# Minimum removals required to convert given array to a Mountain Array

• Difficulty Level : Expert
• Last Updated : 22 Jun, 2022

Given an array arr[] consisting of N integers​​​, the task is to find the minimum number of array elements required to be removed to make the given array a mountain array.

A mountain array has the following properties:

• Length of the array ≥ 3.
• There exists some index i (0-based indexing) with 0 < i < N – 1 such that:
• arr[0] < arr[1] < … < arr[i – 1] < arr[i]
• arr[i] > arr[i + 1] > … > arr[arr.length – 1].

Examples:

Input: arr[] = {1, 3, 1}
Output: 0
Explanation: The array itself is a mountain array. Therefore, no removal is required.

Input: arr[] = {2, 1, 1, 5, 6, 2, 3, 1}
Output: 3
Explanation: Removing arr[0], arr[1] and arr[5] modifies arr[] to {1, 5, 6, 3, 1}, which is a mountain array.

Approach 1:

The idea is to solve this problem using the Bottom-Up Dynamic Programming approach. Follow the steps below to solve the problem:

1. If the length of the given array is less than 3, then the array cannot be converted to a mountain array.
2. Otherwise, traverse the array and for every ith element (0 < i < N), find the length of increasing subsequence in the subarrays {arr[0], …, arr[i – 1]} and store it in an array, say leftIncreasing[].
3. Similarly, find the length of the increasing subsequence in the subarray {arr[i+1], …., arr[N-1]} for every ith element (0 < i < N), and store it in an array, say rightIncreasing[].
4. Find the index i (0 < i < N) which satisfies the following conditions:
1. The first compulsory condition is the peak condition, which is leftIncreasing[i] > 0 and rightIncreasing[i] > 0.
2. Among all indices, If leftIncreasing[i] + rightIncreasing[i] is the maximum, that index is the peak of the mountain array, say X.
5. Return the result as N – (X + 1), adding one to bring the array index to length.

Illustration:

Consider the array arr[] = {4, 3, 6, 4, 5}
Therefore, leftIncreasing[] = {0, 0, 1, 1, 2} & rightIncreasing[] = {2, 1, 1, 0, 0}.
There is only one index i = 2 (0-based indexing), for which leftIncreasing[2] > 0 and rightIncreasing[2] > 0.
Therefore, X = leftIncreasing[2] + rightIncreasing[2] = 2.
Therefore, the required answer = N – (X + 1) = 5 – (2 + 3)= 2.
One of the possible solutions could be {4, 6, 5} i.e. removing 3 (arr[1]) and 4(arr[3]).

Below is the implementation of the above approach:

## C++

 `// C++ program of the above approach``#include ``using` `namespace` `std;` `// Utility function to count array``// elements required to be removed``// to make array a mountain array``int` `minRemovalsUtil(``int` `arr[], ``int` `n)``{``    ``int` `result = 0;``    ``if` `(n < 3) {``        ``return` `-1;``    ``}` `    ``// Stores length of increasing``    ``// subsequence from [0, i-1]``    ``int` `leftIncreasing[n] = {0};` `    ``// Stores length of increasing``    ``// subsequence from [i + 1, n - 1]``    ``int` `rightIncreasing[n] = {0};` `    ``// Iterate for each position up to``    ``// N - 1 to find the length of subsequence``    ``for` `(``int` `i = 1; i < n; i++)``    ``{``        ``for` `(``int` `j = 0; j < i; j++)``        ``{` `            ``// If j is less than i, then``            ``// i-th position has leftIncreasing[j]``            ``// + 1 lesser elements including itself``            ``if` `(arr[j] < arr[i])``            ``{` `                ``// Check if it is the maximum``                ``// obtained so far``                ``leftIncreasing[i]``                    ``= max(leftIncreasing[i],``                          ``leftIncreasing[j] + 1);``            ``}``        ``}``    ``}` `    ``// Search for increasing subsequence from right``    ``for` `(``int` `i = n - 2; i >= 0; i--)``    ``{``        ``for` `(``int` `j = n - 1; j > i; j--)``        ``{``            ``if` `(arr[j] < arr[i])``            ``{``                ``rightIncreasing[i]``                    ``= max(rightIncreasing[i],``                               ``rightIncreasing[j] + 1);``            ``}``        ``}``    ``}` `    ``// Find the position following the peak``    ``// condition and have maximum leftIncreasing[i]``    ``// + rightIncreasing[i]``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``if` `(leftIncreasing[i] != 0``            ``&& rightIncreasing[i] != 0)``        ``{``            ``result = max(result,``                         ``leftIncreasing[i]``                         ``+ rightIncreasing[i]);``        ``}``    ``}``    ``return` `n - (result + 1);``}` `// Function to count elements to be``// removed to make array a mountain array``void` `minRemovals(``int` `arr[], ``int` `n)``{``    ``int` `ans = minRemovalsUtil(arr, n);` `    ``// Print the answer``    ``cout << ans;``}` `// Driver Code``int` `main()``{` `    ``// Given array``    ``int` `arr[] = { 2, 1, 1, 5, 6, 2, 3, 1 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``      ` `    ``// Function Call``    ``minRemovals(arr, n);``    ``return` `0;``}` `// This code is contributed by Dharanendra L V`

