Microsoft Interview Question for Software Developers


Country: United States
Interview Type: Phone Interview




Comment hidden because of low score. Click to expand.
1
of 1 vote

The following implementation construct the iterator lazily and avoid storing the data before it is accessed. so still O(1)

public class SortedIterator<K extends Comparable>  implements Iterator<K> {
    private int currentIdx = 0;
    private K[][] sortedArrays;
    private int size = 0;
    private PriorityQueue<SortedNode> q = new PriorityQueue<SortedNode>((a, b) -> {return a.compareTo(b);});

    public SortedIterator(K[][] sortedArrays){
        this.sortedArrays = sortedArrays;
        initIterator();
    }

    @Override
    public boolean hasNext() {
        return (currentIdx  < size);
    }

    @Override
    public K next() {
        K item = null;
        if(this.sortedArrays != null && hasNext()) {
            SortedNode next = q.poll();
            item = next.val;
            int arrNum = next.arrNum;
            if (next.nextIdx < next.size && this.sortedArrays[arrNum][next.nextIdx] != null) {
                q.offer(new SortedNode(arrNum, this.sortedArrays[arrNum][next.nextIdx], next.nextIdx + 1, next.size));
            }
            currentIdx++;
        }
        return item;
    }

    private void initIterator() {
        if(this.sortedArrays != null) {
            int n = this.sortedArrays.length;
            for (int i = 0; i < n; i++) { //O(n)
                int m = this.sortedArrays[i].length;
                if (m > 0 && this.sortedArrays[i][0] != null) {
                    q.offer(new SortedNode(i, this.sortedArrays[i][0], 1, m));
                }
                size += m;
            }
        }
    }

    public static void main(String[] args) {
        Integer[][] arr = {{1 ,2, 8}, {6,9,10}, {1, 4, 5 , 12 }};
        SortedIterator<Integer> it = new SortedIterator(arr);
        while(it.hasNext()) {
            System.out.println(it.next()+"");
        }
    }

    private class SortedNode implements Comparable<SortedNode>{
        K val;
        int size;
        int nextIdx;
        int arrNum;

        SortedNode(int arrNum, K val, int nextIdx, int size){
            this.val = val;
            this.nextIdx = nextIdx;
            this.size = size;
            this.arrNum = arrNum;
        }

        public int compareTo(SortedNode n){
            if(n == null) return -1;
            if(n.val == null && this.val == null) return 0;
            if(n.val == null && this.val != null) return -1;
            if(n.val != null && this.val == null) return 1;
            return this.val.compareTo(n.val);
        }
    }
}

- azrarabbani February 12, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public class SortedIterator<K extends Comparable>  implements Iterator<K>{

    private List<K> sortedList;
    private int currentIdx = 0;

    public SortedIterator(K[][] sortedArrays){
        iterateListOfLists(sortedArrays);
    }

    @Override
    public boolean hasNext() {
         return (currentIdx  < sortedList.size());
    }

    @Override
    public K next() {
        K item = null;
        if(currentIdx < sortedList.size()){
            item = sortedList.get(currentIdx);
            currentIdx++;
        }
        return item;
    }

    private class SortedNode implements Comparable<SortedNode>{
        K val;
        int size;
        int nextIdx;
        int arrNum;

        SortedNode(int arrNum, K val, int nextIdx, int size){
            this.val = val;
            this.nextIdx = nextIdx;
            this.size = size;
            this.arrNum = arrNum;
        }

        public int compareTo(SortedNode n){
            if(n == null) return -1;
            if(n.val == null && this.val == null) return 0;
            if(n.val == null && this.val != null) return -1;
            if(n.val != null && this.val == null) return 1;
            return this.val.compareTo(n.val);
        }
    }

    private List<K> iterateListOfLists(K[][] arr){
        PriorityQueue<SortedNode> q = new PriorityQueue<SortedNode>((a, b) -> {return a.compareTo(b);});
        int n = arr.length;
        int size = 0;
        for(int i=0; i < n; i++){ //O(n)
            int m = arr[i].length;
            if(m > 0 && arr[i][0] != null) {
                q.offer(new SortedNode(i,arr[i][0],1, m));
            }
            size += m;
        }
        sortedList = new ArrayList<>(size);
        while(!q.isEmpty()) { //O(n + m)
            SortedNode next = q.poll();
            sortedList.add(next.val);
            int arrNum = next.arrNum;
            if(next.nextIdx < next.size && arr[arrNum][next.nextIdx] != null) {
                q.offer(new SortedNode(arrNum, arr[arrNum][next.nextIdx],next.nextIdx+1, next.size ));
            }
       }
       return sortedList;
    }

    public static void main(String[] args) {
        Integer[][] arr = {{1 ,2, 8}, {6,9,10}, {1, 4, 5 , 12 }};
        SortedIterator<Integer> it = new SortedIterator(arr);
        while(it.hasNext()) {
            System.out.println(it.next()+"");
        }
    }
}

- azrarabbani February 10, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public class SortedIterator<K extends Comparable>  implements Iterator<K>{

    private List<K> sortedList;
    private int currentIdx = 0;

    public SortedIterator(K[][] sortedArrays){
        iterateListOfLists(sortedArrays);
    }

    @Override
    public boolean hasNext() {
         return (currentIdx  < sortedList.size());
    }

    @Override
    public K next() {
        K item = null;
        if(currentIdx < sortedList.size()){
            item = sortedList.get(currentIdx);
            currentIdx++;
        }
        return item;
    }

    private class SortedNode implements Comparable<SortedNode>{
        K val;
        int size;
        int nextIdx;
        int arrNum;

