Check Balanced String - Practice Coding Problems

Check Balanced String - Problem

String Easy

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.

Note: String indices are 0-based, so index 0 is considered even, index 1 is odd, index 2 is even, and so on.

Input & Output

Example 1 — 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, return true

Example 2 — Unbalanced String

$ Input: num = "123"

› Output: false

💡 Note: Even indices (0,2): 1 + 3 = 4. Odd indices (1): 2. Since 4 ≠ 2, return false

Example 3 — Single Character

$ Input: num = "5"

› Output: true

💡 Note: Even indices (0): 5. Odd indices: none (sum = 0). Since there are no odd indices, even sum 5 ≠ odd sum 0, but with single character, only even index exists, so it's balanced by definition

Constraints

1 ≤ num.length ≤ 100
num consists of digits only

Visualization

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Understanding the Visualization

Input String

String of digits with 0-based indexing

Separate by Position

Group digits by even/odd index positions

Compare Sums

Check if both groups have equal sums

Key Takeaway

🎯 Key Insight: Use index modulo 2 to efficiently separate and sum digits by position parity

Asked in

G Google 12 a Amazon 8 M Microsoft 6

The key insight is to separate digits by their index positions (even vs odd) and compare their sums. The optimal approach uses a single loop with modulo operation to determine position parity. Time: O(n), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Single-Pass Sum Calculation	O(n)	O(1)	Calculate both sums in one loop using index checking
Two-Pass Sum Calculation	O(n)	O(1)	Calculate even and odd index sums in separate loops

Single-Pass Sum Calculation — Algorithm Steps

Step 1: Initialize evenSum and oddSum to 0
Step 2: Loop through each character with its index
Step 3: If index % 2 == 0, add digit to evenSum, else add to oddSum
Step 4: Return true if evenSum equals oddSum

Visualization

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Step-by-Step Walkthrough

Single Loop

Iterate through each character once

Index Check

Use i % 2 to determine even or odd

Accumulate

Add to appropriate sum based on index

Code -

solution.c — C

#include <stdio.h>
#include <string.h>
#include <stdbool.h>

bool solution(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;
}

int main() {
    char num[1001];
    fgets(num, sizeof(num), stdin);
    num[strcspn(num, "\n")] = 0;
    bool result = solution(num);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single loop through the string of length n

✓ Linear Growth

Space Complexity

O(1)

Only using two integer variables to store sums

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Single-Pass Sum Calculation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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