# Remove minimum elements from either side such that 2*min becomes more than max

Given an unsorted array, trim the array such that twice of minimum is greater than maximum in the trimmed array. Elements should be removed either end of the array.

Number of removals should be minimum.

Examples:

```arr[] = {4, 5, 100, 9, 10, 11, 12, 15, 200}
Output: 4
We need to remove 4 elements (4, 5, 100, 200)
so that 2*min becomes more than max.

arr[] = {4, 7, 5, 6}
Output: 0
We don't need to remove any element as
4*2 > 7 (Note that min = 4, max = 8)

arr[] = {20, 7, 5, 6}
Output: 1
We need to remove 20 so that 2*min becomes
more than max

arr[] = {20, 4, 1, 3}
Output: 3
We need to remove any three elements from ends
like 20, 4, 1 or 4, 1, 3 or 20, 3, 1 or 20, 4, 1```

## We strongly recommend that you click here and practice it, before moving on to the solution.

Naive Solution:
A naive solution is to try every possible case using recurrence. Following is the naive recursive algorithm. Note that the algorithm only returns minimum numbers of removals to be made, it doesn’t print the trimmed array. It can be easily modified to print the trimmed array as well.

```// Returns minimum number of removals to be made in
// arr[l..h]
minRemovals(int arr[], int l, int h)
1) Find min and max in arr[l..h]
2) If 2*min > max, then return 0.
3) Else return minimum of "minRemovals(arr, l+1, h) + 1"
and "minRemovals(arr, l, h-1) + 1"
```

Following is the implementation of above algorithm.

 `// C++ implementation of above approach ` `#include ` `using` `namespace` `std; ` ` `  `// A utility function to find minimum of two numbers ` `int` `min(``int` `a, ``int` `b) {``return` `(a < b)? a : b;} ` ` `  `// A utility function to find minimum in arr[l..h] ` `int` `min(``int` `arr[], ``int` `l, ``int` `h) ` `{ ` `    ``int` `mn = arr[l]; ` `    ``for` `(``int` `i=l+1; i<=h; i++) ` `       ``if` `(mn > arr[i]) ` `         ``mn = arr[i]; ` `    ``return` `mn; ` `} ` ` `  `// A utility function to find maximum in arr[l..h] ` `int` `max(``int` `arr[], ``int` `l, ``int` `h) ` `{ ` `    ``int` `mx = arr[l]; ` `    ``for` `(``int` `i=l+1; i<=h; i++) ` `       ``if` `(mx < arr[i]) ` `         ``mx = arr[i]; ` `    ``return` `mx; ` `} ` ` `  `// Returns the minimum number of removals from either end ` `// in arr[l..h] so that 2*min becomes greater than max. ` `int` `minRemovals(``int` `arr[], ``int` `l, ``int` `h) ` `{ ` `    ``// If there is 1 or less elements, return 0 ` `    ``// For a single element, 2*min > max  ` `    ``// (Assumption: All elements are positive in arr[]) ` `    ``if` `(l >= h) ``return` `0; ` ` `  `    ``// 1) Find minimum and maximum in arr[l..h] ` `    ``int` `mn = min(arr, l, h); ` `    ``int` `mx = max(arr, l, h); ` ` `  `    ``//If the property is followed, no removals needed ` `    ``if` `(2*mn > mx) ` `       ``return` `0; ` ` `  `    ``// Otherwise remove a character from left end and recur, ` `    ``// then remove a character from right end and recur, take ` `    ``// the minimum of two is returned ` `    ``return` `min(minRemovals(arr, l+1, h), ` `               ``minRemovals(arr, l, h-1)) + 1; ` `} ` ` `  `// Driver program to test above functions ` `int` `main() ` `{ ` `  ``int` `arr[] = {4, 5, 100, 9, 10, 11, 12, 15, 200}; ` `  ``int` `n = ``sizeof``(arr)/``sizeof``(arr); ` `  ``cout << minRemovals(arr, 0, n-1); ` `  ``return` `0; ` `} `

