If any, record the length and obtain the minimum value.
Str1 = "daebfacba ";
Str2 = "abc ";
MinLen = len (str1 );
For I = 0: len (str1)
For j = I + 1: len (str1)
If (isContainAllElements (str1, str2, I, j) // record length if the substring from I to j contains all characters
If j-I <minLen
MinLen = j-I;
Considering isContainAllElements (), the complexity is O (n3 ).
 
Solution 2:
Set a variable for each character in str2 to store the latest position in str1. H (x) is used to represent the value.
Take "daebfacba" as an example,
1. When the first a is scanned, H (a) = 1; H (B), H (c) has no value;
2. When the first B is scanned, H (a) = 1; H (B) = 3, H (c) has no value;
3. when the second a is scanned, because C does not have a value, it overwrites the value of a: H (a) = 5; H (B) = 3, H (c) no value;
4. when the first c is scanned, H (a) = 5; H (B) = 3, H (c) = 6. Then, the distance is calculated once, that is, use the maximum value minus the minimum value of 3;
5. continue scanning. H (a) = 5; H (B) = 7, H (c) = 6, and the distance is 7-5 = 2;
6. continue scanning. H (a) = 8; H (B) = 7, H (c) = 6, and the distance is 8-6 = 2;
The minimum value is 2.
 
The complexity of this algorithm is O (n ).
In this algorithm, the corresponding positions of all characters are not found and can be overwritten directly. The simple feasibility of overwriting is as follows:
Assume that a appears twice, and B and c have three relative positions:
(1) a... B... a... c...
In this case, the second a will overwrite the first a, but since c is behind the second a, you only need to calculate the second a. The overwrite is acceptable.
(2) a... B... c......
At this time, both a will participate in the calculation, and there is no coverage problem.
(3) a... B... c...
In this case, it is obvious that only the second a needs to be calculated, and overwriting is acceptable.
Therefore, to sum up, the coverage method is feasible.
The general idea is:
1. Scan the string. If a character in the set is encountered, the position of the character in the set is recorded.
2. If all characters are not found, fill them in or overwrite them;
If all are found, the distance is calculated and compared to the shortest distance. Otherwise, fill in according to the above rules;
3. Scan the string to get the shortest length of the substring, and record the positions of all characters.
 
Expansion:
Question:
Given A string and A string set A, calculate the maximum length of the string containing all characters in A (the characters in A cannot be repeated in the substring ). Www.2cto.com
For example, "ababcda" and "abc"
The rule is as follows:
1. if each character is not filled with a value:
(1) If the value corresponding to the current character is not entered, enter it directly;
(2) If the corresponding value of the current character has been filled (for example, H (a) = 2), The filled value before position a 2 and 2 is cleared, and re-enter the value of a (for example, H (a) = 4); (because there are duplicates, you need to clear the previous)