How do you solve Check Balanced String on LeetCode?

Check Balanced String (LeetCode #3340) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The sums of digits at even and odd indices must be equal for the string to be balanced.

What is the brute force solution for Check Balanced String?

The brute force approach for Check Balanced String is Brute Force, with O(n) time complexity and O(1) space complexity. The brute-force approach involves iterating through the string and calculating the sums of digits at even and odd indices separately. This is straightforward but inefficient, as it requires multiple passes over the string. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Check Balanced String?

Check Balanced String uses the following concepts: String. The recommended patterns to study are: Array, Two Pointers.

What are the interview tips for Check Balanced String?

Clarify the problem statement and constraints before coding. Think about edge cases, such as strings of minimal length. Explain your thought process as you code to demonstrate your understanding.

What are common mistakes when solving Check Balanced String?

Not converting characters to integers before summing. Confusing even and odd indices.

#3340

Check Balanced String

Easy

StringArrayTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution is similar to the brute force but focuses on a single pass through the string, calculating the sums directly without needing to store intermediate results in separate variables.

⚙️

Algorithm

3 steps

1Step 1: Initialize two variables, even_sum and odd_sum, to 0.
2Step 2: Loop through the string. For each index, add the digit to even_sum if the index is even, or to odd_sum if the index is odd.
3Step 3: After the loop, return true if even_sum equals odd_sum, otherwise return false.

solution.py11 lines

1# Full working Python code
2
3def check_balanced_string(num):
4    even_sum = 0
5    odd_sum = 0
6    for i in range(len(num)):
7        if i % 2 == 0:
8            even_sum += int(num[i])
9        else:
10            odd_sum += int(num[i])
11    return even_sum == odd_sum

ℹ

Complexity note: The time complexity remains O(n) as we still traverse the string once. The space complexity is O(1) since we only use a fixed amount of extra space for the sums.

1The sums of digits at even and odd indices must be equal for the string to be balanced.
2The problem can be solved in a single pass through the string.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.