 `// Java implementation of above approach ` `class` `GFG ` `{ ` `// A utility function to find minimum of two numbers ` `static` `int` `min(``int` `a, ``int` `b) {``return` `(a < b)? a : b;} ` ` `  `// A utility function to find minimum in arr[l..h] ` `static` `int` `min(``int` `arr[], ``int` `l, ``int` `h) ` `{ ` `    ``int` `mn = arr[l]; ` `    ``for` `(``int` `i=l+``1``; i<=h; i++) ` `    ``if` `(mn > arr[i]) ` `        ``mn = arr[i]; ` `    ``return` `mn; ` `} ` ` `  `// A utility function to find maximum in arr[l..h] ` `static` `int` `max(``int` `arr[], ``int` `l, ``int` `h) ` `{ ` `    ``int` `mx = arr[l]; ` `    ``for` `(``int` `i=l+``1``; i<=h; i++) ` `    ``if` `(mx < arr[i]) ` `        ``mx = arr[i]; ` `    ``return` `mx; ` `} ` ` `  `// Returns the minimum number of removals from either end ` `// in arr[l..h] so that 2*min becomes greater than max. ` `static` `int` `minRemovals(``int` `arr[], ``int` `l, ``int` `h) ` `{ ` `    ``// If there is 1 or less elements, return 0 ` `    ``// For a single element, 2*min > max  ` `    ``// (Assumption: All elements are positive in arr[]) ` `    ``if` `(l >= h) ``return` `0``; ` ` `  `    ``// 1) Find minimum and maximum in arr[l..h] ` `    ``int` `mn = min(arr, l, h); ` `    ``int` `mx = max(arr, l, h); ` ` `  `    ``//If the property is followed, no removals needed ` `    ``if` `(``2``*mn > mx) ` `    ``return` `0``; ` ` `  `    ``// Otherwise remove a character from left end and recur, ` `    ``// then remove a character from right end and recur, take ` `    ``// the minimum of two is returned ` `    ``return` `min(minRemovals(arr, l+``1``, h), ` `            ``minRemovals(arr, l, h-``1``)) + ``1``; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `int` `arr[] = {``4``, ``5``, ``100``, ``9``, ``10``, ``11``, ``12``, ``15``, ``200``}; ` `int` `n = arr.length; ` `System.out.print(minRemovals(arr, ``0``, n-``1``)); ` `} ` `} ` ` `  `// This code is contributed by Mukul Singh.  `

 `# Python3 implementation of above approach ` `# A utility function to find  ` `# minimum in arr[l..h]  ` `def` `mini(arr, l, h):  ` `    ``mn ``=` `arr[l] ` `    ``for` `i ``in` `range``(l ``+` `1``, h ``+` `1``): ` `        ``if` `(mn > arr[i]): ` `            ``mn ``=` `arr[i]  ` `    ``return` `mn ` ` `  `# A utility function to find  ` `# maximum in arr[l..h]  ` `def` `max``(arr, l, h):  ` `    ``mx ``=` `arr[l]  ` `    ``for` `i ``in` `range``(l ``+` `1``, h ``+` `1``): ` `        ``if` `(mx < arr[i]): ` `            ``mx ``=` `arr[i] ` `    ``return` `mx ` ` `  ` `  `# Returns the minimum number of  ` `# removals from either end in  ` `# arr[l..h] so that 2*min becomes  ` `# greater than max.  ` `def` `minRemovals(arr, l, h): ` `     `  `    ``# If there is 1 or less elements, return 0  ` `    ``# For a single element, 2*min > max  ` `    ``# (Assumption: All elements are positive in arr[])  ` `    ``if` `(l >``=` `h): ` `        ``return` `0` ` `  `    ``# 1) Find minimum and maximum  ` `    ``#    in arr[l..h]  ` `    ``mn ``=` `mini(arr, l, h)  ` `    ``mx ``=` `max``(arr, l, h) ` ` `  `    ``# If the property is followed, ` `    ``# no removals needed  ` `    ``if` `(``2` `*` `mn > mx): ` `        ``return` `0` ` `  `    ``# Otherwise remove a character from  ` `    ``# left end and recur, then remove a  ` `    ``# character from right end and recur,  ` `    ``# take the minimum of two is returned  ` `    ``return` `(``min``(minRemovals(arr, l ``+` `1``, h),  ` `                ``minRemovals(arr, l, h ``-` `1``)) ``+` `1``) ` ` `  `# Driver Code ` `arr ``=` `[``4``, ``5``, ``100``, ``9``, ``10``, ` `       ``11``, ``12``, ``15``, ``200``]  ` `n ``=` `len``(arr) ` `print``(minRemovals(arr, ``0``, n ``-` `1``))  ` ` `  `# This code is contributed  ` `# by sahilshelangia `