## Java

 `// Java program of the above approach` `import` `java.io.*;``import` `java.util.*;` `class` `GFG {` `    ``// Utility function to count array``    ``// elements required to be removed``    ``// to make array a mountain array``    ``public` `static` `int` `minRemovalsUtil(``        ``int``[] arr)``    ``{``        ``int` `result = ``0``;``        ``if` `(arr.length < ``3``) {``            ``return` `-``1``;``        ``}` `        ``// Stores length of increasing``        ``// subsequence from [0, i-1]``        ``int``[] leftIncreasing``            ``= ``new` `int``[arr.length];` `        ``// Stores length of increasing``        ``// subsequence from [i + 1, n - 1]``        ``int``[] rightIncreasing = ``new` `int``[arr.length];` `        ``// Iterate for each position up to``        ``// N - 1 to find the length of subsequence``        ``for` `(``int` `i = ``1``; i < arr.length; i++) {` `            ``for` `(``int` `j = ``0``; j < i; j++) {` `                ``// If j is less than i, then``                ``// i-th position has leftIncreasing[j]``                ``// + 1 lesser elements including itself``                ``if` `(arr[j] < arr[i]) {` `                    ``// Check if it is the maximum``                    ``// obtained so far``                    ``leftIncreasing[i]``                        ``= Math.max(``                            ``leftIncreasing[i],``                            ``leftIncreasing[j] + ``1``);``                ``}``            ``}``        ``}` `        ``// Search for increasing subsequence from right``        ``for` `(``int` `i = arr.length - ``2``; i >= ``0``; i--) {``            ``for` `(``int` `j = arr.length - ``1``; j > i; j--) {``                ``if` `(arr[j] < arr[i]) {``                    ``rightIncreasing[i]``                        ``= Math.max(rightIncreasing[i],``                                   ``rightIncreasing[j] + ``1``);``                ``}``            ``}``        ``}` `        ``// Find the position following the peak``        ``// condition and have maximum leftIncreasing[i]``        ``// + rightIncreasing[i]``        ``for` `(``int` `i = ``0``; i < arr.length; i++) {``            ``if` `(leftIncreasing[i] != ``0``                ``&& rightIncreasing[i] != ``0``) {``                ``result = Math.max(``                    ``result, leftIncreasing[i]``                                ``+ rightIncreasing[i]);``            ``}``        ``}` `        ``return` `arr.length - (result + ``1``);``    ``}` `    ``// Function to count elements to be``    ``// removed to make array a mountain array``    ``public` `static` `void` `minRemovals(``int``[] arr)``    ``{``        ``int` `ans = minRemovalsUtil(arr);` `        ``// Print the answer``        ``System.out.println(ans);``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``// Given array``        ``int``[] arr = { ``2``, ``1``, ``1``, ``5``, ``6``, ``2``, ``3``, ``1` `};` `        ``// Function Call``        ``minRemovals(arr);``    ``}``}`

