LeetCode #1221 — EASY

Split a String in Balanced Strings

Build confidence with an intuition-first walkthrough focused on greedy fundamentals.

Solve on LeetCode

The Problem

Problem Statement

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 2^nd and 5^th 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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Greedy

Example 1

"RLRRLLRLRL"

Example 2

"RLRRRLLRLL"

Example 3

"LLLLRRRR"

Core Insight

What unlocks the optimal approach

Loop from left to right maintaining a balance variable when it gets an L increase it by one otherwise decrease it by one.
Whenever the balance variable reaches zero then we increase the answer by one.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1221: Split a String in Balanced Strings
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;
    }
}

// Accepted solution for LeetCode #1221: Split a String in Balanced Strings
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
}

# Accepted solution for LeetCode #1221: Split a String in Balanced Strings
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

// Accepted solution for LeetCode #1221: Split a String in Balanced Strings
struct Solution;

impl Solution {
    fn balanced_string_split(s: String) -> i32 {
        let mut l = 0;
        let mut r = 0;
        let mut res = 0;
        for c in s.chars() {
            match c {
                'R' => r += 1,
                'L' => l += 1,
                _ => {}
            }
            if l == r {
                res += 1;
            }
        }
        res
    }
}

#[test]
fn test() {
    let s = "RLRRLLRLRL".to_string();
    assert_eq!(Solution::balanced_string_split(s), 4);
    let s = "RLLLLRRRLR".to_string();
    assert_eq!(Solution::balanced_string_split(s), 3);
    let s = "LLLLRRRR".to_string();
    assert_eq!(Solution::balanced_string_split(s), 1);
}

// Accepted solution for LeetCode #1221: Split a String in Balanced Strings
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1221: Split a String in Balanced Strings
// 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;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.