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Problem 1: Balls Problem 2: Josephine Problem 3: Alternating List

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Problem: - Given N balls labeled from 1 to N and M operations, determine the last state after doing M operations. Input The first line contains two integers, N (1<= N <= 1,000) and M (1<= M <= 1,000) The next M lines contain the operations. Sample Input 10 5 A 2 1 A 10 1 A 5 6 B 6 9 R 3

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Operations: 1. A x y: move the ball labeled with number x to the left of the ball labeled with number y. 2. B x y: move the ball labeled with number x to the right of the ball labeled with number y. 3. R x: Remove the ball labeled with number x from our list. Output: Output the final arrangement of the N balls from left to right. Each number is followed by a whitespace.

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1. What kind of data structure can we use to solve this problem?

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1. Hint: use List - Q: What kind of List (Singly-Linked List/ Doubly-Linked List/ Circular-Linked List)? - A: Use Doubly-Linked List Why?

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2. Simplest implementation: -Traversing all the nodes (to find the correct node) for removing and moving Better implementation: - Without traversing through the nodes (will be discussed later)

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ListNode stores: 1. Number/Id 2. Previous Ball 3. Next Ball ID int ID ListNode next ListNode prev

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Remove: 1. Find position of x 2. Update the affected ListNode.

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Rough steps: Iterate though the list to find the node with label x, and remember which node it is (point to it from a variable) Adjust the pointers

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Remove node X from position X next prev Next of X next prev Previous of X next prev

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Remove node X from position X next prev Next of X next prev Previous of X next prev X.next.prev = X.prev X.prev.next = X.next Must handle special case of 1 st and last nodes when X.prev == null or X.next == null

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Move (Simplest Implementation): 1. Find position of X and Y 2. Update the affected ListNode Example: A x y

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Rough steps: Iterate though the list to find the node with labels x and y, and store them in two variables. Remove node X Insert node X left/right of node Y (practice using back the same node removed and insert it back, rather than creating a new node)

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Insert X on the left of Y Y next prev Previous of Y next prev

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Insert X on the left of Y X next prev Y next prev Previous of Y next prev X.prev = Y.prev X.next = Y Y.prev.next = X Y.prev = X Must handle special case of inserting X before the 1 st node when Y.prev == null Do it in the right order – dont lose the previous of Y node

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Insert X on the left of Y X next prev Y next prev Previous of Y next prev X.prev = Y.prev X.next = Y Y.prev.next = X Y.prev = X Must handle special case of inserting X before the 1 st node when Y.prev == null Do it in the right order – dont lose the previous of Y node

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Insert X to the right of Y Next of Y next prev Y next prev

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Insert X to the right of Y X next prev Next of Y next prev Y next prev X.next = Y.next X.prev = Y Y.next.prev = X Y.next = X Must handle special case of inserting X after the last node when Y.next == null Do it in the right order – dont lose the previous of Y node

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Insert X to the right of Y X next prev Next of Y next prev Y next prev X.next = Y.next X.prev = Y Y.next.prev = X Y.next = X Must handle special case of inserting X after the last node when Y.next == null Do it in the right order – dont lose the previous of Y node

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Y PYNY PX NX X

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X Y PY PX NX

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An idea to improve performance: 1. We store a reference of all balls (ListNodes) in an array, so that we dont need to traverse the Linked List to find the balls every time head 12345

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Problem: - Given N candidates in circle, we want to keep removing the K-th candidate until we find the number of people in the circle equals to 1. Output the removed candidate for each remove operation.

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Input The first line consists of T, the number of test cases, T <= 100. The following T lines describe T test cases, each line containing two integers, N and K. Output Output the final arrangement of the N balls from left to right. Each number is followed by a whitespace. Sample Input Sample Output

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1. What kind of data structure can we use to solve this problem?

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1. Hint: - Use Circular Linked List Why? Simulates the problem!

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Keep removing K-th person by iterating K times from current position. Update the State of ListNode. More or less similar to Previous Problem but using different type of LinkedList.

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Removing a node: Since you dont have the previous pointer in a singly-linked circular linked list, you need to keep track of the previous node when you traverse the list! 12435head

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Problem: Given a list of integers, and a list of operations, determine whether it is alternating after each operation Definition of Alternating: -Adjacent integers have different signs List of one or zero elements is alternating #3: A LTERNATING LIST

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Input 1 <= N <= <= Q <= 100 N is the size of the original linked list Q the number of operations Output For each operation print YES if the updated linked list is alternating, otherwise print NO. #3: A LTERNATING LIST Sample Input M 1 3 A R 2 2 A Sample Output YES NO YES

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1. What kind of data structure can we use to solve this problem?

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1. Hint: use List - Q: What kind of List (Singly-Linked List/ Doubly-Linked List/ Circular-Linked List)? - A: Use (Singly) Linked List Why?

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M [index] [size] Move size number of elements starting from index to the end of list Index M 2 3 Index

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Pseudo-code: function Move (int index, int size) { for i=1 to size { temp remove element at index insert temp to the end of list } } Why removing at the index only works?

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R [index] [size] Remove size number of elements starting from index. Index R 2 3 Index15 -42

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Pseudo-code: fun Remove (int index, int size) { for i=1 to size { remove element at index } } Can you use this to implement Move?

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A [index] [size] [value] add the elements between index [index] and [index + size - 1] (inclusive) with [value]. Index A Index

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Pseudo-code: fun AddValue (int index, int size, int value) { for i=1 to size { temp remove value at index+i-1 insert (temp+value) to index+i-1 } } How do we get the (index+i-1) ? (or we can cheat and just modify the number in listNode…)

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List of 1 element : YES Loop through list: once two adjacent elements with same sign found, return NO i.e. Keep track of previousSign, and then check every iteration if previousSign == currentSign Return YES if loop completes without returning

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Any Questions?

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