 `// C# implementation of above approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `// A utility function to find minimum of two numbers ` `static` `int` `min(``int` `a, ``int` `b) {``return` `(a < b)? a : b;} ` ` `  `// A utility function to find minimum in arr[l..h] ` `static` `int` `min(``int``[] arr, ``int` `l, ``int` `h) ` `{ ` `    ``int` `mn = arr[l]; ` `    ``for` `(``int` `i=l+1; i<=h; i++) ` `    ``if` `(mn > arr[i]) ` `        ``mn = arr[i]; ` `    ``return` `mn; ` `} ` ` `  `// A utility function to find maximum in arr[l..h] ` `static` `int` `max(``int``[] arr, ``int` `l, ``int` `h) ` `{ ` `    ``int` `mx = arr[l]; ` `    ``for` `(``int` `i=l+1; i<=h; i++) ` `    ``if` `(mx < arr[i]) ` `        ``mx = arr[i]; ` `    ``return` `mx; ` `} ` ` `  `// Returns the minimum number of removals from either end ` `// in arr[l..h] so that 2*min becomes greater than max. ` `static` `int` `minRemovals(``int``[] arr, ``int` `l, ``int` `h) ` `{ ` `    ``// If there is 1 or less elements, return 0 ` `    ``// For a single element, 2*min > max  ` `    ``// (Assumption: All elements are positive in arr[]) ` `    ``if` `(l >= h) ``return` `0; ` ` `  `    ``// 1) Find minimum and maximum in arr[l..h] ` `    ``int` `mn = min(arr, l, h); ` `    ``int` `mx = max(arr, l, h); ` ` `  `    ``//If the property is followed, no removals needed ` `    ``if` `(2*mn > mx) ` `    ``return` `0; ` ` `  `    ``// Otherwise remove a character from left end and recur, ` `    ``// then remove a character from right end and recur, take ` `    ``// the minimum of two is returned ` `    ``return` `min(minRemovals(arr, l+1, h), ` `            ``minRemovals(arr, l, h-1)) + 1; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main() ` `{ ` `int``[] arr = {4, 5, 100, 9, 10, 11, 12, 15, 200}; ` `int` `n = arr.Length; ` `Console.Write(minRemovals(arr, 0, n-1)); ` `} ` `} ` ` `  `// This code is contributed by Akanksha Rai `

 ` ``\$arr``[``\$i``]) ` `        ``\$mn` `= ``\$arr``[``\$i``]; ` `    ``return` `\$mn``; ` `} ` ` `  `// A utility function to find ` `// maximum in arr[l..h] ` `function` `max_1(&``\$arr``, ``\$l``, ``\$h``) ` `{ ` `    ``\$mx` `= ``\$arr``[``\$l``]; ` `    ``for` `(``\$i` `= ``\$l` `+ 1; ``\$i` `<= ``\$h``; ``\$i``++) ` `    ``if` `(``\$mx` `< ``\$arr``[``\$i``]) ` `        ``\$mx` `= ``\$arr``[``\$i``]; ` `    ``return` `\$mx``; ` `} ` ` `  `// Returns the minimum number of removals  ` `// from either end in arr[l..h] so that ` `// 2*min becomes greater than max. ` `function` `minRemovals(&``\$arr``, ``\$l``, ``\$h``) ` `{ ` `    ``// If there is 1 or less elements,  ` `    ``// return 0. For a single element,  ` `    ``// 2*min > max. (Assumption: All  ` `    ``// elements are positive in arr[]) ` `    ``if` `(``\$l` `>= ``\$h``) ``return` `0; ` ` `  `    ``// 1) Find minimum and maximum in arr[l..h] ` `    ``\$mn` `= min_1(``\$arr``, ``\$l``, ``\$h``); ` `    ``\$mx` `= max_1(``\$arr``, ``\$l``, ``\$h``); ` ` `  `    ``// If the property is followed,  ` `    ``// no removals needed ` `    ``if` `(2 * ``\$mn` `> ``\$mx``) ` `    ``return` `0; ` ` `  `    ``// Otherwise remove a character from left ` `    ``// end and recur, then remove a character  ` `    ``// from right end and recur, take the  ` `    ``// minimum of two is returned ` `    ``return` `min(minRemovals(``\$arr``, ``\$l` `+ 1, ``\$h``), ` `               ``minRemovals(``\$arr``, ``\$l``, ``\$h` `- 1)) + 1; ` `} ` ` `  `// Driver Code ` `\$arr` `= ``array``(4, 5, 100, 9, 10,  ` `             ``11, 12, 15, 200); ` `\$n` `= sizeof(``\$arr``); ` `echo` `minRemovals(``\$arr``, 0, ``\$n` `- 1); ` ` `  `// This code is contributed  ` `// by ChitraNayal ` `?> `

Output:
` 4 `

Time complexity: Time complexity of the above function can be written as following

`  T(n) = 2T(n-1) + O(n) `

An upper bound on solution of above recurrence would be O(n x 2n).

