Reduce the array by deleting elements which are greater than all elements to its left

Given an array arr[] of N integers, the task is to delete the element from the given array if element to it’s left is smaller than it. Keep on deleting the elements from the array until no element has a smaller adjacent left element. Print the resultant array after above operation.

Examples:

Input: arr[] = {2, 4, 1, 3, 4}
Output: 2 1
Explanation:
Since 4 is greater than 2 remove 4, and arr become {2, 1, 3, 4}.
Now 3 is greater than 1 so remove 3 and arr become {2, 1, 4}.
Now 4 is greater than 1 so remove 4 and arr become {2, 1}.
Now no elements satisfy the removing criteria so the resultant array is {2, 1}.

Input: arr[] = {5, 4, 3, 2, 1}
Output: 5 4 3 2 1

Approach: The idea is to use the concept of Merge Sort.



  1. Divide the input array into sub-arrays till the size of each sub-array becomes 1.
  2. Start merging the element.
  3. While merging, delete elements from the left subarray till it’s rightmost element, which have a value greater than the leftmost element of the right subarray.
  4. Repeat the above steps in each merging step such all elements with value smaller to it’s left have been deleted.
  5. Finally print the resultant array.

Below is the implementation of the above approach:

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# Python program for the above approach
  
# Function to delete all elements which
# satisfy the condition A[i] > A[i-1] 
def mergeDel(l):
  
    # Divide array into its subarray
    if len(l) == 1:
  
        return l
  
    m = int( len(l) / 2)
  
    # Getting back merged array with all 
    # its right element greater than left one.
    return merge(mergeDel(l[ 0 : m ]),
                 mergeDel(l[ m : len(l)]) )
  
  
# Function to implement merging of arr[]
def merge(x, y):
  
    for i in y:
  
        if x[-1] > i :
  
            x = x + [i]
  
    return x
  
# Function defination for main()
def main():
  
# Given array arr[]
    arr = [5, 4, 3, 2, 1]
    print(mergeDel(arr))
  
# Driver Code
main()

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Output:

[5, 4, 3, 2, 1]

Time Complexity: O(N*log N)
Auxiliary Space: O(1)

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