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Check Balanced String

Difficulty: Easy

Problem Description

You are given a string num consisting of only digits. A string of digits is called balanced if the sum of the digits at even indices is equal to the sum of digits at odd indices. Return true if num is balanced, otherwise return false.

Key Insights

  • The problem requires checking the sums of digits based on their indices (even vs. odd).
  • The string will always have at least 2 characters, ensuring that we will have both even and odd indices to compare.
  • Simple iteration over the string while maintaining two accumulators for the sums is efficient.

Space and Time Complexity

Time Complexity: O(n), where n is the length of the string num. Space Complexity: O(1), as we are using a constant amount of space for the sums.

Solution

To solve the problem, we will iterate through the string and maintain two separate sums: one for the digits at even indices and one for the digits at odd indices. We can use a simple loop to check the index of each character in the string and add the numeric value of the character to the appropriate sum. Finally, we will compare the two sums and return the result.

Code Solutions

def check_balanced_string(num: str) -> bool:
    even_sum = 0
    odd_sum = 0
    
    for i in range(len(num)):
        digit = int(num[i])  # Convert character to integer
        if i % 2 == 0:       # Even index
            even_sum += digit
        else:                # Odd index
            odd_sum += digit
            
    return even_sum == odd_sum  # Compare the two sums
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