Dynamic Programming:
The above recursive code exhibits many overlapping subproblems. For example minRemovals(arr, l+1, h-1) is evaluated twice. So Dynamic Programming is the choice to optimize the solution. Following is Dynamic Programming based solution.

 `// C++ program of above approach ` `#include ` `using` `namespace` `std; ` ` `  `// A utility function to find minimum of two numbers ` `int` `min(``int` `a, ``int` `b) {``return` `(a < b)? a : b;} ` ` `  `// A utility function to find minimum in arr[l..h] ` `int` `min(``int` `arr[], ``int` `l, ``int` `h) ` `{ ` `    ``int` `mn = arr[l]; ` `    ``for` `(``int` `i=l+1; i<=h; i++) ` `       ``if` `(mn > arr[i]) ` `         ``mn = arr[i]; ` `    ``return` `mn; ` `} ` ` `  `// A utility function to find maximum in arr[l..h] ` `int` `max(``int` `arr[], ``int` `l, ``int` `h) ` `{ ` `    ``int` `mx = arr[l]; ` `    ``for` `(``int` `i=l+1; i<=h; i++) ` `       ``if` `(mx < arr[i]) ` `         ``mx = arr[i]; ` `    ``return` `mx; ` `} ` ` `  `// Returns the minimum number of removals from either end ` `// in arr[l..h] so that 2*min becomes greater than max. ` `int` `minRemovalsDP(``int` `arr[], ``int` `n) ` `{ ` `    ``// Create a table to store solutions of subproblems ` `    ``int` `table[n][n], gap, i, j, mn, mx; ` ` `  `    ``// Fill table using above recursive formula. Note that the table ` `    ``// is filled in diagonal fashion (similar to http://goo.gl/PQqoS), ` `    ``// from diagonal elements to table[n-1] which is the result. ` `    ``for` `(gap = 0; gap < n; ++gap) ` `    ``{ ` `        ``for` `(i = 0, j = gap; j < n; ++i, ++j) ` `        ``{ ` `            ``mn = min(arr, i, j); ` `            ``mx = max(arr, i, j); ` `            ``table[i][j] = (2*mn > mx)? 0: min(table[i][j-1]+1, ` `                                              ``table[i+1][j]+1); ` `        ``} ` `    ``} ` `    ``return` `table[n-1]; ` `} ` ` `  `// Driver program to test above functions ` `int` `main() ` `{ ` `  ``int` `arr[] = {20, 4, 1, 3}; ` `  ``int` `n = ``sizeof``(arr)/``sizeof``(arr); ` `  ``cout << minRemovalsDP(arr, n); ` `  ``return` `0; ` `} `

 `// Java program of above approach ` `class` `GFG { ` ` `  `// A utility function to find minimum of two numbers ` `    ``static` `int` `min(``int` `a, ``int` `b) { ` `        ``return` `(a < b) ? a : b; ` `    ``} ` ` `  `// A utility function to find minimum in arr[l..h] ` `    ``static` `int` `min(``int` `arr[], ``int` `l, ``int` `h) { ` `        ``int` `mn = arr[l]; ` `        ``for` `(``int` `i = l + ``1``; i <= h; i++) { ` `            ``if` `(mn > arr[i]) { ` `                ``mn = arr[i]; ` `            ``} ` `        ``} ` `        ``return` `mn; ` `    ``} ` ` `  `// A utility function to find maximum in arr[l..h] ` `    ``static` `int` `max(``int` `arr[], ``int` `l, ``int` `h) { ` `        ``int` `mx = arr[l]; ` `        ``for` `(``int` `i = l + ``1``; i <= h; i++) { ` `            ``if` `(mx < arr[i]) { ` `                ``mx = arr[i]; ` `            ``} ` `        ``} ` `        ``return` `mx; ` `    ``} ` ` `  `// Returns the minimum number of removals from either end ` `// in arr[l..h] so that 2*min becomes greater than max. ` `    ``static` `int` `minRemovalsDP(``int` `arr[], ``int` `n) { ` `        ``// Create a table to store solutions of subproblems ` `        ``int` `table[][] = ``new` `int``[n][n], gap, i, j, mn, mx; ` ` `  `        ``// Fill table using above recursive