You are given a circular array 'a' of length 'n'.
A circular array is an array in which we consider the first element is next of the last element. That is, the next element of 'a[n - 1]' is 'a[0]'.
Find the Next Greater Element(NGE) for every element.
The Next Greater Element for an element 'x' is the first element on the right side of 'x' in the array, which is greater than 'x'.
If no greater elements exist to the right of 'x', consider the next greater element as -1.
Input: 'a' = [1, 5, 3, 4, 2]
Output: NGE = [5, -1, 4, 5, 5]
Explanation: For the given array,
- The next greater element for 1 is 5.
- There is no greater element for 5 on the right side. So we consider NGE as -1.
- The next greater element for 3 is 4.
- The next greater element for 4 is 5, when we consider the next elements as 4 -> 2 -> 1 -> 5.
- The next greater element for 2 is 5, when we consider the next elements as 2 -> 1 -> 5.
The first line contains an integer 'n' denoting the size of the array 'a'.
The second line contains 'n' single space-separated integers, elements of the array 'a'.
Find the Next Greater Element(NGE) for every element in the circular array.
You don't need to print anything. It has already been taken care of. Just implement the given function.
5
1 5 3 4 2
5 -1 4 5 5
For the given array,
- The next greater element for 1 is 5.
- There is no greater element for 5 on the right side. So we consider NGE as -1.
- The next greater element for 3 is 4.
- The next greater element for 4 is 5, when we consider the next elements as 4 -> 2 -> 1 -> 5.
- The next greater element for 2 is 5, when we consider the next elements as 2 -> 1 -> 5.
5
5 5 5 5 5
-1 -1 -1 -1 -1
The expected time complexity is O(n).
1 <= 'n' <= 10^5
1 <= 'a[i]' <= 10^9
Time Limit: 1 sec
Check all possibilities.
In this approach, we will check individually for each element of the array.
For each element of the array, we will linearly traverse the whole array circularly starting from the next index, and if we find any greater element then it will be our answer, otherwise, if we reach the starting index without any greater element, our answer is -1.
The steps are as follows:-
function nextGreaterElementII(int ‘a[]’):
O(n ^ 2), Where 'n' is the size of the given array 'a'.
We are running two nested for loops, each of O(n) complexity.
Hence the time complexity is O(n ^ 2).
O(n), Where 'n' is the size of the given array 'a'.
We are using an array of size ‘n’ to store the Next Greater Element for each element.
Hence space complexity is O(n).