## Adobe Interview Question for Computer Scientists

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Country: United States
Interview Type: In-Person

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``````// insertion sort can also work well for partially sorted arrays. Below, is a technique that
// uses k -sized min heaps

/**
*
* @param a a k-sorted array, each element is at most k spaces away from its final position after sorting
* @param k value of k, k<= array size
* @return fully sorted array
*/
static int [] ksort(int[] a, int k) {
final int windowSize = k + 1;
if (a.length < windowSize) throw new IllegalArgumentException();
PriorityQueue<Integer> q = new PriorityQueue<>();
int i = 0, j = 0;
//first k + 1 items are put in heap
for (; i < windowSize; i++) q.add(a[i]);
// we extra elements from the heap and add them to sorted array
// for each item lost from the heap, one is added to the heap from the
// unsorted part of the array
for (; i < a.length; i++, j++) {
a[j] = q.poll();
q.add(a[i]);
}
// the right edge of window k+1 elements has merged with the right
// end of the array. There are no more elements to add. We delete
// elements from the heap till it is empty
while (!q.isEmpty()) {
a[j++] = q.poll();
}
return a;
}``````

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@markarand, is that O(nlogn)?

Maybe I'm misinterpreting the problem, if each element is K length away from its sorted position, wouldn't a first easy solution be, find minimum element index, its index is K, then we just have to shift each element K position.

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@ tnutty2k8 - this is O (n Log k), assuming that k<< N, this is more efficient

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``````static void sortedArray(int[] arr, int k) {

int size = arr.length;
PriorityQueue<Integer> queue = new PriorityQueue<Integer>();
for(int i=0; i<k; i++) {
queue.add(arr[i]);
}

int m =0;
int n = k;

while(!queue.isEmpty()){

Integer value = queue.poll();
arr[m++] = value;
if(n < size) {
queue.add(arr[n++]);
}

}

System.out.println(" "+Arrays.toString(arr));

}``````

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@Kapil: You've considered each number is k positions ahead of its appropriate position. It may equally be possible that it's behind its actual position. So in that situation, you code won't work nor Makarand's code.

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@Sameer Oak The solution given by @Makarand is correct. He's building a min heap using the first (k + 1) elements. For the following input, the solution will give the correct output.

2 1 4 3
k = 1

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