formula. Note that the table ` `        ``// is filled in diagonal fashion (similar to http://goo.gl/PQqoS), ` `        ``// from diagonal elements to table[n-1] which is the result. ` `        ``for` `(gap = ``0``; gap < n; ++gap) { ` `            ``for` `(i = ``0``, j = gap; j < n; ++i, ++j) { ` `                ``mn = min(arr, i, j); ` `                ``mx = max(arr, i, j); ` `                ``table[i][j] = (``2` `* mn > mx) ? ``0` `: min(table[i][j - ``1``] + ``1``, ` `                        ``table[i + ``1``][j] + ``1``); ` `            ``} ` `        ``} ` `        ``return` `table[``0``][n - ``1``]; ` `    ``} ` ` `  `// Driver program to test above functions ` `    ``public` `static` `void` `main(String[] args) { ` `        ``int` `arr[] = {``20``, ``4``, ``1``, ``3``}; ` `        ``int` `n = arr.length; ` `        ``System.out.println(minRemovalsDP(arr, n)); ` ` `  `    ``} ` `} ` `// This code contributed by 29AJayKumar `

 `# Python3 program of above approach ` ` `  `# A utility function to find  ` `# minimum in arr[l..h] ` `def` `min1(arr, l, h): ` `    ``mn ``=` `arr[l]; ` `    ``for` `i ``in` `range``(l ``+` `1``,h``+``1``): ` `        ``if` `(mn > arr[i]): ` `            ``mn ``=` `arr[i]; ` `    ``return` `mn; ` ` `  `# A utility function to find  ` `# maximum in arr[l..h] ` `def` `max1(arr, l, h): ` `    ``mx ``=` `arr[l]; ` `    ``for` `i ``in` `range``(l ``+` `1``, h ``+` `1``): ` `        ``if` `(mx < arr[i]): ` `            ``mx ``=` `arr[i]; ` `    ``return` `mx; ` ` `  `# Returns the minimum number of removals ` `# from either end in arr[l..h] so that  ` `# 2*min becomes greater than max. ` `def` `minRemovalsDP(arr, n): ` `     `  `    ``# Create a table to store ` `    ``# solutions of subproblems ` `    ``table ``=` `[[``0` `for` `x ``in` `range``(n)] ``for` `y ``in` `range``(n)]; ` `     `  `    ``# Fill table using above recursive formula. ` `    ``# Note that the table is filled in diagonal fashion ` `    ``# (similar to http:#goo.gl/PQqoS), from diagonal elements ` `    ``# to table[n-1] which is the result. ` `    ``for` `gap ``in` `range``(n): ` `        ``i ``=` `0``; ` `        ``for` `j ``in` `range``(gap,n): ` `            ``mn ``=` `min1(arr, i, j); ` `            ``mx ``=` `max1(arr, i, j); ` `            ``table[i][j] ``=` `0` `if` `(``2` `*` `mn > mx) ``else` `min``(table[i][j ``-` `1``] ``+` `1``,table[i ``+` `1``][j] ``+` `1``); ` `            ``i ``+``=` `1``; ` `    ``return` `table[``0``][n ``-` `1``]; ` ` `  `# Driver Code ` `arr ``=` `[``20``, ``4``, ``1``, ``3``]; ` `n ``=` `len``(arr); ` `print``(minRemovalsDP(arr, n)); ` ` `  `# This code is contributed by mits `

 `// C# program of above approach ` `using` `System; ` ` `  `public` `class` `GFG { ` `  `  `    ``// A utility function to find minimum of two numbers ` `    ``static` `int` `min(``int` `a, ``int` `b) { ` `        ``return` `(a < b) ? a : b; ` `    ``} ` `  `  `    ``// A utility function to find minimum in arr[l..h] ` `    ``static` `int` `min(``int` `[]arr, ``int` `l, ``int` `h) { ` `        ``int` `mn = arr[l]; ` `        ``for` `(``int` `i = l + 1; i <= h; i++) { ` `            ``if` `(mn > arr[i]) { ` `                ``mn = arr[i]; ` `            ``} ` `        ``} ` `        ``return` `mn; ` `    ``} ` `  `  `// A utility function to find maximum in arr[l..h] ` `    ``static` `int` `max(``int` `[]arr, ``int` `l, ``int` `h) { ` `        ``int` `mx = arr[l]; ` `        ``for` `(``int` `i = l + 1; i <= h; i++) { ` `            ``if` `(mx < arr[i]) { ` `                ``mx = arr[i]; ` `            ``} ` `        ``} ` `        ``return` `mx; ` `    ``} ` `  `  `    ``// Returns the minimum number of removals from either end ` `    ``// in arr[l..h] so that 2*min becomes greater than max. ` `    ``static` `int` `minRemovalsDP(``int` `[]arr, ``int` `n) { ` `        ``// Create a table to store solutions of subproblems ` `        ``int` `[,]table = ``new` `int``[n,n]; ` `        ``int` `gap, i, j, mn, mx; ` `  `  `        ``// Fill table using above recursive formula. Note that the table ` `        ``// is filled in diagonal fashion (similar to http://goo.gl/PQqoS), ` `        ``// from diagonal elements to table[n-1] which is the result. ` `        ``for` `(gap = 0; gap < n; ++gap) { ` `            ``for` `(i = 0, j = gap; j < n; ++i, ++j) { ` `                ``mn = min(arr, i, j); ` `                ``mx = max(arr, i, j); ` `                ``table[i,j] = (2 * mn > mx) ? 0 : min(table[i,j - 1] + 1, ` `                        ``table[i + 1,j] + 1); ` `            ``} ` `        ``} ` `        ``return` `table[0,n - 1]; ` `    ``} ` `  `  `    ``// Driver program to test above functions ` `    ``public` `static` `void` `Main() { ` `        ``int` `[]arr = {20, 4, 1, 3}; ` `        ``int` `n = arr.Length; ` `        ``Console.WriteLine(minRemovalsDP(arr, n)); ` `  `  `    ``} ` `} ` `// This code contributed by 29AJayKumar `

 ` ``\$arr``[``\$i``]) ` `        ``\$mn` `= ``\$arr``[``\$i``]; ` `    ``return` `\$mn``; ` `} ` ` `  `// A utility function to find  ` `// maximum in arr[l..h] ` `function` `max1(``\$arr``, ``\$l``, ``\$h``) ` `{ ` `    ``\$mx` `= ``\$arr``[``\$l``]; ` `    ``for` `(``\$i` `= ``\$l` `+ 1; ``\$i` `<= ``\$h``; ``\$i``++) ` `    ``if` `(``\$mx` `< ``\$arr``[``\$i``]) ` `        ``\$mx` `= ``\$arr``[``\$i``]; ` `    ``return` `\$mx``; ` `} ` ` `  `// Returns the minimum number of removals ` `// from either end in arr[l..h] so that  ` `// 2*min becomes greater than max. ` `function` `minRemovalsDP(``\$arr``, ``\$n``) ` `{ ` `     `  `    ``// Create a table to store  ` `    ``// solutions of subproblems ` `    ``\$table` `= ``array_fill``(0, ``\$n``,  ` `             ``array_fill``(0, ``\$n``, 0)); ` ` `  `    ``// Fill table using above recursive formula.  ` `    ``// Note that the table is filled in diagonal fashion  ` `    ``// (similar to http://goo.gl/PQqoS), from diagonal elements  ` `    ``// to table[n-1] which is the result. ` `    ``for` `(``\$gap` `= 0; ``\$gap` `< ``\$n``; ++``\$gap``) ` `    ``{ ` `        ``for` `(``\$i` `= 0, ``\$j` `= ``\$gap``; ``\$j` `< ``\$n``; ++``\$i``, ++``\$j``) ` `        ``{ ` `            ``\$mn` `= min1(``\$arr``, ``\$i``, ``\$j``); ` `            ``\$mx` `= max1(``\$arr``, ``\$i``, ``\$j``); ` `            ``\$table``[``\$i``][``\$j``] = (2 * ``\$mn` `> ``\$mx``) ? 0 :  ` `                              ``min(``\$table``[``\$i``][``\$j` `- 1] + 1, ` `                                  ``\$table``[``\$i` `+ 1][``\$j``] + 1); ` `        ``} ` `    ``} ` `    ``return` `\$table``[``\$n` `- 1]; ` `} ` ` `  `// Driver Code ` `\$arr` `= ``array``(20, 4, 1, 3); ` `\$n` `= ``count``(``\$arr``); ` `echo` `minRemovalsDP(``\$arr``, ``\$n``); ` ` `  `// This code is contributed by mits ` `?> `

Output:
` 3`

Time Complexity: O(n3) where n is the number of elements in arr[].

