Basic Calculator || | Leetcode

In this video, we break down a C++ program to evaluate basic arithmetic expressions involving addition, subtraction, multiplication, and division. Learn how the program:
Time stamps:
✓ Problem Algorithm and Solution Approach (0:00)
✓ Live Coding (07:00)
✅ Parses expressions like "3+2*2"
✅ Uses a stack to handle operator precedence efficiently
✅ Processes the input in a single pass for optimal performance
Time Complexity: O(N)
Space Complexity: O(N)
Perfect for coding interviews and competitive programming! Watch now to master this clean and effective solution. 💻✨
🔗 Code Walkthrough and Dry Run:
Example Input: "3-5/2"
Step-by-Step Dry Run
1. Initialization:
s = "3-5/2+" (a + is appended to process the last number).
num = 0, ans = 0, sign = '+', stack
2. Processing Characters in the Loop:
2.1. Character 3:
It is a digit, so num = 0 * 10 + (3 - '0') = 3.
2.2 Character -:
sign is +, so st.push(3) (push current num to the stack).
Reset num = 0, update sign = '-'.
Stack: [3]
2.3 Character 5:
It is a digit, so num = 0 * 10 + (5 - '0') = 5.
2.4 Character /:
sign is -, so st.push(-5) (push -num to the stack).
Reset num = 0, update sign = '/'.
Stack: [3, -5]
2.5 Character 2:
It is a digit, so num = 0 * 10 + (2 - '0') = 2.
2.6 Character +:
sign is /, so:
Pop the top of the stack: temp = -5.
Compute temp / num = -5 / 2 = -2 (integer division truncates toward 0).
Push the result back to the stack: st.push(-2).
Reset num = 0, update sign = '+'.
Stack: [3, -2]
Final Accumulation:
While the stack is not empty:
Pop 3, add to ans: ans = 0 + 3 = 3.
Pop -2, add to ans: ans = 3 + (-2) = 1.
Result:
The final result is 1.
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