Multiply Strings

Multiply Strings - Problem

Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2, also represented as a string.

Note: You must not use any built-in BigInteger library or convert the inputs to integer directly.

Input & Output

Example 1 — Basic Multiplication

$ Input: num1 = "2", num2 = "3"

› Output: "6"

💡 Note: Simple single digit multiplication: 2 × 3 = 6

Example 2 — Multi-digit Numbers

$ Input: num1 = "123", num2 = "456"

› Output: "56088"

💡 Note: Grade school multiplication: 123 × 456 = 56088 using digit-by-digit multiplication with carries

Example 3 — Zero Case

$ Input: num1 = "0", num2 = "9133"

› Output: "0"

💡 Note: Any number multiplied by zero equals zero

Constraints

1 ≤ num1.length, num2.length ≤ 200
num1 and num2 consist of digits only
Both num1 and num2 do not contain any leading zero, except the number 0 itself

Visualization

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Asked in

G Google 15 f Facebook 12 a Amazon 8 M Microsoft 6

The key insight is to simulate grade school multiplication using arrays to handle large numbers without integer conversion. Best approach uses position-based multiplication where digits at positions i and j contribute to result position i+j. Time: O(n×m), Space: O(n+m)

Common Approaches

✓ Direct Conversion (Invalid)

⏱️ Time: O(n + m) Space: O(1)

This approach would convert the string numbers to integers, perform multiplication, and convert the result back to string. However, this violates the problem constraint of not converting inputs to integers directly.

Grade School Multiplication

⏱️ Time: O(n × m) Space: O(n + m)

This approach simulates the grade school multiplication algorithm where we multiply each digit of num2 with each digit of num1, handling carries and positioning correctly. We use arrays to store intermediate results and avoid integer conversion.

Direct Conversion (Invalid) — Algorithm Steps

Convert num1 string to integer
Convert num2 string to integer
Multiply the integers
Convert result back to string

Visualization

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Step-by-Step Walkthrough

Convert Strings

Convert num1 and num2 to integers

Multiply

Perform integer multiplication

Problem

Integer overflow occurs with large numbers

Code -

solution.c — C

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

void reverseString(char* str) {
    int len = strlen(str);
    for (int i = 0; i < len / 2; i++) {
        char temp = str[i];
        str[i] = str[len - 1 - i];
        str[len - 1 - i] = temp;
    }
}

char* solution(char* num1, char* num2) {
    if (strcmp(num1, "0") == 0 || strcmp(num2, "0") == 0) {
        char* result = malloc(2);
        strcpy(result, "0");
        return result;
    }
    
    int m = strlen(num1);
    int n = strlen(num2);
    int* result = calloc(m + n, sizeof(int));
    
    // Reverse both strings for easier processing
    reverseString(num1);
    reverseString(num2);
    
    // Grade school multiplication
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            int mul = (num1[i] - '0') * (num2[j] - '0');
            int p1 = i + j;
            int p2 = i + j + 1;
            
            mul += result[p1];
            result[p1] = mul % 10;
            result[p2] += mul / 10;
        }
    }
    
    // Convert result array to string
    // Reverse result array
    for (int i = 0; i < (m + n) / 2; i++) {
        int temp = result[i];
        result[i] = result[m + n - 1 - i];
        result[m + n - 1 - i] = temp;
    }
    
    // Remove leading zeros
    int start = 0;
    while (start < m + n && result[start] == 0) {
        start++;
    }
    
    if (start == m + n) {
        free(result);
        char* res = malloc(2);
        strcpy(res, "0");
        return res;
    }
    
    char* res = malloc(m + n - start + 1);
    for (int i = start; i < m + n; i++) {
        res[i - start] = result[i] + '0';
    }
    res[m + n - start] = '\0';
    
    free(result);
    return res;
}

int main() {
    char num1[1000], num2[1000];
    fgets(num1, sizeof(num1), stdin);
    fgets(num2, sizeof(num2), stdin);
    num1[strcspn(num1, "\n")] = 0;
    num2[strcspn(num2, "\n")] = 0;
    char* result = solution(num1, num2);
    printf("%s\n", result);
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n + m)

String to integer conversion takes O(n+m) time

✓ Linear Growth

Space Complexity

O(1)

Only storing integer values, no extra data structures

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Direct Conversion (Invalid) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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