Complex Number Multiplication - Practice Coding Problems

Complex Number Multiplication - Problem

A complex number can be represented as a string on the form "real+imaginaryi" where:

real is the real part and is an integer in the range [-100, 100]
imaginary is the imaginary part and is an integer in the range [-100, 100]
i² == -1

Given two complex numbers num1 and num2 as strings, return a string of the complex number that represents their multiplication.

Input & Output

Example 1 — Basic Multiplication

$ Input: num1 = "1+1i", num2 = "1+1i"

› Output: "0+2i"

💡 Note: Using formula (a+bi)(c+di) = (ac-bd) + (ad+bc)i: (1+1i)(1+1i) = (1×1 - 1×1) + (1×1 + 1×1)i = 0 + 2i

Example 2 — With Negative Components

$ Input: num1 = "1+1i", num2 = "1-1i"

› Output: "2+0i"

💡 Note: (1+1i)(1-1i) = (1×1 - 1×(-1)) + (1×(-1) + 1×1)i = (1+1) + (-1+1)i = 2+0i

Example 3 — Both Parts Negative

$ Input: num1 = "1-1i", num2 = "1-1i"

› Output: "0-2i"

💡 Note: (1-1i)(1-1i) = (1×1 - (-1)×(-1)) + (1×(-1) + (-1)×1)i = (1-1) + (-1-1)i = 0-2i

Constraints

num1 and num2 are valid complex numbers in the form "a+bi" or "a-bi"
a and b are integers in the range [-100, 100]
The output format should match the input format

Visualization

Tap to expand

Understanding the Visualization

Input

Two complex numbers as strings: "1+1i" and "1+1i"

Process

Parse components and apply multiplication formula

Output

Result as formatted string: "0+2i"

Key Takeaway

🎯 Key Insight: Complex multiplication follows the distributive property with the key rule that i² = -1

Asked in

f Facebook 15 G Google 12 a Amazon 8

The key insight is to parse complex numbers into their real and imaginary components, then apply the standard multiplication formula: (a+bi)(c+di) = (ac-bd) + (ad+bc)i. Best approach uses regex for robust parsing. Time: O(n), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Regex-Based Parsing	O(n)	O(1)	Use regular expressions for cleaner parsing of complex number components
Parse and Apply Formula	O(n)	O(1)	Parse the complex numbers and apply the standard multiplication formula

Regex-Based Parsing — Algorithm Steps

Step 1: Define regex pattern to match complex number format
Step 2: Extract real and imaginary parts using regex groups
Step 3: Apply multiplication formula (ac-bd) + (ad+bc)i
Step 4: Format result string with proper sign handling

Visualization

Tap to expand

Step-by-Step Walkthrough

Regex Pattern

Use ([+-]?\d+)([+-]\d+)i to extract components

Extract Values

Parse matched groups into integers

Apply Formula

Calculate result with proper sign handling

Code -

solution.c — C

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

void parseComplex(char* s, int* real, int* imag) {
    char temp[100];
    strcpy(temp, s);
    
    char* iPos = strstr(temp, "i");
    *iPos = '\0';
    
    char* signPos = NULL;
    for (int i = 1; i < strlen(temp); i++) {
        if (temp[i] == '+' || temp[i] == '-') {
            signPos = &temp[i];
        }
    }
    
    *signPos = '\0';
    *real = atoi(temp);
    *imag = atoi(signPos + 1);
    
    if (*(signPos + 1 - temp + s) == '-') {
        *imag = -*imag;
    }
}

char* solution(char* num1, char* num2) {
    static char result[100];
    int a, b, c, d;
    
    parseComplex(num1, &a, &b);
    parseComplex(num2, &c, &d);
    
    int realPart = a * c - b * d;
    int imagPart = a * d + b * c;
    
    if (imagPart >= 0) {
        sprintf(result, "%d+%di", realPart, imagPart);
    } else {
        sprintf(result, "%d%di", realPart, imagPart);
    }
    
    return result;
}

int main() {
    char num1[100], num2[100];
    fgets(num1, sizeof(num1), stdin);
    fgets(num2, sizeof(num2), stdin);
    
    num1[strcspn(num1, "\n")] = 0;
    num2[strcspn(num2, "\n")] = 0;
    
    printf("%s\n", solution(num1, num2));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Linear time for regex matching and parsing both strings

✓ Linear Growth

Space Complexity

O(1)

Only storing extracted integer values, regex uses constant space

✓ Linear Space

28.5K Views

Medium Frequency

~15 min Avg. Time

856 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Regex-Based Parsing — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler