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  2. Merge two arrays and sort the final one

Merge two arrays and sort the final one

由 拥有回忆 提交于 2019-12-01 13:46:23

You can try following approach:

Logic

  • Create a new array that will be returned.
  • Check for first element in arrayTwo and keep it in a variable say val.
  • Loop over arrayOne and check if current value is greater than val, push it in array and decrement value of i by 1 to check next value as well with current element.
  • Now check for current element. If it is less than 0, ignore it, else push value to array.
  • Return this array.

function mergeAndSort(a1, a2) {
  var matchCount = 0;
  var ret = [];

  for (var i = 0; i < a1.length; i++) {
    var val = a2[matchCount];
    if (a1[i] > val) {
      ret.push(val)
      matchCount++
      i--;
      continue;
    }
    if (a1[i] > 0) {
      ret.push(a1[i]);
    }
  }
  console.log(ret.join())
  return ret;
}


var arrayOne = [3, 6, -1, 11, 15, -1, 23, 34, -1, 42]
var arrayTwo = [7, 19, 38];
var arrayThree = [1, 9, 28];
var arrayFour = [1,2,5]

mergeAndSort(arrayOne, arrayTwo)
mergeAndSort(arrayOne, arrayThree)
mergeAndSort(arrayOne, arrayFour)
.as-console-wrapper {
  max-height: 100% !important;
  top: 0;
}

Note: Not putting check for number of elements in arrayTwo as its clearly mentioned in question that it will be same.

There is a clean O(N) in-place solution.

First "pack" arrayOne by moving all -1 (-- below) to the front. This takes a single backward pass.

Then perform a merge by iteratively moving the smallest element among arrayTwo and the tail of arrayOne and overwriting the next --. The gap will narrow down but there will always remain room for the elements of arrayTwo.

 3,  6, --, 11, 15, --, 23, 34, --, 42
 1,  9, 28

Packing:

 3,  6, --, 11, 15, --, 23, 34, --, 42

 3,  6, --, 11, 15, --, 23, --, 34, 42

 3,  6, --, 11, 15, --, --, 23, 34, 42

 3,  6, --, 11, --, --, 15, 23, 34, 42

 3,  6, --, --, --, 11, 15, 23, 34, 42

 3, --, --, --,  6, 11, 15, 23, 34, 42

 --, --, --, 3,  6, 11, 15, 23, 34, 42

Merging:

  --, --, --,  3,  6, 11, 15, 23, 34, 42
   1,  9, 28

   1, --, --,  3,  6, 11, 15, 23, 34, 42
  --,  9, 28

   1,  3, --, --,  6, 11, 15, 23, 34, 42
  --,  9, 28

   1,  3,  6, --, --, 11, 15, 23, 34, 42
  --,  9, 28

   1,  3,  6,  9, --, 11, 15, 23, 34, 42
  --, --, 28

   1,  3,  6,  9, 11, --, 15, 23, 34, 42
  --, --, 28

   1,  3,  6,  9, 11, 15, --, 23, 34, 42
  --, --, 28

   1,  3,  6,  9, 11, 15, 23, --, 34, 42
  --, --, 28

   1,  3,  6,  9, 11, 15, 23, 28, 34, 42
  --, --, --

You should do something like insertion sort. As both the arrays are sorted (except -1s), the smallest number in array2 will be placed somewhere between first element and the first -1, 2nd element of array2 will be placed somewhere anywhere after the 1st -1 in array1 and before or at the 2nd -1 in array1 and so on.

So you have to insert a particular element of array2 in only a segment of array1 and not the whole array. Also each element of array2 have to consider different segment or subarray of array1. So, effective time complexity of this logic will be O(n+m) where n is the length of array1 and m is the length of array2.

Couldn't it be something like

compare each item of arrayTwo with each item of arrayOne If it comes to bigger that that of arrayOne, insert the item and while iterating arrayOne delete all the -1 .

That's actually an interesting question. There are many sorting algorithms, and the most efficient ones always starts from one "unchangeable" array, so without changing the values inside that. Yet here your goal is to change the value when it encounters -1, so that the value is taken from the second array.

So, you need a sorting algorithm that doesn't divide the array in pieces because if the last element of your second array is 1 (the lowest), it has to be moved to the start. If you're using a sorting algorithm that breaks the array in pieces (the divide-and-conquer tactic like quick sort) or that uses recursion, it can be problematic because it cannot be moved to the start of your main array. Unless you are aware of the main array.

