Basic Calculator II

Basic Calculator II - Problem

Given a string s which represents a mathematical expression, evaluate this expression and return its value.

The integer division should truncate toward zero.

You may assume that the given expression is always valid. All intermediate results will be in the range of [-2³¹, 2³¹ - 1].

Note: You are not allowed to use any built-in function which evaluates strings as mathematical expressions, such as eval().

Input & Output

Example 1 — Basic Expression

$ Input: s = "3+2*2"

› Output: 7

💡 Note: Following order of operations: 2*2 = 4, then 3+4 = 7

Example 2 — Division with Truncation

$ Input: s = " 3/2 "

› Output: 1

💡 Note: 3 divided by 2 equals 1.5, but we truncate toward zero to get 1

Example 3 — Complex Expression

$ Input: s = " 3+5 / 2 "

› Output: 5

💡 Note: Division first: 5/2 = 2 (truncated), then 3+2 = 5

Constraints

1 ≤ s.length ≤ 3 × 10⁵
s consists of integers and operators ('+', '-', '*', '/') separated by some number of spaces
s represents a valid expression
All intermediate results are in the range [-2³¹, 2³¹ - 1]
The divisor is never zero

Visualization

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Asked in

G Google 15 f Facebook 12 a Amazon 10 M Microsoft 8

The key insight is handling operator precedence by processing multiplication and division immediately while deferring addition and subtraction. The stack approach naturally handles this by storing intermediate results. Time: O(n), Space: O(n) for stack or O(1) for optimized approach.

Common Approaches

✓ No Extra Space

⏱️ Time: O(n) Space: O(1)

Instead of using a stack, keep track of the running result and the last number. For multiplication and division, update the last number immediately. For addition and subtraction, add the last number to result.

Stack-Based Evaluation

⏱️ Time: O(n) Space: O(n)

Process the expression from left to right, using a stack to store numbers. For +/- operations, push numbers to stack. For */÷ operations, immediately calculate with the top of stack.

Two-Pass Evaluation

⏱️ Time: O(n) Space: O(n)

First pass handles multiplication and division to respect precedence. Second pass handles addition and subtraction on the simplified expression.

No Extra Space — Algorithm Steps

Track result, last number, and current operation
For +/-, add last number to result
For */÷, calculate with last number immediately
Handle the final number at end

Visualization

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Step-by-Step Walkthrough

Initialize

result=0, lastNum=0, sign='+'

Process

Update result for +/-, lastNum for */

Final

Add lastNum to result

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>

int solution(char* s) {
    int result = 0;
    int lastNum = 0;
    int num = 0;
    char sign = '+';
    int len = strlen(s);
    
    for (int i = 0; i < len; i++) {
        char c = s[i];
        if (isdigit(c)) {
            num = num * 10 + (c - '0');
        }
        
        if (strchr("+-*/", c) || i == len - 1) {
            if (sign == '+' || sign == '-') {
                result += lastNum;
                lastNum = (sign == '+') ? num : -num;
            } else if (sign == '*') {
                lastNum *= num;
            } else if (sign == '/') {
                lastNum /= num;
            }
            
            sign = c;
            num = 0;
        }
    }
    
    return result + lastNum;
}

int main() {
    char s[1000];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    printf("%d\n", solution(s));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the string

✓ Linear Growth

Space Complexity

O(1)

Only using variables for result, last number, and current number

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

No Extra Space — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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