Matrix Multiplication

Hard Matrix

Multiply two square matrices and print the resulting matrix.

Sample input

A=[[1,2],[3,4]], B=[[5,6],[7,8]]

Sample output

19 22 43 50

Solution

a = [[1, 2], [3, 4]]
b = [[5, 6], [7, 8]]
n = len(a)
result = [[0] * n for _ in range(n)]
for i in range(n):
    for j in range(n):
        for k in range(n):
            result[i][j] += a[i][k] * b[k][j]
for row in result:
    print(*row)

const a = [[1, 2], [3, 4]];
const b = [[5, 6], [7, 8]];
const n = a.length;
const result = Array.from({ length: n }, () => new Array(n).fill(0));
for (let i = 0; i < n; i++) {
  for (let j = 0; j < n; j++) {
    for (let k = 0; k < n; k++) {
      result[i][j] += a[i][k] * b[k][j];
    }
  }
}
for (const row of result) console.log(row.join(" "));

import java.util.Arrays;
import java.util.stream.Collectors;

public class Main {
    public static void main(String[] args) {
        int[][] a = {{1, 2}, {3, 4}};
        int[][] b = {{5, 6}, {7, 8}};
        int n = a.length;
        int[][] result = new int[n][n];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                for (int k = 0; k < n; k++) {
                    result[i][j] += a[i][k] * b[k][j];
                }
            }
        }
        for (int[] row : result) {
            System.out.println(Arrays.stream(row).mapToObj(String::valueOf).collect(Collectors.joining(" ")));
        }
    }
}

fun main() {
    val a = arrayOf(intArrayOf(1, 2), intArrayOf(3, 4))
    val b = arrayOf(intArrayOf(5, 6), intArrayOf(7, 8))
    val n = a.size
    val result = Array(n) { IntArray(n) }
    for (i in 0 until n) {
        for (j in 0 until n) {
            for (k in 0 until n) {
                result[i][j] += a[i][k] * b[k][j]
            }
        }
    }
    for (row in result) {
        println(row.joinToString(" "))
    }
}

let a = [[1, 2], [3, 4]]
let b = [[5, 6], [7, 8]]
let n = a.count
var result = Array(repeating: Array(repeating: 0, count: n), count: n)
for i in 0..<n {
    for j in 0..<n {
        for k in 0..<n {
            result[i][j] += a[i][k] * b[k][j]
        }
    }
}
for row in result {
    print(row.map(String.init).joined(separator: " "))
}

void main() {
  final a = [[1, 2], [3, 4]];
  final b = [[5, 6], [7, 8]];
  final n = a.length;
  final result = List.generate(n, (_) => List.filled(n, 0));
  for (var i = 0; i < n; i++) {
    for (var j = 0; j < n; j++) {
      for (var k = 0; k < n; k++) {
        result[i][j] += a[i][k] * b[k][j];
      }
    }
  }
  for (final row in result) {
    print(row.join(' '));
  }
}

#include <iostream>
#include <vector>
using namespace std;

int main() {
    vector<vector<int>> a = {{1, 2}, {3, 4}};
    vector<vector<int>> b = {{5, 6}, {7, 8}};
    int n = a.size();
    vector<vector<int>> result(n, vector<int>(n, 0));
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            for (int k = 0; k < n; k++) {
                result[i][j] += a[i][k] * b[k][j];
            }
        }
    }
    for (auto& row : result) {
        for (int j = 0; j < n; j++) {
            cout << row[j];
            if (j < n - 1) cout << " ";
        }
        cout << endl;
    }
    return 0;
}

#include <stdio.h>

int main() {
    int a[2][2] = {{1, 2}, {3, 4}};
    int b[2][2] = {{5, 6}, {7, 8}};
    int n = 2;
    int result[2][2] = {{0, 0}, {0, 0}};
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            for (int k = 0; k < n; k++) {
                result[i][j] += a[i][k] * b[k][j];
            }
        }
    }
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            printf("%d", result[i][j]);
            if (j < n - 1) printf(" ");
        }
        printf("\n");
    }
    return 0;
}