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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 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 <= 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 $s$.

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

• 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;
  };
  
  

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