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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 $s$.

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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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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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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;
    }
};

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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;
};

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