1221. Split a String in Balanced StringsBalanced 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.Returnthe maximum number of balanced strings you can obtain.Example 1:Input:s “RLRRLLRLRL”Output:4Explanation:s can be split into “RL”, “RRLL”, “RL”, “RL”, each substring contains same number of ‘L’ and ‘R’.Example 2:Input:s “RLRRRLLRLL”Output:2Explanation: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:1Explanation:s can be split into “LLLLRRRR”.Constraints:2 s.length 1000s[i] is either ‘L’ or ‘R’.s is a balanced string.From: LeetCodeLink: 1221. Split a String in Balanced StringsSolution:Ideas:use a counter. Add 1 for R, -1 for L. Every time counter becomes 0, we found one balanced substring.Code:intbalancedStringSplit(char*s){intbalance0;intcount0;for(inti0;s[i]!\0;i){if(s[i]R){balance;}else{balance--;}if(balance0){count;}}returncount;}
1221. Split a String in Balanced StringsBalanced 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.Returnthe maximum number of balanced strings you can obtain.Example 1:Input:s “RLRRLLRLRL”Output:4Explanation:s can be split into “RL”, “RRLL”, “RL”, “RL”, each substring contains same number of ‘L’ and ‘R’.Example 2:Input:s “RLRRRLLRLL”Output:2Explanation: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:1Explanation:s can be split into “LLLLRRRR”.Constraints:2 s.length 1000s[i] is either ‘L’ or ‘R’.s is a balanced string.From: LeetCodeLink: 1221. Split a String in Balanced StringsSolution:Ideas:use a counter. Add 1 for R, -1 for L. Every time counter becomes 0, we found one balanced substring.Code:intbalancedStringSplit(char*s){intbalance0;intcount0;for(inti0;s[i]!\0;i){if(s[i]R){balance;}else{balance--;}if(balance0){count;}}returncount;}