## Python3

 `# Python3 program of the above approach` `# Utility function to count array``# elements required to be removed``# to make array a mountain array``def` `minRemovalsUtil(arr):``    ` `    ``result ``=` `0``    ` `    ``if` `(``len``(arr) < ``3``):``        ``return` `-``1` `    ``# Stores length of increasing``    ``# subsequence from [0, i-1]``    ``leftIncreasing ``=` `[``0``] ``*` `len``(arr)` `    ``# Stores length of increasing``    ``# subsequence from [i + 1, n - 1]``    ``rightIncreasing ``=` `[``0``] ``*` `len``(arr)` `    ``# Iterate for each position up to``    ``# N - 1 to find the length of subsequence``    ``for` `i ``in` `range``(``1``, ``len``(arr)):``        ``for` `j ``in` `range``(i):` `            ``# If j is less than i, then``            ``# i-th position has leftIncreasing[j]``            ``# + 1 lesser elements including itself``            ``if` `(arr[j] < arr[i]):` `                ``# Check if it is the maximum``                ``# obtained so far``                ``leftIncreasing[i] ``=` `max``(leftIncreasing[i],``                                        ``leftIncreasing[j] ``+` `1``);``                    ` `    ``# Search for increasing subsequence from right``    ``for` `i ``in` `range``(``len``(arr) ``-` `2` `, ``-``1``, ``-``1``):``        ``j ``=` `len``(arr) ``-` `1``        ` `        ``while` `j > i:``            ``if` `(arr[j] < arr[i]) :``                ``rightIncreasing[i] ``=` `max``(rightIncreasing[i],``                                         ``rightIncreasing[j] ``+` `1``)``                                        ` `            ``j ``-``=` `1` `    ``# Find the position following the peak``    ``# condition and have maximum leftIncreasing[i]``    ``# + rightIncreasing[i]``    ``for` `i ``in` `range``(``len``(arr)):``        ``if` `(leftIncreasing[i] !``=` `0` `and``           ``rightIncreasing[i] !``=` `0``):``            ``result ``=` `max``(result, leftIncreasing[i] ``+``                                ``rightIncreasing[i]);``    ` `    ``return` `len``(arr) ``-` `(result ``+` `1``)` `# Function to count elements to be``# removed to make array a mountain array``def` `minRemovals(arr):``    ` `    ``ans ``=` `minRemovalsUtil(arr)` `    ``# Print the answer``    ``print``(ans)` `# Driver Code``if` `__name__ ``=``=` `"__main__"` `:``        ` `    ``# Given array``    ``arr ``=` `[ ``2``, ``1``, ``1``, ``5``, ``6``, ``2``, ``3``, ``1` `]` `    ``# Function Call``    ``minRemovals(arr)``    ` `# This code is contributed by AnkThon`

## C#

 `// C# program of the above approach``using` `System;` `class` `GFG``{` `    ``// Utility function to count array``    ``// elements required to be removed ``    ``// to make array a mountain array``    ``public` `static` `int` `minRemovalsUtil(``int``[] arr)``    ``{``        ``int` `result = 0;``        ``if` `(arr.Length < 3)``        ``{``            ``return` `-1;``        ``}` `        ``// Stores length of increasing``        ``// subsequence from [0, i-1]``        ``int``[] leftIncreasing``            ``= ``new` `int``[arr.Length];` `        ``// Stores length of increasing``        ``// subsequence from [i + 1, n - 1]``        ``int``[] rightIncreasing = ``new` `int``[arr.Length];` `        ``// Iterate for each position up to``        ``// N - 1 to find the length of subsequence``        ``for` `(``int` `i = 1; i < arr.Length; i++)``        ``{``            ``for` `(``int` `j = 0; j < i; j++)``            ``{` `                ``// If j is less than i, then``                ``// i-th position has leftIncreasing[j]``                ``// + 1 lesser elements including itself``                ``if` `(arr[j] < arr[i])``                ``{` `                    ``// Check if it is the maximum``                    ``// obtained so far``                    ``leftIncreasing[i]``                        ``= Math.Max(``                            ``leftIncreasing[i],``                            ``leftIncreasing[j] + 1);``                ``}``            ``}``        ``}` `        ``// Search for increasing subsequence from right``        ``for` `(``int` `i = arr.Length - 2; i >= 0; i--)``        ``{``            ``for` `(``int` `j = arr.Length - 1; j > i; j--)``            ``{``                ``if` `(arr[j] < arr[i])``                ``{``                    ``rightIncreasing[i]``                        ``= Math.Max(rightIncreasing[i],``                                   ``rightIncreasing[j] + 1);``                ``}``            ``}``        ``}` `        ``// Find the position following the peak``        ``// condition and have maximum leftIncreasing[i]``        ``// + rightIncreasing[i]``        ``for` `(``int` `i = 0; i < arr.Length; i++)``        ``{``            ``if` `(leftIncreasing[i] != 0``                ``&& rightIncreasing[i] != 0)``            ``{``                ``result = Math.Max(result, leftIncreasing[i]``                                ``+ rightIncreasing[i]);``            ``}``        ``}``        ``return` `arr.Length - (result + 1);``    ``}` `    ``// Function to count elements to be``    ``// removed to make array a mountain array``    ``public` `static` `void` `minRemovals(``int``[] arr)``    ``{``        ``int` `ans = minRemovalsUtil(arr);` `        ``// Print the answer``        ``Console.WriteLine(ans);``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main(String[] args) ``    ``{``        ``// Given array``        ``int``[] arr = {2, 1, 1, 5, 6, 2, 3, 1};` `        ``// Function Call``        ``minRemovals(arr);``    ``}``}`` ` `// This code is contributed by shikhasingrajput.`

