[Code Signal][js] matrix Elements Sum

🎆문제

After becoming famous, the CodeBots decided to move into a new building together. Each of the rooms has a different cost, and some of them are free, but there's a rumour that all the free rooms are haunted! Since the CodeBots are quite superstitious, they refuse to stay in any of the free rooms, or any of the rooms below any of the free rooms.

Given matrix, a rectangular matrix of integers, where each value represents the cost of the room, your task is to return the total sum of all rooms that are suitable for the CodeBots (ie: add up all the values that don't appear below a 0).

Example

For

matrix = [[0, 1, 1, 2],
          [0, 5, 0, 0],
          [2, 0, 3, 3]]
the output should be
matrixElementsSum(matrix) = 9.

example 1

There are several haunted rooms, so we'll disregard them as well as any rooms beneath them. Thus, the answer is 1 + 5 + 1 + 2 = 9.

For

matrix = [[1, 1, 1, 0],
          [0, 5, 0, 1],
          [2, 1, 3, 10]]
the output should be
matrixElementsSum(matrix) = 9.

example 2

Note that the free room in the final column makes the full column unsuitable for bots (not just the room directly beneath it). Thus, the answer is 1 + 1 + 1 + 5 + 1 = 9.

🎇풀이

function matrixElementsSum(matrix) {
    let rooms = matrix;
    let result = [];

    for(i = 0; i < rooms.length; i++) {
        for(j = 0; j < rooms[0].length; j++) {
            if(rooms[i][j] === 0 && i < rooms.length - 1) {
                rooms[i+1][j] = 0;
            } else if (rooms[i][j] !== 0){
                result.push(rooms[i][j]);
            }
        }
    }
    return result.reduce((sum,curr) => {
        return sum + curr;
    }, 0)
}
/**
 * ex 1)
 * matrix[0][0] === 0이므로, 그 아래층 matrix[1][0], matrix[2][0] 은 탈락
 * matrix[1][2] === 0이므로, 그 아래층 matrix[2][2]는 탈락
 * 즉, matrix[i][j] === 0이라고 할 때 matrix[i+1][j], matrix[i+2][j]는 탈락이다.
 *
 * 한 층에 유령이 들어간 방을 찾고, x표시를 한다.
 */

🧨다른 풀이

function matrixElementsSum(matrix) {
    var total = 0;
    for (var i = 0; i < matrix[0].length; i++) {
        for (var j = 0; j < matrix.length; j++) {
            if (matrix[j][i] === 0) {
                break;
            }
            total += matrix[j][i];
        }
    }
    return total;
}

좀 더 효율적인 풀이이다. 다음 알고리즘 문제 때 참고하자! :

1. 일일히 내 풀이처럼 0으로 표시하지 않고, break로 구문 중단
2. 새로운 배열에 푸시한 후 reduce로 전체 배열 요소 합계를 구하는 대신 total+=matrix[j][i]사용