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# Minimize length of an array by repeatedly removing elements that are smaller than the next element

• Last Updated : 13 Jul, 2021

Given an array arr[] consisting of N integers, the task is to repeatedly remove elements that are smaller than the next element.

Examples:

Input: arr[] = {20, 10, 25, 30, 40}
Output: 40
Explanation:
Below are the operations that are performed:

1. Current array: arr[] = {20, 10, 25, 30, 40}
Currently, arr[1](= 10) is less than arr[2](= 25). Therefore, removing arr[1] modifies the array to {20, 25, 30, 40}.
2. Currently arr[0](= 20) is less than arr[1](= 25). Therefore, removing arr[0] modifies the array to {25, 30, 40}.
3. Currently, arr[0](= 25) is less than arr[1](= 30). Therefore, removing arr[0] modifies the array to {30, 40}.
4. Currrently, arr[0](= 30) is less than arr[1](= 40). Therefore, removing arr[0] modifies the array to {40}.

Therefore, the remaining array is arr[] = {40}.

Input: arr[] = {2, 5, 1, 0}
Output: 5 1 0

Naive Approach: The simplest approach to solve the problem is to traverse the array and remove the element if there are any greater elements in the range [i + 1, N – 1].

Time Complexity: O(N2)
Auxiliary Space: O(1)

Efficient Approach – Iterative: The above approach can be optimized by traversing the array from the end and keeping track of the largest element found and deleting the current element if it is lesser than the largest element. Follow the steps below to solve the problem:

• Initialize a vector, say res, to store the resultant array.
• Initialize a variable, say maxi as INT_MIN, to store the maximum value from the end.
• Traverse the array arr[] in a reverse manner and perform the following steps:
• If the value of arr[i] is less than the maxi then continue.
• Otherwise, push the element arr[i] in res and update the value of maxi to the maximum of maxi and arr[i].
• Reverse the vector res.
• After completing the above steps, print the vector res as the result.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to minimize length of an``// array by repeatedly removing elements``// that are smaller than the next element``void` `DeleteSmaller(``int` `arr[], ``int` `N)``{``    ``// Stores the maximum value in``    ``// the range [i, N]``    ``int` `maxi = INT_MIN;` `    ``// Stores the resultant array``    ``vector<``int``> res;` `    ``// Iterate until i is atleast 0``    ``for` `(``int` `i = N - 1; i >= 0; i--) {` `        ``// If arr[i] is less than maxi``        ``if` `(arr[i] < maxi)``            ``continue``;` `        ``// Push the arr[i] in res``        ``res.push_back(arr[i]);` `        ``// Update the value of maxi``        ``maxi = max(maxi, arr[i]);``    ``}` `    ``// Reverse the vector res``    ``reverse(res.begin(), res.end());` `    ``// Print the vector res``    ``for` `(``auto` `x : res)``        ``cout << x << ``" "``;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 2, 5, 1, 0 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``DeleteSmaller(arr, N);` `    ``return` `0;``}`

## Python3

 `# Python3 program for the above approach` `# Function to minimize length of an``# array by repeatedly removing elements``# that are smaller than the next element``def` `DeleteSmaller(arr, N):``  ` `    ``# Stores the maximum value in``    ``# the range [i, N]``    ``maxi ``=``-``10``*``*``9` `    ``# Stores the resultant array``    ``res ``=` `[]` `    ``# Iterate until i is atleast 0``    ``for` `i ``in` `range``(N ``-` `1``, ``-``1``, ``-``1``):``        ``# If arr[i] is less than maxi``        ``if` `(arr[i] < maxi):``            ``continue` `        ``# Push the arr[i] in res``        ``res.append(arr[i])` `        ``# Update the value of maxi``        ``maxi ``=` `max``(maxi, arr[i])` `    ``# Reverse the vector res``    ``res ``=` `res[::``-``1``]``    ` `    ``# Prthe vector res``    ``print``(``*``res)` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``arr ``=` `[``2``, ``5``, ``1``, ``0``]``    ``N ``=` `len``(arr)``    ``DeleteSmaller(arr, N)` `# This code is contributed by mohit kumar 29.`

