Find the Noble Integer
- Noble Integer: InterviewBit
Given an integer array, find if an integer p exists in the array such that the number of integers greater than p in the array equals to p.If such an integer is found return 1 else return -1.
- Category: Arrays
This problem is not very difficult but to solve it we have to apply a well known technique to simplify the given inputs.
Let’s take a small example:
2, 6, 1, 3
This input should return 1 as 2 is a noble integer. We know that by counting the number of integers greater than 2 (3 and 4).
Yet, how to solve this problem without having an implementation in O(n²)? The answer is to simplify the inputs.
What if in our case the array was sorted?
1, 2, 3, 6
The integer 2 is at index 1, whereas the array size is 4. The success condition is the following:
size(array) - index - 1 == array[i]
Once the problem is solved using a simplification, we need to check the implications in terms of complexity. If our solution is acceptable, we generalize to the initial problem.
In our case, we have to:
- Sort the array => O(n log(n))
- Iterate over each element and check the previous condition => O(n)
It means the solution is O(n log(n)).
Bear in mind, sorting an array can’t be done with a better solution than a O(n log(n)) (like a merge sort for example).
Also, we have to make sure our solution covers all corner cases.
What about the following example, after being sorted:
1, 2, 2, 3
At index 1, the condition is going to be true. Yet, this should not be a match.
In our case, we also have to cover duplicate by checking if the next integer is equals to the current one.
A possible implementation in Java:
- Solving Algo publication
- Java sort implementation: https://docs.oracle.com/javase/8/docs/api/java/util/List.html#sort-java.util.Comparator-
- Big-O Cheat Sheet: http://bigocheatsheet.com/