Yes. On the input you have an array of frequencies for all possible numbers (0-200) for the gap of values from basic array (this gap size is set by d input) .

you create another one array and start to build it from frequencies array the following way: you sum frequency of every number and frequencies of previous numbers.

What you really need to understand: this second array values will give you last indexes at which every number of frequencies would appear in sorted array.

That is, imagine frequency array [0]2, [1]1, [2]0, [3]4. Building second array would give you [0]2, [1]1+2 =3, [2]0+3=3, [3]3+4=7.

Now, imagine building sorted array from input frequencies array: 0 0 1 3 3 3 3 [0][1][2][3][4][5][6]

so now you take just half of a 'd' and iterate through second array, looking for number which is more huge or equals to d. Meeting such number means that you're currently pointing at index whose value would be in the middle of sorted array, and this is just what you need.