## C#

 `// C# program for the above approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG {` `    ``// Function to minimize length of an``    ``// array by repeatedly removing elements``    ``// that are smaller than the next element``    ``static` `void` `DeleteSmaller(``int``[] arr, ``int` `N)``    ``{` `        ``// Stores the maximum value in``        ``// the range [i, N]``        ``int` `maxi = Int32.MinValue;` `        ``// Stores the resultant array``        ``List<``int``> res = ``new` `List<``int``>();` `        ``// Iterate until i is atleast 0``        ``for` `(``int` `i = N - 1; i >= 0; i--) {` `            ``// If arr[i] is less than maxi``            ``if` `(arr[i] < maxi)``                ``continue``;` `            ``// Push the arr[i] in res``            ``res.Add(arr[i]);` `            ``// Update the value of maxi``            ``maxi = Math.Max(maxi, arr[i]);``        ``}` `        ``// Reverse the vector res``        ``res.Reverse();` `        ``// Print the vector res``        ``foreach``(``int` `x ``in` `res)``            ``Console.Write(x + ``" "``);``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{``        ``int``[] arr = { 2, 5, 1, 0 };``        ``int` `N = arr.Length;` `        ``DeleteSmaller(arr, N);``    ``}``}` `// This code is contributed by SURENDRA_GANGWAR`

## Javascript

 ``

Output:
`5 1 0`

Time Complexity: O(N)
Auxiliary Space: O(N)

Efficient Approach – Recursive: The above approach can also be implemented using recursion. Follow the steps below to solve the problem:

• Initialize a vector, say res to store the resultant array.
• Define a recursive function, say RecursiveDeleteFunction(int i) to delete element smaller than next element:
• If the value of i is equal to N then return INT_MIN.
• Call recursively call the function RecursiveDeleteFunction(i+1) and store the returned value in a variable say curr.
• If the value of arr[i] is at least curr then push arr[i] in vector res and update the value of curr as the maximum of curr and arr[i].
• Now, return the curr.
• Call the function RecursiveDeleteFunction(0) to remove the smaller elements to next elements and then, reverse the vector res.
• After completing the above steps, print the vector res as the result.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Recursive function to remove``// all elements which are smaller``// than the next element``int` `RecursiveDeleteFunction(``    ``int` `arr[], ``int` `N, ``int` `i,``    ``vector<``int``>& res)``{``    ``// If i is equal to N``    ``if` `(i == N)``        ``return` `INT_MIN;` `    ``// Recursive call to the next``    ``// element``    ``int` `curr = RecursiveDeleteFunction(``        ``arr, N, i + 1, res);` `    ``// If arr[i] is greater than the``    ``// or equal to curr``    ``if` `(arr[i] >= curr) {` `        ``// Push the arr[i] in res``        ``res.push_back(arr[i]);` `        ``// Update the value curr``        ``curr = max(arr[i], curr);``    ``}` `    ``// Return the value of curr``    ``return` `curr;``}` `// Function to minimize length of an``// array by repeatedly removing elements``// that are smaller than the next element``void` `DeleteSmaller(``int` `arr[], ``int` `N)``{``    ``// Stores maximum value in the``    ``// the range [i, N]``    ``int` `maxi = INT_MIN;` `    ``// Stores the resultant array``    ``vector<``int``> res;` `    ``// Recursive call to function``    ``// RecursiveDeleteFunction``    ``RecursiveDeleteFunction(arr, N, 0, res);` `    ``// Reverse the vector res``    ``reverse(res.begin(), res.end());` `    ``// Print the vector res``    ``for` `(``auto` `x : res)``        ``cout << x << ``" "``;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 2, 5, 1, 0 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``DeleteSmaller(arr, N);` `    ``return` `0;``}`
Output:
`5 1 0`

Time Complexity: O(N)
Auxiliary Space: O(N)

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