Multiply Two Polynomials

Multiply Two Polynomials - Problem

Array Math Hard

You are given two integer arrays poly1 and poly2, where the element at index i in each array represents the coefficient of xⁱ in a polynomial.

Let A(x) and B(x) be the polynomials represented by poly1 and poly2, respectively.

Return an integer array result of length (poly1.length + poly2.length - 1) representing the coefficients of the product polynomial R(x) = A(x) * B(x), where result[i] denotes the coefficient of xⁱ in R(x).

Input & Output

Example 1 — Basic Multiplication

$ Input: poly1 = [1,2], poly2 = [3,4]

› Output: [3,10,8]

💡 Note: Polynomial A(x) = 1 + 2x, B(x) = 3 + 4x. A(x) * B(x) = (1)(3) + (1)(4x) + (2x)(3) + (2x)(4x) = 3 + 4x + 6x + 8x² = 3 + 10x + 8x²

Example 2 — Single Coefficients

$ Input: poly1 = [1,2,1], poly2 = [2]

› Output: [2,4,2]

💡 Note: A(x) = 1 + 2x + x², B(x) = 2. A(x) * B(x) = 2(1 + 2x + x²) = 2 + 4x + 2x²

Example 3 — Zero Coefficients

$ Input: poly1 = [1,0,1], poly2 = [1,1]

› Output: [1,1,1,1]

💡 Note: A(x) = 1 + 0x + x² = 1 + x², B(x) = 1 + x. A(x) * B(x) = (1 + x²)(1 + x) = 1 + x + x² + x³

Constraints

1 ≤ poly1.length, poly2.length ≤ 100
-100 ≤ poly1[i], poly2[i] ≤ 100

Visualization

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Asked in

G Google 15 f Facebook 12 M Microsoft 8

The key insight is recognizing that polynomial multiplication follows the rule x^i × x^j = x^(i+j). For each coefficient pair poly1[i] and poly2[j], their product contributes to the coefficient at position i+j in the result. Best approach uses direct accumulation with Time: O(n×m), Space: O(n+m).

Common Approaches

✓ Brute Force Multiplication

⏱️ Time: O(n×m) Space: O(n+m)

For each coefficient in poly1, multiply it with every coefficient in poly2. When multiplying coefficients at positions i and j, the result contributes to position i+j in the result array.

Direct Coefficient Accumulation

⏱️ Time: O(n×m) Space: O(n+m)

Uses the mathematical property that when multiplying x^i and x^j, the result is x^(i+j). We directly accumulate coefficients at the correct positions without additional processing.

Brute Force Multiplication — Algorithm Steps

Step 1: Create result array of size (poly1.length + poly2.length - 1)
Step 2: For each term i in poly1 and term j in poly2, add poly1[i] * poly2[j] to result[i+j]
Step 3: Return the result array

Visualization

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Step-by-Step Walkthrough

Setup

Initialize result array of size n+m-1

Multiply

For each pair (i,j), add poly1[i]*poly2[j] to result[i+j]

Result

Final coefficient array

Code -

solution.c — C

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

int* solution(int* poly1, int poly1Size, int* poly2, int poly2Size, int* returnSize) {
    *returnSize = poly1Size + poly2Size - 1;
    int* result = (int*)calloc(*returnSize, sizeof(int));
    
    for (int i = 0; i < poly1Size; i++) {
        for (int j = 0; j < poly2Size; j++) {
            result[i + j] += poly1[i] * poly2[j];
        }
    }
    
    return result;
}

int main() {
    char line1[1000], line2[1000];
    fgets(line1, sizeof(line1), stdin);
    fgets(line2, sizeof(line2), stdin);
    
    int poly1[100], poly2[100];
    int poly1Size = 0, poly2Size = 0;
    
    // Parse first array
    char* token = strtok(line1, "[],\n");
    while (token != NULL) {
        poly1[poly1Size++] = atoi(token);
        token = strtok(NULL, "[],\n");
    }
    
    // Parse second array
    token = strtok(line2, "[],\n");
    while (token != NULL) {
        poly2[poly2Size++] = atoi(token);
        token = strtok(NULL, "[],\n");
    }
    
    int returnSize;
    int* result = solution(poly1, poly1Size, poly2, poly2Size, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf("%d", result[i]);
        if (i < returnSize - 1) printf(",");
    }
    printf("]\n");
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n×m)

Nested loops where outer runs n times and inner runs m times

✓ Linear Growth

Space Complexity

O(n+m)

Result array of size n+m-1 to store coefficients

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force Multiplication — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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