Given an array arr[] consisting of N distinct positive integers, the task is to repeatedly find the minimum peak element from the given array and remove that element until all the array elements are removed.
Peak Element: Any element in the array is know as the peak element based on the following conditions:
If arr[i – 1] < arr[i] > arr[i + 1], where 1 < i < N – 1, then arr[i] is the peak element.
If arr[0] > arr[1], then arr[0] is the peak element, where N is the size of the array.
If arr[N – 2] < arr[N – 1], then arr[N – 1] is the peak element, where N is the size of the array.
If more than one peak element exists in the array, then the minimum value among them needs to be printed.
Examples:
Input: arr[] = {1, 9, 7, 8, 2, 6}
Output: [6, 8, 9, 7, 2, 1]
Explanation:
First min peak = 6, as 2 < 6.
The array after removing min peak will be [1, 9, 7, 8, 2].
Second min peak = 8, as 7 < 8 > 2.
The array after removing min peak will be [1, 9, 7, 2]
Third min peak = 9, as 1 < 9 > 7.
The array after removing min peak will be [1, 7, 2]
Fourth min peak = 7, as 1 < 7 > 2.
The array after removing min peak will be [1, 2]
Fifth min peak = 2, as 1 < 2.
The array after removing min peak will be [1]
Sixth min peak = 1.
Therefore, the list of minimum peak is [6, 8, 9, 7, 2, 1].
Input: arr []= {1, 5, 3, 7, 2}
Output: [5, 7, 3, 2, 1]
Explanation:
First min peak = 5, as 1 < 5 > 3.
The array after removing min peak will be [1, 3, 7, 2]
Second min peak = 7, as 3 < 7 > 2.
The array after removing min peak will be [1, 3, 2]
Third min peak = 3, as 1 < 3 > 2.
The array after removing min peak will be [1, 2]
Fourth min peak = 2, as 1 < 2.
The array after removing min peak will be [1]
Fifth min peak = 1.
Therefore, the list of minimum peak is [5, 7, 3, 2, 1]
Approach: The idea is to find the minimum peak element of the array by iterating over the array using two nested loops, where the outer loop points to the current element and the inner loop execute to find the index of min peak element, remove that peak element from the array and store the current peak element in the resultant list. After completing the above steps, print all the minimum peak elements stored in the list.
Below is the implementation of the above approach:
// Java program for the above approach
import java.util.*;
import java.lang.*;
class GFG {
// Function to return the list of
// minimum peak elements
static void
minPeaks(ArrayList<Integer> list)
{
// Length of original list
int n = list.size();
// Initialize resultant list
ArrayList<Integer> result
= new ArrayList<>();
// Traverse each element of list
for (int i = 0; i < n; i++) {
int min = Integer.MAX_VALUE;
int index = -1;
// Length of original list after
// removing the peak element
int size = list.size();
// Traverse new list after removal
// of previous min peak element
for (int j = 0; j < size; j++) {
// Update min and index,
// if first element of
// list > next element
if (j == 0 && j + 1 < size) {
if (list.get(j) > list.get(j + 1)
&& min > list.get(j)) {
min = list.get(j);
index = j;
}
}
else if (j == size - 1
&& j - 1 >= 0) {
// Update min and index,
// if last elemnt of
// list > previous one
if (list.get(j)
> list.get(j - 1)
&& min
> list.get(j)) {
min = list.get(j);
index = j;
}
}
// Update min and index, if
// list has single element
else if (size == 1) {
min = list.get(j);
index = j;
}
// Update min and index,
// if current element >
// adjacent elements
else if (list.get(j)
> list.get(j - 1)
&& list.get(j)
> list.get(j + 1)
&& min
> list.get(j)) {
min = list.get(j);
index = j;
}
}
// Remove current min peak
// element from list
list.remove(index);
// Insert min peak into
// resultant list
result.add(min);
}
// Print resultant list
System.out.println(result);
}
// Driver Code
public static void main(String[] args)
{
// Given array arr[]
ArrayList<Integer> arr = new ArrayList<>(
Arrays.asList(1, 9, 7, 8, 2, 6));
// Function Call
minPeaks(arr);
}
}
Output:
[6, 8, 9, 7, 2, 1]
Time Complexity: O(N2)
Auxiliary Space: O(N)
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