K reverse a linked list with increasing K

Problem

Reverse k elements in the linked list such that k goes on increasing

Example
Eg.        1 - 2 - 3 - 4 - 5 - 6 - 7
output - 1 - 3 - 2 - 6 - 5- 4 - 7

Solution

You can take this problem here. Here we are just increasing k.

   public static ListNode<Integer> reverseSubLinkLists(ListNode<Integer> headerNode) {
        
        ListNode<Integer> nextNode = headerNode.next;
        ListNode<Integer> startNode = null;
        ListNode<Integer> endNode = null;
        
        int k = 2;
        while (nextNode != null) {
            
            startNode = nextNode;
            endNode = nextNode;
            
            for (int i = 1; i < k; i++) {
                endNode = endNode.next;
                if (endNode == null) {
                    break;
                }
            }



            if (endNode != null) {
                // Save the node next to endNode
                nextNode = endNode.next;
                // Unlink the endNode
                endNode.next = null;
                // Reverse the list starting from startNode

                startNode = reverseListIterative(startNode);
                k++;
            } else {
                nextNode = null;
//              // Save the node next to endNode
//              nextNode = endNode.next;
//              // Unlink the endNode
//              endNode.next = null;
                // Reverse the list starting from startNode

                startNode = reverseListIterative(startNode);
            }

            if (headerNode == null) {
                headerNode = startNode;
            } else {
                headerNode.next = startNode;
                ListNode<Integer> temp = startNode;
                while(temp!=null){
                    if(temp.next==null){
                        break;
                    }
                    temp = temp.next;
                }
                headerNode = temp;
            }
        }

        return headerNode;
    }

    public static ListNode<Integer> reverseListIterative(ListNode<Integer> headerNode) {
        ListNode<Integer> prevNode = null;
        ListNode<Integer> currNode = headerNode;
        ListNode<Integer> nextNode = null;

        while (currNode != null) {
            nextNode = currNode.next;
            currNode.next = prevNode;
            prevNode = currNode;
            currNode = nextNode;
        }

        return prevNode;
    }
    

Time complexity - O(n)

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Maximum number of chickens you cannot order

Problem

There is a non vegetarian restaurant which sells chicken in orders of 6, 9 and 20.

Calculate the maximum number of chicken pieces you cannot order from that restaurant ?

Solution

43

If you analyze, then you will find that all the 6 numbers divisible by 3 can be ordered. Why? Because you can break them own as the sum of 6 and 9.
Now after 26, all the numbers that are divisible by 3 if subtracted by 40 can be obtained.
After 46, all the numbers will fit into one of the above categories and thus all such numbers can be obtained. But 43 will be last number that will not fall into one of the above categories.
44 = 20 + 6 * 4, 45 = 6 * 6 + 9

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