Empresas
Empleos
  • Sobre nosotros
  • Soluciones
    • Publicación de vacantes
      Publica tu vacante y recibe candidatos calificados en 48h.
    • Evaluación de candidatos
      500+ pruebas técnicas y psicológicas, más anti-fraude.
    • Headhunting
      Búsqueda ejecutiva a la medida de principio a fin.
    • Nómina + EOR
      Dispersión de nómina y EOR en más de 15 países de LATAM.
  • Precios
  • Empleos

0

128
Vistas
how to get sorted index of min values of 2d array

having 2d number array like;

const arr = [
  [1, 5, 9],
  [2, 7, 8],
  [3, 0, 6],
];

what is the simplest way to get sorted array of array indexes where sort critera is values of original 2d array?

result should be:

  `[2,1]`, // (value=0)
  `[0,0]`, // (value=1)
  `[2,0]`, // (value=2)
  `[0,1]`, // (value=3)
  ...

btw, actual values are floats not that it matters.
but complexity does matter as loop runs on each frame.

about 4 years ago · Juan Pablo Isaza
2 Respuestas
Responde la pregunta

0

You could get the indices first and sort them by the value of the matrix.

const
    array = [[1, 5, 9], [2, 7, 8], [3, 0, 6]],
    result = array
        .flatMap((a, i) => a.map((_, j) => [i, j]))
        .sort((a, b) => array[a[0]][a[1]] - array[b[0]][b[1]]);

console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }

about 4 years ago · Juan Pablo Isaza Denunciar

0

Something like this?

const arr = [
  [1, 5, 9],
  [2, 7, 8],
  [3, 0, 6],
];

const map = []

arr.forEach((row, rowIndex) => row.forEach((col, colIndex) => {
  map.push({ c: colIndex, r: rowIndex, value: col });
}))

console.log(map)
const sorted = map.sort((a, b) => {
  if(a.value === b.value) return 0;
  return a.value > b.value ? 1 : -1;
})

console.log(sorted)
console.log(sorted.map(v => [v.r, v.c]))

it's working version, not optimalized

EDIT: little optimalization with execution time measure:

const arr = [
  [1, 5, 9],
  [2, 7, 8],
  [3, 0, 6],
];

const sortFn = (a, b) => a.value === b.value ? 0 : a.value > b.value ? 1 : -1;

let count = 3;
function process(arr) {
  const map = []
  arr.forEach((row, rowIndex) => row.forEach((col, colIndex) => {
    map.push({ row: rowIndex, col: colIndex, value: col });
  }))
  return map.sort(sortFn).map(v => [v.row, v.col]);
}

const interval = setInterval(() => {
  console.time("process");
  console.log(process(arr))
  console.timeEnd("process");
  if(--count === 0) clearInterval(interval);
}, 100)

Output:

[
  [ 2, 1 ], [ 0, 0 ],
  [ 1, 0 ], [ 2, 0 ],
  [ 0, 1 ], [ 2, 2 ],
  [ 1, 1 ], [ 1, 2 ],
  [ 0, 2 ]
]
process: 3.822ms
[
  [ 2, 1 ], [ 0, 0 ],
  [ 1, 0 ], [ 2, 0 ],
  [ 0, 1 ], [ 2, 2 ],
  [ 1, 1 ], [ 1, 2 ],
  [ 0, 2 ]
]
process: 1.661ms
[
  [ 2, 1 ], [ 0, 0 ],
  [ 1, 0 ], [ 2, 0 ],
  [ 0, 1 ], [ 2, 2 ],
  [ 1, 1 ], [ 1, 2 ],
  [ 0, 2 ]
]
process: 1.665ms

EDIT 2: (inspired by another answer with flatMap)

const arr = [
  [1, 5, 9],
  [2, 7, 8],
  [3, 0, 6],
];

const sortFn = (a, b) => a.value === b.value ? 0 : a.value > b.value ? 1 : -1;
arr.flatMap((_row, row) => _row.map((value, col) => { return { row, col, value } })).sort(sortFn).map(v => [v.row, v.col]);

let count = 3;
const process = (arr) => arr.flatMap((_row, row) => _row.map((value, col) => { return { row, col, value } })).sort(sortFn).map(v => [v.row, v.col]);

const interval = setInterval(() => {
  console.time("process");
  console.log(process(arr))
  console.timeEnd("process");
  if(--count === 0) clearInterval(interval);
}, 100)

Result:

[
  [ 2, 1 ], [ 0, 0 ],
  [ 1, 0 ], [ 2, 0 ],
  [ 0, 1 ], [ 2, 2 ],
  [ 1, 1 ], [ 1, 2 ],
  [ 0, 2 ]
]
process: 10.979ms
[
  [ 2, 1 ], [ 0, 0 ],
  [ 1, 0 ], [ 2, 0 ],
  [ 0, 1 ], [ 2, 2 ],
  [ 1, 1 ], [ 1, 2 ],
  [ 0, 2 ]
]
process: 1.346ms
[
  [ 2, 1 ], [ 0, 0 ],
  [ 1, 0 ], [ 2, 0 ],
  [ 0, 1 ], [ 2, 2 ],
  [ 1, 1 ], [ 1, 2 ],
  [ 0, 2 ]
]
process: 2.043ms
about 4 years ago · Juan Pablo Isaza Denunciar
Responde la pregunta
Encuentra empleos remotos

¡Descubre la nueva forma de encontrar empleo!

Top de empleos
Top categorías de empleo
Empresas
Publicar vacante Precios Comercial
Legal
Términos y condiciones Política de privacidad
© 2026 PeakU Inc. All Rights Reserved.
Andres GPT
Recomiéndame algunas ofertas
Necesito ayuda