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Compute the distance between elements in a bi-dimensional array in Javascript

Having the following input string: 923857614

This is represented into a matrix like this:

9 2 3
8 5 7
6 1 4 

Having a moving sequence like this: 423692, this means that we start in point 4, we move to 2, then to 3, then to 6, then to 9 and finally to 2.

It must be computed the length of the road. At beginning it starts from 0, if the next step is adjacent to the current one, add 1, if it isn't adjacent, add 2.

How I tried to do it:

function computeRoadLength(keypad, movingSequence) {
  // build the matrix
 const arr = [[keypad[0], keypad[1], keypad[2]],
              [keypad[3], keypad[4], keypad[5]], 
              [keypad[6], keypad[7], keypad[8]]];
  let roadLength = 0;
  for (i = 0; i < movingSequence.length; i++) {
    // some way to compute the distance here
    if (arr[i] > arr[i+1]) roadLength = roadLength + 1;
    if (arr[i] < arr[i+1]) roadLength = roadLength + 2;
  }

  return roadLength;
}

computeRoadLength(923857614, 423692); // 2 + 1 + 2  + 2  + 1 = 8, should return 8
about 4 years ago · Juan Pablo Isaza
1 Respuestas
Responde la pregunta

0

You could take a different approach by using an object of positions of all keypad values and take the absolute delta of the positions.

For adding to movingSequence add one or max two.

function computeRoadLength(keypad, movingSequence) {
    const positions = {};
    
    for (let i = 0; i < keypad.length; i++) {
        positions[keypad[i]] = [Math.floor(i / 3), i % 3];
    }
    
    let roadLength = 0,
        last = positions[movingSequence[0]];

    for (let i = 1; i < movingSequence.length; i++) {
        const
            item = positions[movingSequence[i]],
            sum = Math.abs(last[0] - item[0]) + Math.abs(last[1] - item[1]);
                
        roadLength += Math.min(sum, 2);        
        last = item;
    }

    return roadLength;
}

console.log(computeRoadLength('923857614', '423692')); // 2 + 1 + 2 + 2 + 1 = 8

about 4 years ago · Juan Pablo Isaza Denunciar
Responde la pregunta
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