        SortedNode(int arrNum, K val, int nextIdx, int size){
            this.val = val;
            this.nextIdx = nextIdx;
            this.size = size;
            this.arrNum = arrNum;
        }

        public int compareTo(SortedNode n){
            if(n == null) return -1;
            if(n.val == null && this.val == null) return 0;
            if(n.val == null && this.val != null) return -1;
            if(n.val != null && this.val == null) return 1;
            return this.val.compareTo(n.val);
        }
    }

    private List<K> iterateListOfLists(K[][] arr){
        PriorityQueue<SortedNode> q = new PriorityQueue<SortedNode>((a, b) -> {return a.compareTo(b);});
        int n = arr.length;
        int size = 0;
        for(int i=0; i < n; i++){ //O(n)
            int m = arr[i].length;
            if(m > 0 && arr[i][0] != null) {
                q.offer(new SortedNode(i,arr[i][0],1, m));
            }
            size += m;
        }
        sortedList = new ArrayList<>(size);
        while(!q.isEmpty()) { //O(n + m)
            SortedNode next = q.poll();
            sortedList.add(next.val);
            int arrNum = next.arrNum;
            if(next.nextIdx < next.size && arr[arrNum][next.nextIdx] != null) {
                q.offer(new SortedNode(arrNum, arr[arrNum][next.nextIdx],next.nextIdx+1, next.size ));
            }
       }
       return sortedList;
    }

    public static void main(String[] args) {
        Integer[][] arr = {{1 ,2, 8}, {6,9,10}, {1, 4, 5 , 12 }};
        SortedIterator<Integer> it = new SortedIterator(arr);
        while(it.hasNext()) {
            System.out.println(it.next()+"");
        }
    }
}

- azrarabbani February 10, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public class SortedIterator<K extends Comparable>  implements Iterator<K>{

    private List<K> sortedList;
    private int currentIdx = 0;

    public SortedIterator(K[][] sortedArrays){
        iterateListOfLists(sortedArrays);
    }

    @Override
    public boolean hasNext() {
         return (currentIdx  < sortedList.size());
    }

    @Override
    public K next() {
        K item = null;
        if(currentIdx < sortedList.size()){
            item = sortedList.get(currentIdx);
            currentIdx++;
        }
        return item;
    }

    private class SortedNode implements Comparable<SortedNode>{
        K val;
        int size;
        int nextIdx;
        int arrNum;

        SortedNode(int arrNum, K val, int nextIdx, int size){
            this.val = val;
            this.nextIdx = nextIdx;
            this.size = size;
            this.arrNum = arrNum;
        }

        public int compareTo(SortedNode n){
            if(n == null) return -1;
            if(n.val == null && this.val == null) return 0;
            if(n.val == null && this.val != null) return -1;
            if(n.val != null && this.val == null) return 1;
            return this.val.compareTo(n.val);
        }
    }

    private List<K> iterateListOfLists(K[][] arr){
        PriorityQueue<SortedNode> q = new PriorityQueue<SortedNode>((a, b) -> {return a.compareTo(b);});
        int n = arr.length;
        int size = 0;
        for(int i=0; i < n; i++){ //O(n)
            int m = arr[i].length;
            if(m > 0 && arr[i][0] != null) {
                q.offer(new SortedNode(i,arr[i][0],1, m));
            }
            size += m;
        }
        sortedList = new ArrayList<>(size);
        while(!q.isEmpty()) { //O(n + m)
            SortedNode next = q.poll();
            sortedList.add(next.val);
            int arrNum = next.arrNum;
            if(next.nextIdx < next.size && arr[arrNum][next.nextIdx] != null) {
                q.offer(new SortedNode(arrNum, arr[arrNum][next.nextIdx],next.nextIdx+1, next.size ));
            }
       }
       return sortedList;
    }

    public static void main(String[] args) {
        Integer[][] arr = {{1 ,2, 8}, {6,9,10}, {1, 4, 5 , 12 }};
        SortedIterator<Integer> it = new SortedIterator(arr);
        while(it.hasNext()) {
            System.out.println(it.next()+"");
        }
    }
}

- azrarabbani February 10, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

What's the space complexity of this algo according to you? I was asked to submit something with O(1) space complexity.

- sanjos February 11, 2019 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 votes

The space complexity is O(1) since the priority queue will keep only 1 item of each list at a time that should be negligible as compared to O(n).

e.g. if there are 3 lists {2, 3, 7, ,6} , {1,3,5} , {8,7,9}
In this scenario: There will be 3 items in the priority queue at max at a time, which will decrease to 2 as soon as we poll the third element of the 2nd list that is

- azrarabbani February 11, 2019 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 votes

that is 5

- azrarabbani February 11, 2019 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 votes

@azrarabbani isn't that O(n) memory nonetheless because the priority queue does store n items (once we poll an item from each list) if there are n lists.

- Abinash March 10, 2019 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

Create a minheap of size M storing the currentIndex of each List but sorted by the value at the index. This way the as we remove the root values, we update the root key cuurentindex, thus trigerring heapify.
When we exhaust all indices of a list, the node for the list is removed from heap.
Finally heap will go empty when algo terminates

This is space O(M) -> and this will be treated constant

- Sayantan February 14, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

we can build a min heap using the giving list, cause a heap is also based on an array.

- Yingrui March 06, 2019 | Flag Reply


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