## Javascript

 ``

Output

`3`

Time Complexity: O(N2), where N is the number of elements in the array
In the worst case every time we have to compare with all previous elements again.
Auxiliary Space: O(N)
For the left and right increasing array

Approach 2 : (Efficient Code)

The idea is same but by doing a slight change in the previous code we can reduce the redundant work.
The algorithm is basically works on finding the largest bitonic subsequence and after finding it subtract from the total length of the array. That will be the required answer. Below explanation is for the following.
We were making the left increasing and right increasing subsequence array and for each we are doing the same work twice. That can be done by a single function and reverse the result to our need., i.e. a slight observation upon the current scenario is, we basically need a longest increasing subsequence (LIS) and longest decreasing subsequence (LDS). And taking both of them in right direction.

For understanding it, if given array is [2 1 1 5 6 2 3 1] then the LIS array would look something like, [1 1 1 2 3 2 3 1] and if we would find the LDS array that would look like [2 1 1 3 3 2 2 1]. It can be easily achieved by the LIS function by passing the reversed array.

Below is the implementation of the algorithm :

## C++

 `// C++ program of the above approach``#include ``using` `namespace` `std;` `// The LIS function will return the LIS of passed array``vector<``int``> giveLIS(vector<``int``>& nums)``{``    ``int` `n = nums.size();``    ``vector<``int``> lis(n, 1);` `    ``for` `(``int` `i = 1; i < n; i++) {``        ``int` `canAns = 1;``        ``for` `(``int` `j = i - 1; j >= 0; j--) {``            ``if` `(nums[j] < nums[i])``                ``canAns = max(canAns, lis[j] + 1);``        ``}` `        ``lis[i] = canAns;``    ``}` `    ``return` `lis;``}` `void` `minRemovals(vector<``int``>& nums, ``int` `n)``{``    ``vector<``int``> lis = giveLIS(nums);` `    ``// find the lds using lis by just reversing``    ``reverse(nums.begin(), nums.end());``    ``vector<``int``> lds = giveLIS(nums);``    ``reverse(lds.begin(), lds.end());` `    ``int` `maxi = 0;` `    ``// ignoring the edge elements``    ``for` `(``int` `i = 0; i < n; i++) {``        ``// it can't be taken as we don't want the mountain``        ``// to have a steep``        ``if` `(lis[i] == 1 or lds[i] == 1)``            ``continue``;` `        ``maxi = max(maxi, lis[i] + lds[i] - 1);``    ``}` `    ``// maxi is holding the length of largest bitonic subseq``    ``// i.e. that the largest length of mountain possible is``    ``// 'maxi' so the minimum number of points removal is``    ``// "length - maxi"` `    ``int` `ans = (n - maxi);` `    ``// Print the answer``    ``cout << ans;``}` `int` `main()``{``    ``// Given array``    ``vector<``int``> arr = { 2, 1, 1, 5, 6, 2, 3, 1 };``    ``int` `n = arr.size();` `    ``// Function call``    ``minRemovals(arr, n);``    ``return` `0;``}`

