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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.

• Java
• C++
• Python
• Go
• Javascript

• 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;
      }
  }
  

• // 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;
      }
  };
  

• 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;
  };
  
  

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