Description
Balanced strings are those that have an equal quantity of 'L' and 'R' characters.
Given a balanced string s, split it into some number of substrings such that:
- Each substring is balanced.
Return the maximum number of balanced strings you can obtain.
Example 1:
Input: s = "RLRRLLRLRL" Output: 4 Explanation: s can be split into "RL", "RRLL", "RL", "RL", each substring contains same number of 'L' and 'R'.
Example 2:
Input: s = "RLRRRLLRLL" Output: 2 Explanation: s can be split into "RL", "RRRLLRLL", each substring contains same number of 'L' and 'R'. Note that s cannot be split into "RL", "RR", "RL", "LR", "LL", because the 2nd and 5th substrings are not balanced.
Example 3:
Input: s = "LLLLRRRR" Output: 1 Explanation: s can be split into "LLLLRRRR".
Constraints:
2 <= s.length <= 1000s[i]is either'L'or'R'.sis a balanced string.
Solutions
Solution 1: Greedy
We use a variable l to maintain the current balance of the string, i.e., the value of l is the number of 'L's minus the number of 'R's in the current string. When the value of l is 0, we have found a balanced string.
We traverse the string s. When we traverse to the i-th character, if s[i] = L, then the value of l is increased by 1, otherwise, the value of l is decreased by 1. When the value of l is 0, we increase the answer by 1.
The time complexity is O(n), and the space complexity is O(1). Here, n is the length of the string s.
class Solution: def balancedStringSplit(self, s: str) -> int: ans = l = 0 for c in s: if c == 'L': l += 1 else: l -= 1 if l == 0: ans += 1 return ans(code-box)
