Question

Given n>=0, create an array with the pattern {1,1, 2,1, 2, 3, ... 1, 2, 3 .. n}.

As an example if you given n=3, your method should return array as {1,1,2,1,2,3}.

My solution is here….

public int[] upSeries(int n) {

    int var1 = n + 1;        
    int var2 = n;
    int var3 = (var1*var2) / 2;
    int arr_length = var3;
    int value = 1;
    int index = 0;
    int[] arr = new int[arr_length];
    for (int j = 0; j < arr.length; j++) {
        for (int p = 0; p < j + 1; p++) {
            arr[index] = value;
            value++;

            if (index == arr.length - 1) {
                arr[index] = n;
                break;
            } else {
                index++;
            }
        }
        value = 1;
    }
    return arr;
}

What will be the best solution?

Answer 1

I cleaned it up a bit. But its the same general idea. You had a few redundancies that you could have removed.

public int[] upSeries(int i) {
    assert i >= 0;        

    int[] array = new int[(i * (i+1)) / 2)];  // Standard Sum
    int seriesnum = 1;
    int seriesmax= 1;
    for (int index=0; index < array.length; index++) {
        array[index] = seriesnum;            

        if (seriesnum++ == seriesmax) {
            seriesnum = 1;
            seriesmax++;
        }
    }

    return array;
}

This solution is better by a constant multiple. If you are looking for Big-Oh they are both O(i^2) and it is not possible to do better.

If you need a proof to why it is not possible to do better. First prove that the size of the returned array is ((i) * (i+1)) /2). Then make the argument that you have to fill every position in the array. If you do better than O(i^2) you have not filled every position in the array.