Posted: 16 Mar, 2021
Implement the given PeekingIterator class consisting of four functions:
Peeking Iterator (int arr): Initialize the object with the given array element. int next(): Returns the next element and moves the pointer to the next element. bool hasNext(): Return true if there is still an element in the array that otherwise returns false. int peek(): Return the next element in the array but don’t move the pointer to the next element.
You will be given ‘Q’ queries consisting of 'next', 'hasNext', and 'peek' operations. In each query input is of the following type:
0 ‘array/list’ - Initialize the object with the given array element. 1 - Return the next element and move the pointer to the next element. 2 - Return true if there are elements in the array, otherwise return false. 3 - Perform a peek operation in the array.
0 3 7 9 10 1 2 3 Here we have 4 queries: Firstly the query is of type 0. Sp, an object will be created with the list [7,9,10], and the pointer will be at 7. Then the next query is 1 so we will return 7 and the pointer will move to 9. Then the next query is 2 and there are elements in the array so we will return 'True'. The last query is 3. So, we will just return 9 by not moving the pointer ahead.
Only the first query will be of type 0. You can't directly access the array, instead, you will be given access to a separate iterator class.
The first line of input contains an integer ‘T’ denoting the number of test cases to run. Then the test case follows. The first line of each test case contains an integer ‘Q’ denoting the number of queries to be answered. In the next ‘Q’ lines, inputs are 0,1,2 or 3 as mentioned above.
For each test case, for all the calls of the query of type: 1: Return the next element 2: Return true or false 3: Return element at the pointer position Note for each call of query, the output is printed in a separate line.
1 <= T <= 5 1 <= size of arr <= 1000 1 <= arr[i] <= 1000 1 <= Q <= 1000 All queries are valid. Time Limit: 1 second
- Try to implement dynamic array/linked list as it has two attributes with it one is the data field and the other is the pointer pointing to the next node which is all we need in this problem.
- Initialize it with the given array.
- Initialize a head pointer that points to the first node of List.
- For the next() query, first, check if the current node we are at is the last node of the list or not. If it is not the last node of the list, we can shift to the next node by means of our iterator and return its data field.
- For hasnext(), simply it can be checked if we currently point to the last node of the list or not which can tell if next() exists or not.
- For peek simply return the data field of the current node you are pointing to.
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