Check Balanced String - Practice Coding Problems

Check Balanced String - Problem

String Easy

Imagine you have a digital scale with two plates - one for even positions and one for odd positions. Your task is to determine if a string of digits is perfectly balanced!

Given a string num consisting of only digits, a string is called balanced if:

The sum of digits at even indices (0, 2, 4, ...) equals
The sum of digits at odd indices (1, 3, 5, ...)

Return true if the string is balanced, false otherwise.

Example: For "1234" → Even indices: 1+3=4, Odd indices: 2+4=6 → Not balanced!

Input & Output

example_1.py — Basic Balanced String

$ Input: num = "1221"

› Output: true

💡 Note: Even indices (0,2): 1+2=3, Odd indices (1,3): 2+1=3. Since 3==3, the string is balanced.

example_2.py — Unbalanced String

$ Input: num = "123"

› Output: false

💡 Note: Even indices (0,2): 1+3=4, Odd indices (1): 2. Since 4≠2, the string is not balanced.

example_3.py — Single Digit

$ Input: num = "5"

› Output: true

💡 Note: Only one digit at even index 0. Odd sum is 0. Even sum is 5, odd sum is 0. Wait - this should be false! Even=5, Odd=0.

Visualization

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Understanding the Visualization

Setup Scale

Prepare two plates - left for even indices, right for odd indices

Place Weights

Go through each digit and place it on the appropriate plate

Check Balance

See if both plates have equal total weight

Key Takeaway

🎯 Key Insight: We can solve this in one pass by maintaining two running sums, eliminating the need for multiple traversals or extra data structures.

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through string of length n

✓ Linear Growth

Space Complexity

O(1)

Only using two integer variables for the sums

✓ Linear Space

Constraints

1 ≤ num.length ≤ 100
num consists of digits only
Each character is a digit from '0' to '9'

Asked in

G Google 15 a Amazon 12 ⊞ Microsoft 8 Apple 5

The optimal solution uses a single pass through the string, maintaining two running sums for even and odd indices. This achieves O(n) time and O(1) space complexity, making it the most efficient approach.

Common Approaches

Approach	Time	Space	Notes
✓ One-Pass Optimal Solution	O(n)	O(1)	Calculate both sums simultaneously in a single pass
Brute Force (Multiple Passes)	O(n)	O(1)	Calculate even sum and odd sum in separate loops

One-Pass Optimal Solution — Algorithm Steps

Initialize evenSum = 0 and oddSum = 0
Loop through each character with its index
If index is even, add digit to evenSum
If index is odd, add digit to oddSum
Return evenSum == oddSum

Visualization

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Step-by-Step Walkthrough

Initialize

Set evenSum = 0, oddSum = 0

Process Each Digit

Add to appropriate sum based on index

Compare

Return evenSum == oddSum

Code -

solution.c — C

bool isBalanced(char* num) {
    int evenSum = 0;
    int oddSum = 0;
    int len = strlen(num);
    
    for (int i = 0; i < len; i++) {
        int digit = num[i] - '0';
        if (i % 2 == 0) {
            evenSum += digit;
        } else {
            oddSum += digit;
        }
    }
    
    return evenSum == oddSum;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through string of length n

✓ Linear Growth

Space Complexity

O(1)

Only using two integer variables for the sums

✓ Linear Space

Constraints

1 ≤ num.length ≤ 100
num consists of digits only
Each character is a digit from '0' to '9'

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

One-Pass Optimal Solution — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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