Code Samples

Remove Duplicates from a Sorted List

This is a medium level question from LeetCode

static ListNode deleteDuplicates(ListNode head) {
        ListNode current = head;
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode previous = dummy;
        while (current != null && current.next != null) {
            if (current.val == current.next.val) {
                while (current.next != null && current.val == current.next.val) {
                    current = current.next;
                }
                previous.next = current.next;
            } else {
                previous = current;
            }
            current = current.next;
        }
        return dummy.next;
    }

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Keys and Rooms

Question - https://leetcode.com/problems/keys-and-rooms/

    public boolean canVisitAllRooms(List<List<Integer>> rooms) {
        //This is a depth first search problem 
        //Solved by using a Stack and keeping track of each node visited or not in order to avoid infinite loop
        if ( rooms == null || rooms.isEmpty() ) return true;
        final int N = rooms.size();
        //keep track of each node if visited or not with each node set to false initially
        boolean visits[] = new boolean[N];
        //Since by problem definition first one is unlocked so can visit without key
        visits[0] = true;
        Stack<List<Integer>> roomStack = new Stack<>();
        //Add the list of rooms for which keys are available on first node
        roomStack.push(rooms.get(0));
        while (! roomStack.isEmpty() ){
            List<Integer> keysInRoom = roomStack.pop();
            //Iterate through each key found in the node
            for(Integer key : keysInRoom){
                if ( ! visits[ key ] )  { 
                    visits[key] = true;  
                    roomStack.push( rooms.get( key) ); 
                }             
            }
        }
        //check if all visited is true or not
        for(boolean visit : visits)
            if (!visit) return false;
        return true;
    }

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