Leetcode 984: String Without AAA or BBB
Given two integers a and b, generate a string of length a + b such that the string contains exactly a ‘a’ letters and exactly b ‘b’ letters, while ensuring the substring ‘aaa’ and ‘bbb’ do not appear.
Problem
Approach
Steps
Complexity
Input: You are given two integers a and b, representing the number of occurrences of 'a' and 'b' in the desired string.
Example: a = 2, b = 3
Constraints:
• 0 <= a, b <= 100
Output: Return a string s that satisfies the conditions outlined in the problem statement.
Example: Output: 'abbab'
Constraints:
• The output string should have exactly a 'a' letters and b 'b' letters.
Goal: The goal is to construct a string that follows the specified conditions without violating the 'aaa' and 'bbb' substrings constraint.
Steps:
• Start by adding 'a' characters in a balanced way with 'b' characters to avoid consecutive 'a' or 'b' characters.
• Ensure that at no point do the number of 'a' characters exceed those of 'b' by more than 2 or vice versa.
• Interleave 'a' and 'b' characters as needed, making sure that the string does not contain 'aaa' or 'bbb'.
Goal: The constraints ensure that the problem can be solved within the given bounds.
Steps:
• 0 <= a, b <= 100
Assumptions:
• The string must be constructed in such a way that it adheres to the rules of avoiding 'aaa' and 'bbb'.
• Input: a = 2, b = 3
• Explanation: In this case, a valid output is 'abbab', which avoids 'aaa' and 'bbb' while ensuring the correct number of 'a' and 'b' characters.
Approach: The solution involves constructing a string by alternating between 'a' and 'b' while ensuring the constraints on consecutive occurrences of 'a' and 'b' are met.
Observations:
• We need to construct the string without causing the substring 'aaa' or 'bbb' to appear.
• We can start by interleaving 'a' and 'b', and then adjust the string as necessary to ensure the constraints are maintained.
• A greedy approach can work well, ensuring that we alternate between 'a' and 'b' without violating the constraints.
Steps:
• Check if a > b, if so, swap the values of a and b.
• Alternately add 'a' and 'b' to the result string, making sure that no three consecutive 'a's or 'b's are added.
Empty Inputs:
• If a or b is 0, the string should just consist of the remaining characters.
Large Inputs:
• If a and b are large (close to 100), the approach should still ensure that no 'aaa' or 'bbb' substrings appear.
Special Values:
• Consider cases where a and b are equal, or where one is much larger than the other.
Constraints:
• Ensure the solution handles cases with both small and large values of a and b.
string strWithout3a3b(int A, int B, char a = 'a', char b = 'b', string res = "") {
if(B > A) return strWithout3a3b(B, A, b, a);
while(A-- > 0) {
res += a;
if(A > B) res += a, A--;
if(B-->0) res += b;
}
return res;
}
1 : Function Definition
string strWithout3a3b(int A, int B, char a = 'a', char b = 'b', string res = "") {
Defines the function `strWithout3a3b`, which takes two integers `A` and `B`, representing the counts of characters `a` and `b`, respectively, and ensures no three consecutive `a`s or `b`s in the result string.
2 : Recursive Call for Swapping
if(B > A) return strWithout3a3b(B, A, b, a);
If `B` is greater than `A`, the function swaps the roles of `a` and `b` (and their counts) to ensure the function always processes the character with the larger count first.
3 : Main Loop Start
while(A-- > 0) {
A loop that runs while there are still `a`s to add to the result string. It decreases the count of `a` with each iteration.
4 : Add Character A
res += a;
Appends the character `a` to the result string `res`.
5 : Check If More A's Should Be Added
if(A > B) res += a, A--;
If there are more `a`s remaining than `b`s, the function appends one more `a` to the result and decrements `A`.
6 : Add Character B
if(B-->0) res += b;
If `B` (count of `b`s) is greater than 0, append a `b` to the result string and decrement `B`.
7 : Return Result
return res;
Returns the final result string `res` that contains no more than two consecutive `a`s or `b`s.
Best Case: O(a + b)
Average Case: O(a + b)
Worst Case: O(a + b)
Description: The time complexity is linear in the total number of 'a' and 'b' characters, which is the length of the resulting string.
Best Case: O(a + b)
Worst Case: O(a + b)
Description: The space complexity is O(a + b), as we need to store the result string of length a + b.
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