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Formatted question description: https://leetcode.ca/all/1221.html
1221. Split a String in Balanced Strings (Easy)
Balanced strings are those who have equal quantity of 'L' and 'R' characters.
Given a balanced string s
split it in the maximum amount of balanced strings.
Return the maximum amount of splitted balanced strings.
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 = "RLLLLRRRLR" Output: 3 Explanation: s can be split into "RL", "LLLRRR", "LR", each substring contains same number of 'L' and 'R'.
Example 3:
Input: s = "LLLLRRRR" Output: 1 Explanation: s can be split into "LLLLRRRR".
Example 4:
Input: s = "RLRRRLLRLL" Output: 2 Explanation: s can be split into "RL", "RRRLLRLL", since each substring contains an equal number of 'L' and 'R'
Constraints:
1 <= s.length <= 1000
s[i] = 'L' or 'R'
Related Topics:
String, Greedy
Solution 1.
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class Solution { public int balancedStringSplit(String s) { int count = 0; int position = 0; int length = s.length(); for (int i = 0; i < length; i++) { char c = s.charAt(i); if (c == 'R') position++; else position--; if (position == 0) count++; } return count; } } ############ 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; } }
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// OJ: https://leetcode.com/problems/split-a-string-in-balanced-strings/ // Time: O(N) // Space: O(1) class Solution { public: int balancedStringSplit(string s) { int cnt = 0, ans = 0; for (int i = 0; i < s.size(); ++i) { cnt += s[i] == 'L' ? 1 : -1; if (cnt == 0) ++ans; } return ans; } };
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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
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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 }
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/** * @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; };