[LeetCode] Daily LeetCode: 1221

1221. Split a String in Balanced Strings

Problem type - Greedy

🧩 Problem

🎯 Strategy

Greedy Algorithm

To solve this problem, we need to design this problem by using greedy algorithm.

Set variables

We need two variables: the counter to counting the letters and the result variable. When the condition matches, the result variable goes up by one.

In order to sort the number, we should change the number into a string first.

 def balancedStringSplit(self, s: str) -> int:
 	count = 0
    result = 0

Use the counter

Now we need to set the logic, which can be by iterating over the string s.

The function will increase the counter if the letter is R. The function will decrease the counter if the letter is L.

When the count == 0, the result should increase since we made RL the balanced string.

def balancedS... -> int:
	...
    for letter in s:
    	if letter == "R":
        	count += 1
        elif letter == "L":
        	count -= 1

        if count == 0:
        	result += 1

     return result

If you know the stack method, this logic is similar to the Python DFS method. As this reminds me of the stack method, let’s try the stack method.

First, we set up the stack and result. Then iterate over the string as previously we did, and if the stack is empty, the result goes up by one(same as if the count == 0)

Next, we will append the letter to the stack.

def balanced.. -> int:
	stack, result = [], 0

    for letter in s:
    	if stack == []:
        	result += 1
            stack.append(letter)

Second, since we appended the letter in the stack, we are checking the second letter here. We will pass over the first if statement since there is a letter in the stack.

If the current letter is equal to the appended letter in the stack, which means if they are the same letter, append the letter in the stack and go to the third letter.

If the current letter is not equal to the appended letter, then just pop() the letter from the stack

        elif letter == stack[-1]:
        	stack.append(letter)

        else stack.pop()

	return result

If all the letter pops out from the stack, we check the second balanced string. The result goes up by one at this point, and in the end, the for loop loops until the end of the string.

📌 Thoughts

It was a good question regarding practicing a greedy method. I struggled to manage the counter. Next time, I should remember the counter can be used in various ways.

💻 Solution

Normal Greedy

 def balancedStringSplit(self, s: str) -> int:
        count = 0
        result = 0

        for letter in s:
        	if letter == "R":
            	count += 1
            if letter == "L":
            	count -= 1

            if count == 0:
            	result += 1

		return result

Stack Method

 def balancedStringSplit(self, s: str) -> int:
        stack, result = [], 0

        for letter in s:
        	if stack == []:
            	result += 1
            	stack.append(letter)

            elif letter == stack[-1]:
            	stack.append(letter)

            else:
            	stack.pop()

        return result