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.

// 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;
    }
};

Java

  • 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;
        }
    }
    
  • // 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;
        }
    };
    
  • print("Todo!")
    

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