LeetCode 3340 - Check Balanced String | Prefix Sum Logic | C++ Full Solution

📌 Problem Statement:
You are given a string num consisting of digits only.

Return true if the sum of digits at even indices is equal to the sum of digits at odd indices. Otherwise, return false.

🧩 What You'll Learn:
✔ Traversing strings efficiently
✔ Difference between even and odd indices
✔ Digit to integer conversion in C++
✔ Simple prefix sum style logic
✔ Time Complexity: O(n)
✔ Space Complexity: O(1)

🛠️ Approach - Step by Step:
STEP 1 → Initialize evenSum and oddSum as 0
STEP 2 → Traverse each character of the string
STEP 3 → Convert character digit into integer value
STEP 4 → If index is even, add digit to evenSum
STEP 5 → Otherwise add digit to oddSum
STEP 6 → Compare both sums and return result

💡 Example:
Input:
num = "1234"

Even indices:
1 + 3 = 4

Odd indices:
2 + 4 = 6

Since 4 is not equal to 6 → Answer is false

⏱️ Timestamps:
0:00 - Problem Statement
1:00 - Understanding Even & Odd Indices
2:30 - Digit Conversion Logic
4:00 - C++ Code Walkthrough
6:00 - Dry Run Example
7:30 - Complexity Analysis

💻 Difficulty: Easy
🏷️ Topics: String, Math, Simulation

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Видео LeetCode 3340 - Check Balanced String | Prefix Sum Logic | C++ Full Solution канала JALAK PALAN