행렬의 곱셈 (프로그래머스)

• basically multiply matrices (dot product)
• https://school.programmers.co.kr/learn/courses/30/lessons/12949

solution

  public static int[][] solution(int[][] arr1, int[][] arr2) {
    int aRow = arr1.length;
    int aCol = arr1[0].length;
    // bRow = arr2.length; // aCol과 동일
    int bCol = arr2[0].length;
    int[][] res = new int[aRow][bCol];
 
    for (int i = 0; i < aRow; i++) { // arr1의 row
      for (int j = 0; j < bCol; j++) { // arr2의 col
        for (int k = 0; k < aCol; k++) { // 순회하면서 곱셈, bRow = aCol
          res[i][j] += arr1[i][k] * arr2[k][j];
        }
      }
    }
    return res;
  }

time complexity

• O(M * K * N)
  • M = row1 (number of rows in arr1)
  • N = col2 (number of cols in arr2)
  • K = col1 (shared dimension: columns of arr1/rows of arr2)