CODEWITHSUNDEEP
javascriptmatrixtranspose
Natalia Grzywacz
Natalia Grzywacz

Reputation: 53

Matrix transpose JS

Functions that output the transpose of a matrix - a new matrix where the columns and rows of the original are swapped. I'm trying to figure out why the second function outputs incorrect results?

function transpose(matrix) {
    let res = [];
    for(let i = 0;  i < matrix[0].length; i++) {
        res[i] = [];
        for(let j = 0;  j < matrix.length; j++) {
            res[i][j] = matrix[j][i];
        }
    }
    return res; 
}

function transpose(matrix) {
    let res = Array(matrix[0].length).fill([]);
    for(let i = 0;  i < res.length; i++) {
        for(let j = 0;  j < matrix.length; j++) {
            res[i][j] = matrix[j][i];
        }
    }
    return res; 
}

Upvotes: 0

Views: 1388

Answers (1)

lklklklkpopopo
lklklklkpopopo

Reputation: 81

I think the second function fails because of the let res line. When you use .fill([]), you are filling with the same array as givemesnacks mentions in this comment

If you do console.log(transpose([[1, 2, 3], [7, 4, 3]])), you'll notice that the elements in the array are the same.

To fix this, you can try let res = Array(matrix[0].length).fill().map(() => []);, which should add a new empty array for every row.

function transpose(matrix) {
  let res = Array(matrix[0].length).fill().map(() => []);
  for (let i = 0; i < res.length; i++) {
    for (let j = 0; j < matrix.length; j++) {
      res[i][j] = matrix[j][i];
    }
  }
  return res;
}

let m = [
  [0, 1, 2],
  [3, 4, 5],
  [6, 7, 8]
]

console.log(transpose(m))

Upvotes: 2

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