Discussion on Flipping the Matrix Challenge

8 years ago+ 0 comments

SPOILER ALLERT: this post contains a solution to the problem.

Just to share a fun functional apporach in JS, and yes it could be even simplier with python.

// Matrix library
const transpose = (m) => m[0].map((_, i) => m.map((x) => x[i]));
const hmirror = (m) => m.map((mr) => mr.map((_, i, a) => a[a.length - 1 - i]));
const vmirror = (m) => transpose(hmirror(transpose(m)));
const flatten = (m) => [].concat.apply([], m);
// Eval and output the max sum
const evalMtx = (m, n) => {
  // split the matrix into quadrants
  const first = m.map((r) => r.slice(0, n)).slice(0, n);
  const second = m.map((r) => r.slice(n)).slice(0, n);
  const third = m.slice(n).map((r) => r.slice(0, n));
  const last = m.slice(n).map((r) => r.slice(n));
  // mirror the quadrants to meet the first quadrant and flatten them all
  const m_prime = [first, hmirror(second), vmirror(third), vmirror(hmirror(last))]
    .map((s) => flatten(s));
  // console.log(m_prime);
  return m_prime[0].reduce((sum, _, i) => sum + 
    m_prime.map((q) => q[i]).reduce((max, v) => v > max ? v : max, 0)
    , 0);
}

// input parsing and main program
const inputBuffer = input.split('\n').reverse()
const q = parseInt(inputBuffer.pop());
Array.from({ length: q }).forEach((_) => {
  const n = parseInt(inputBuffer.pop());
  const mtx = Array.from({ length: n + n })
    .map((_) => inputBuffer.pop().split(' ').map((i) => parseInt(i)));
  // console.log(mtx)
  console.log(evalMtx(mtx, n))
})

To understand this appoach you'd have to realize that each cell in the first quadrant can be 1 of the 4 cells from each quadrant in the fixed positions.

Array.from({ length: n }) creates an array of n elements.

Ragarding to the reduce functions, the outter one is a sum and the inner one is a max function.

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