## Java

 `import` `java.util.*;``import` `java.io.*;` `// Java program for the above approach``class` `GFG{` `  ``// The LIS function will return the LIS of passed array``  ``static` `ArrayList giveLIS(ArrayList nums)``  ``{``    ``int` `n = nums.size();``    ``ArrayList lis = ``new` `ArrayList();``    ``for``(``int` `i = ``0` `; i < n ; i++){``      ``lis.add(``1``);``    ``}` `    ``for` `(``int` `i = ``1` `; i < n ; i++) {``      ``int` `canAns = ``1``;``      ``for` `(``int` `j = i - ``1` `; j >= ``0` `; j--) {``        ``if` `(nums.get(j) < nums.get(i))``          ``canAns = Math.max(canAns, lis.get(j) + ``1``);``      ``}` `      ``lis.set(i, canAns);``    ``}` `    ``return` `lis;``  ``}` `  ``static` `void` `minRemovals(ArrayList nums, ``int` `n)``  ``{``    ``ArrayList lis = giveLIS(nums);` `    ``// find the lds using lis by just reversing``    ``Collections.reverse(nums);``    ``ArrayList lds = giveLIS(nums);``    ``Collections.reverse(lds);` `    ``int` `maxi = ``0``;` `    ``// ignoring the edge elements``    ``for` `(``int` `i = ``0` `; i < n ; i++) {``      ``// it can't be taken as we don't want the mountain``      ``// to have a steep``      ``if` `(lis.get(i) == ``1` `|| lds.get(i) == ``1``)``        ``continue``;` `      ``maxi = Math.max(maxi, lis.get(i) + lds.get(i) - ``1``);``    ``}` `    ``// maxi is holding the length of largest bitonic subseq``    ``// i.e. that the largest length of mountain possible is``    ``// 'maxi' so the minimum number of points removal is``    ``// "length - maxi"` `    ``int` `ans = (n - maxi);` `    ``// Print the answer``    ``System.out.println(ans);``  ``}` `  ``public` `static` `void` `main(String args[])``  ``{``    ` `    ``// Given array``    ``ArrayList arr = ``new` `ArrayList(``      ``List.of(``        ``2``, ``1``, ``1``, ``5``, ``6``, ``2``, ``3``, ``1``      ``)``    ``);``    ``int` `n = arr.size();` `    ``// Function call``    ``minRemovals(arr, n);``  ``}``}` `// This code is contributed by subhamgoyal2014.`

## C#

 `// C# program to implement above approach``using` `System;``using` `System.Collections;``using` `System.Collections.Generic;` `class` `GFG``{` `  ``// The LIS function will return the LIS of passed array``  ``static` `List<``int``> giveLIS(List<``int``> nums)``  ``{``    ``int` `n = nums.Count;``    ``List<``int``> lis = ``new` `List<``int``>();``    ``for``(``int` `i = 0 ; i < n ; i++){``      ``lis.Add(1);``    ``}` `    ``for` `(``int` `i = 1 ; i < n ; i++) {``      ``int` `canAns = 1;``      ``for` `(``int` `j = i - 1 ; j >= 0 ; j--) {``        ``if` `(nums[j] < nums[i])``          ``canAns = Math.Max(canAns, lis[j] + 1);``      ``}` `      ``lis[i] = canAns;``    ``}` `    ``return` `lis;``  ``}` `  ``static` `void` `minRemovals(List<``int``> nums, ``int` `n)``  ``{``    ``List<``int``> lis = giveLIS(nums);` `    ``// find the lds using lis by just reversing``    ``nums.Reverse();``    ``List<``int``> lds = giveLIS(nums);``    ``lds.Reverse();` `    ``int` `maxi = 0;` `    ``// ignoring the edge elements``    ``for` `(``int` `i = 0 ; i < n ; i++) {``      ``// it can't be taken as we don't want the mountain``      ``// to have a steep``      ``if` `(lis[i] == 1 || lds[i] == 1)``        ``continue``;` `      ``maxi = Math.Max(maxi, lis[i] + lds[i] - 1);``    ``}` `    ``// maxi is holding the length of largest bitonic subseq``    ``// i.e. that the largest length of mountain possible is``    ``// 'maxi' so the minimum number of points removal is``    ``// "length - maxi"` `    ``int` `ans = (n - maxi);` `    ``// Print the answer``    ``Console.WriteLine(ans);``  ``}` `  ``// Driver code``  ``public` `static` `void` `Main(``string``[] args){` `    ``// Given array``    ``List<``int``> arr = ``new` `List<``int``>{2, 1, 1, 5, 6, 2, 3, 1};``    ``int` `n = arr.Count;` `    ``// Function call``    ``minRemovals(arr, n);` `  ``}``}` `// This code is contributed by entertain2022.`

Output

`3`

Time Complexity:  O(N^2), where N is the number of elements in the array
In the worst case every time we have to compare with all previous elements again.
Auxiliary Space:  O(N)
For the LIS and LDS array we are making to calculate

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