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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.

• 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;
};