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1221. Split a String in Balanced Strings
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 2^{nd} and 5^{th} substrings are not balanced.
Example 3:
Input: s = "LLLLRRRR" Output: 1 Explanation: s can be split into "LLLLRRRR".
Constraints:
2 <= s.length <= 1000
s[i]
is either'L'
or'R'
.s
is 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 { public int balancedStringSplit(String s) { int ans = 0, l = 0; for (char c : s.toCharArray()) { if (c == 'L') { ++l; } else { l; } if (l == 0) { ++ans; } } return ans; } }

class Solution { public: int balancedStringSplit(string s) { int ans = 0, l = 0; for (char c : s) { if (c == 'L') ++l; else l; if (l == 0) ++ans; } return ans; } };

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

func balancedStringSplit(s string) int { ans, l := 0, 0 for _, c := range s { if c == 'L' { l++ } else { l } if l == 0 { ans++ } } return ans }

/** * @param {string} s * @return {number} */ var balancedStringSplit = function (s) { let ans = 0; let l = 0; for (let c of s) { if (c == 'L') { ++l; } else { l; } if (l == 0) { ++ans; } } return ans; };