What you need is an algorithm that performs a step-by-step algorithm.

The algorithm that I've used is a bubble sort, which checks each element step by step. It's then easier to replace the value if it's -1 and move its position correctly to the array. However, it is not so efficient. Maybe I will edit my post to see if I can improve that.

function mergeAndSort(arr1, arrMergeIn) {
  // merge + sort using arr1 as large one
  var mergeIndex = 0;

  for (var i = 0; i < arr1.length; ++i) {
    if (arr1[i] === -1) arr1[i] = arrMergeIn[mergeIndex++];
    var j = i;
    while (j > 0 && arr1[j - 1] > arr1[j]) {
      var tmp = arr1[j - 1];
      arr1[j - 1] = arr1[j];
      arr1[j] = tmp;
      j--
    }
  }
  return arr1;
}

// one liner console output
function showArray(arr) {
console.log(arr.join(','));
}


showArray(mergeAndSort([3, 6, -1, 11, 15, -1, 23, 34, -1, 42], [7, 19, 38]));

showArray(mergeAndSort([3, 36, -1, 1, 10, -1, 9, 34, -1, 42], [17, 9, 38]));

showArray(mergeAndSort([3, 36, -1, 1, 10, -1, 9, 34, -1, 42], [17, 9, 1]));

showArray(mergeAndSort([-1, 36, -1, 1, 10, -1, 9, 34, -1, 42], [17, 9, 100, 1]));

showArray(mergeAndSort([-1, -1, 1, 100, -1, 9, 34, -1], [17, 9, 9, 1]));

Or, you can use another strategy: replace the "-1" elements with the elements from that another array and perform an efficient algorithm on that. Although in worst-case scenario's, the "-1"s are at the end of the array, which means that there is an ~N operation + additional average complexity of a sorting algorithm (the efficient ones are of ~N*log(N))

You could iterate arrayOne in a single loop and then arrayTwo.

The idea is to separate the target index from the actual index. The first loop ignores -1 and and keep the target index.

If an actual value is greater then the first value of arrayTwo, both values swapped and in arrayTwo takes a sorting place by iterating and swapping with grater values.

Then the actual item is assigned to the target index.

Both indices gets incremented.

At the end all items of arrayTwo are added to arrayOne.

function order(arrayOne, arrayTwo) {
    var i = 0, j, l = 0;
    while (i < arrayOne.length) {
        if (arrayOne[i] === -1) {
            i++;
            continue;
        }
        if (arrayTwo[0] < arrayOne[i]) {
            [arrayOne[i], arrayTwo[0]] = [arrayTwo[0], arrayOne[i]];
            j = 0;
            while (arrayTwo[j] > arrayTwo[j + 1]) {
                [arrayTwo[j], arrayTwo[j + 1]] = [arrayTwo[j + 1], arrayTwo[j]];
                j++;
            }
        }
        arrayOne[l++] = arrayOne[i++];
    }
    j = 0;
    while (l < arrayOne.length) {
        arrayOne[l++] = arrayTwo[j++];
    }
    return arrayOne;
}

console.log(order([3, 6, -1, 11, 15, -1, 23, 34, -1, 42], [7, 19, 38]));
console.log(order([3, 6, -1, 11, 15, -1, 23, 34, -1, 42], [1, 9, 28]));
console.log(order([3, 6, -1, 11, 15, -1, 23, 34, -1, 42], [1, 2, 5]));
console.log(order([3, 6, -1, 11, 15, -1, 23, 34, -1, 42], [43, 44, 45]));
.as-console-wrapper { max-height: 100% !important; top: 0; }

Here is a simple and compact implementation of the merge sort algorithm. There will only be the same amount of operations as there are elements.

I achieve this by creating an array with the same number of elements as both arrays combined, then iterating that array.

Upon each iteration:

  • If both arrays have elements (and the smallest element is not -1) it will move the smallest element into the result.
  • If only one of the arrays still has elements (and the next element is not -1), it will move the first element from that array into the result.

