Split a String in Balanced Strings
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Split a String in Balanced Strings

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  • 1Balanced strings contain an equal number of 'L' and 'R' characters.
  • 2The maximum number of balanced strings can be determined by counting 'R' and 'L' characters.
  • 3The provided solution has a time complexity of O(n) and a space complexity of O(1).

AI-generated summary · May not capture all nuances

Key Insight
AskGif

"Balanced strings contain an equal number of 'L' and 'R' characters."

Split a String in Balanced Strings

Balanced strings are those who have an 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 split 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'

Solution:

using System;
using System.Collections.Generic;
using System.Text;

namespace LeetCode.AskGif.Easy.String
{
 class BalancedStringSplitSol
 {
 public void execute()
 {
 var res = BalancedStringSplit("RLRRLLRLRL");
 }

 public int BalancedStringSplit(string s)
 {
 int val = 0;
 int ans = 0;
 for(int i = 0; i < s.Length; i++)
 {
 if (s[i] == 'R')
 {
 val++; 
 }
 else
 {
 val--;
 }
 if (val == 0)
 ans++;
 }

 return ans;
 }
 }
}

Time Complexity: O(n)

Space Complexity: O(1)

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AskGif

Published on 3 May 2020 · 1 min read · 214 words

Part of AskGif Blog · coding

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