In how many ways can this function be simplified?

Question

I'm trying to simplify/generalize a function foo_one that traverses through a multidimensional array and sets all sub-array elements apart from the first one to have null values.

I was given only the first function, for which I should then find more ways to generalize it to still accomplish the intended purpose of foo_one.

Any suggestions would be highly appreciated.

I've already implemented 2 other variations of the function and I like to have more if possible.

var arr = [
  [2, null, 2, null],
  [2, null, 2, null],
  [null, null, null, null],
  [null, null, null, null]
];

function foo_one() {
  for (let y = 0; y < arr.length - 1; y++) {
    for (let x = 0; x < arr.length - 1; x++) {
      if (arr[x + 1][y] != null) {
        if (arr[x + 1][y] == arr[x][y]) {
          arr[x][y] = arr[x][y] * 2;
          arr[x + 1][y] = null;
        }
        if (arr[x][y] == null) {
          arr[x][y] = arr[x + 1][y];
          arr[x + 1][y] = null;
        }
      }
    }
  }
}

function foo_two() {
  for (let y = 0; y < arr.length - 1; y++) {
    for (let x = 0; x < arr.length - 1; x++) {
      if (arr[x + 1][y] != null && arr[x + 1][y] == arr[x][y]) {
        arr[x][y] = arr[x][y] * 2;
        arr[x + 1][y] = null;
      }
    }
  }
}

function foo_three() {
  for (let y = 0; y < arr.length - 1; y++) {
    for (let x = 0; x < arr[y].length - 1; x++) {
      if (arr[x + 1][y] != null && arr[x + 1][y] == arr[x][y]) {
        arr[x][y] = arr[x][y] * 2;
        arr[x + 1][y] = null;
      }
    }
  }
}

// Output
[ [ 4, null, 4, null ],
  [ null, null, null, null ],
  [ null, null, null, null ],
  [ null, null, null, null ] ]

What is all that attempting to do, can you describe in English? — CertainPerformance, Apr 07 '19 at 05:19
@CertainPerformance I'm just trying to generalize the first function into simpler versions. That's all — moodygeek, Apr 07 '19 at 05:24
you don't specify what that function tries to do actually! please explain what is its role? — S.Serpooshan, Apr 07 '19 at 05:30
May be you can "club nested if statement" in single using && operator. — TheKingPinMirza, Apr 07 '19 at 05:54
Why are you ignoring the last column? with "let x = 0; x < arr.length - 1; x++"? That means that if there are any 2's in the last column that match, they will be ignore? — Rickard Elimää, Apr 07 '19 at 08:07
Also, the first two foo methods are wrong because you don't take the second array's, within the array, length into account. — Rickard Elimää, Apr 07 '19 at 08:08
@CertainPerformance I've edited the question description to bring more clarity. — moodygeek, Apr 07 '19 at 12:08
let arr = [ [2, null, 2, null], [2, null, 2, null], [null, null, null, null], [null, null, null, null] ]; function foo(){ x = []; for (line in arr){ for (items in line){ if (arr[line][items]!= NaN) { x.push(parseInt(arr[line][items])*2)}}} }; foo(); console.log(x) — PythonProgrammi, Apr 07 '19 at 14:07

Rex Hsu · Answer 1 · 2019-04-08T05:24:26.177

The following rewriting is more efficient than the original post.

let arr = [
    [2, null, 2, null],
    [2, null, 2, null],
    [null, null, null, null],
    [null, null, null, null]
];

function foo_four(arr) {
    for (let y = 0, yl = arr.length - 1; y < yl; y++) {
        let currRow = arr[y];
        let nextRow = arr[y + 1];

        for (let x = 0, xl = nextRow.length - 1; x < xl; x++) {
            let currValue = currRow[x];

            if (currValue && currValue === nextRow[x]) {
                currRow[x] = currValue * 2;
                nextRow[x] = null;
            }
        }
    }
}

foo_four(arr);
console.log(arr);
// [
//     [4, null, 4, null],
//     [null, null, null, null],
//     [null, null, null, null],
//     [null, null, null, null]
// ];

thanks. this is efficient, but do couldn't there be a way to not use `nested if`? — moodygeek, Apr 07 '19 at 12:27

nice_dev · Answer 2 · 2019-04-07T07:50:36.333

You can make the processing inside the nested loops more concise. Also, I don't see why you need to ignore the last column.

function foo_four() {
  for (let y = 0; y < arr.length; y++) {
    for (let x = 0; x < arr.length - 1; x++) {
      arr[x][y] = arr[x][y] === null ? arr[x+1][y] : arr[x][y] * 2;
      arr[x + 1][y] = null;
    }
  }
}

var arr = [
  [2, null, 2, null],
  [2, null, 2, null],
  [null, null, null, null],
  [null, null, null, null]
];

foo_four();
console.log(arr);

Rickard Elimää · Accepted Answer · 2019-04-07T08:54:39.767

function foo_four(arr) {
   let yLoopCount = arr.length - 1,
       xArrLength = 0,
       currentRowValue = 0,
       nextRowValue = 0;

   for (let y = 0; y < yLoopCount; y++) {
     xArrLength = arr[y].length;  // arr[y].length - 1

     for (let x = 0; x < xArrLength; x++) {
       currentRowValue = arr[y][x];
       nextRowValue  = arr[y+1][x];

       if (currentRowValue && currentRowValue == nextRowValue) {
         arr[y][x] = currentRowValue * 2;
         arr[y+1][x] = null;
       }
    }

  return arr;
}


var arr = [
  [2, null, 2, null],
  [2, null, 2, null],
  [null, null, null, null],
  [null, null, null, null]
];

foo_four();

// Output
// [
//     [4, null, 4, null],
//     [null, null, null, null],
//     [null, null, null, null],
//     [null, null, null, null]
// ];

a) I'm looping through the last row as well.
b) It's a minor optimization if you create the variables outside the loop, and then reuse them.
c) It's a minor optimization if you're using variables to shorten any kind of deep reference. Diving into each subproperty takes time, like referencing arr[y][x] several times instead of currentRowValue.
d) I prefer to send the array as a parameter, and also return it, even if I change the array. Cleaner code in my book, and easier to debug.
e) Use variable (or method) names to explain what's going on.
f) By using variables, it's easier to debug.
g) I usually give all variables a starting value to describe what kind of type they are (empty array, object, number, string) so the next person can look at them and see what they are without having to read and process what the variable names means.

thank you. this is very clear and concise and I love you explanations — moodygeek, Apr 07 '19 at 12:24

In how many ways can this function be simplified?

3 Answers3