Finally assign the result to the first array as per your spec

const mergeSort = (a, b) => Object.assign(a, 
  new Int32Array(a.length + b.length).reduce(m => {
    const el = [
      [a, b][+!(a[0] <= b[0])], 
      [a, b][+!a.length]][+!(a.length && b.length)
    ].shift()
    if(el !== -1) m.push(el)
    return m
  }, [])
)

const arr1 = [1,3,5,7,9]
const arr2 = [0,2,4]

mergeSort(arr1, arr2)

console.log(arr1) // [0,1,2,3,4,5,7,9]

Let me re-phrase three most important aspects of the task:

  • Both arrays are sorted and include positive integers (except -1 place-holders in Array1)
  • The -1 is meaningless placeholder (its position in Array1 is random)
  • Array1 length equals output array length

Pseudo logic:

  • parse both array1 and array2 from the first position to end of array
  • ignore -1 values in array1
  • if array1[current position] <= array2[current position] then write array1[current position] into merged array and move to next position in array1; otherwise apply same for array2

Code example:

            public static void AMerge()
        {
            int[] array1 = new int[] { 3, 6, -1, 11, 15, -1, 23, 34, -1, 42 };
            int[] array2 = new int[] { 1, 9, 28 };
            int[] arrayMerged = new int[array1.Length];

            int array1Index = 0;
            int array2Index = 0;

            for (int arrayMergedIndex = 0; arrayMergedIndex < array1.Length; arrayMergedIndex++)
            {
                while (array1[array1Index] == -1) array1Index++; //ignore -1 placeholders
                if ((array1Index < array1.Length && array2Index >= array2.Length) 
                    || array1[array1Index] <= array2[array2Index])  //choose which array will provide current merged value
                {
                    arrayMerged[arrayMergedIndex] = array1[array1Index];
                    array1Index++;
                }
                else
                {
                    arrayMerged[arrayMergedIndex] = array2[array2Index];
                    array2Index++;
                }
            }

            char[] charsToTrim = { ',', ' '};
            string arrayMergedString = "{";
            foreach (int f in arrayMerged) arrayMergedString += f.ToString() + " ";
            arrayMergedString = arrayMergedString.TrimEnd(charsToTrim) + "}";
            Console.WriteLine(arrayMergedString);
            Console.ReadKey();
        }
    }

Note:

  • Optimizing for speed => creating new arrayMerged; optimization for space would require moving array1 elements in the else branch

You should use merge function from Merge sort, and modify it so that it doesn't create a new array, but instead uses array1, and perform translation, after insertion of an element from array2, that will shift elements to the right until the next -1, and thus overwrite the -1.

Mavi Domates

function merger(a1, a2) {
  var i = 0;
  var j = 0;

  while (i < a1.length && j < a2.length) {
    if (a1[i] > a2[j]) {
      // Swap values
      var temp = a2[j];
      a2[j] = a1[i];
      a1[i] = temp;
      i++;
    } else if (a1[i] !== -1 && a1[i] <= a2[j]) {
      i++;
    } else {
      var temp = a2[j];
      a2[j] = a1[i];
      a1[i] = temp;
      i++;
      j++;
    }
  }

  return a1;
}

var arrayOne = [3, 5, -1, 11, 15, -1, 23, 34, -1, 42];
var arrayTwo = [6, 19, 38];

console.log(merger(arrayOne, arrayTwo))

With certain pre-conditions (no other numbers with <0, a1 should be smaller than a2 etc - which could all be handled) this should solve the problem in JS.

Since they are both sorted, the order of arrayTwo's items should match the order of -1s in arrayOne. Then the job becomes simple and can be implemented as follows;

function replaceMissing(a,b){
  var i = 0;
  return a.map(n => n < 0 ? b[i++] : n);
}
var arrayOne = [3,6,-1,11,15,-1,23,34,-1,42],
    arrayTwo = [7,19,38];
      result = replaceMissing(arrayOne,arrayTwo);
console.log(result);

Edit: I believe the upper solution does make more sense in the general logic of the question. If the position of -1s does not mean anything then what use do they have? Let's just delete the -1's and do a simple insertion of arrayTwo items at proper indices in arrayOne. This can very simply be done as follows.

function replaceMissing(a,b){
  var j = b.length-1;
  return b.concat(a.reduceRight((r,m,i) => (m < 0 ? r.splice(i,1)
                                                  : m < b[j] && r.splice(i+1,0,b.splice(j--,1)[0]),
                                            r), a.slice()));
}
var arrayOne = [3,6,-1,11,15,-1,23,34,-1,42],
    arrayTwo = [1,25,38];
      result = replaceMissing(arrayOne,arrayTwo);
console.log(result);
标签
javascript
arrays
algorithm
sorting
puzzle
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