Further Optimizations:
The above code can be optimized in many ways.
1) We can avoid calculation of min() and/or max() when min and/or max is/are not changed by removing corner elements.

2) We can pre-process the array and build segment tree in O(n) time. After the segment tree is built, we can query range minimum and maximum in O(Logn) time. The overall time complexity is reduced to O(n2Logn) time.

A O(n^2) Solution
The idea is to find the maximum sized subarray such that 2*min > max. We run two nested loops, the outer loop chooses a starting point and the inner loop chooses ending point for the current starting point. We keep track of longest subarray with the given property.

Following is the implementation of the above approach. Thanks to Richard Zhang for suggesting this solution.

 `// A O(n*n) solution to find the minimum of elements to ` `// be removed ` `#include ` `#include ` `using` `namespace` `std; ` ` `  `// Returns the minimum number of removals from either end ` `// in arr[l..h] so that 2*min becomes greater than max. ` `int` `minRemovalsDP(``int` `arr[], ``int` `n) ` `{ ` `    ``// Initialize starting and ending indexes of the maximum ` `    ``// sized subarray with property 2*min > max ` `    ``int` `longest_start = -1, longest_end = 0; ` ` `  `    ``// Choose different elements as starting point ` `    ``for` `(``int` `start=0; start max) max = val; ` ` `  `            ``// If the property is violated, then no ` `            ``// point to continue for a bigger array ` `            ``if` `(2 * min <= max) ``break``; ` ` `  `            ``// Update longest_start and longest_end if needed ` `            ``if` `(end - start > longest_end - longest_start || ` `                ``longest_start == -1) ` `            ``{ ` `                ``longest_start = start; ` `                ``longest_end = end; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// If not even a single element follow the property, ` `    ``// then return n ` `    ``if` `(longest_start == -1) ``return` `n; ` ` `  `    ``// Return the number of elements to be removed ` `    ``return` `(n - (longest_end - longest_start + 1)); ` `} ` ` `  `// Driver program to test above functions ` `int` `main() ` `{ ` `    ``int` `arr[] = {4, 5, 100, 9, 10, 11, 12, 15, 200}; ` `    ``int` `n = ``sizeof``(arr)/``sizeof``(arr); ` `    ``cout << minRemovalsDP(arr, n); ` `    ``return` `0; ` `}`

 `// A O(n*n) solution to find the minimum of elements to  ` `// be removed  ` ` `  `class` `GFG { ` ` `  `// Returns the minimum number of removals from either end  ` `// in arr[l..h] so that 2*min becomes greater than max.  ` `    ``static` `int` `minRemovalsDP(``int` `arr[], ``int` `n) { ` `        ``// Initialize starting and ending indexes of the maximum  ` `        ``// sized subarray with property 2*min > max  ` `        ``int` `longest_start = -``1``, longest_end = ``0``; ` ` `  `        ``// Choose different elements as starting point  ` `        ``for` `(``int` `start = ``0``; start < n; start++) { ` `            ``// Initialize min and max for the current start  ` `            ``int` `min = Integer.MAX_VALUE, max = Integer.MIN_VALUE; ` ` `  `            ``// Choose different ending points for current start  ` `            ``for` `(``int` `end = start; end < n; end++) { ` `                ``// Update min and max if necessary  ` `                ``int` `val = arr[end]; ` `                ``if` `(val < min) { ` `                    ``min = val; ` `                ``} ` `                ``if` `(val > max) { ` `                    ``max = val; ` `                ``} ` ` `  `                ``// If the property is violated, then no  ` `                ``// point to continue for a bigger array  ` `                ``if` `(``2` `* min <= max) { ` `                    ``break``; ` `                ``} ` ` `  `                ``// Update longest_start and longest_end if needed  ` `                ``if` `(end - start > longest_end - longest_start ` `                        ``|| longest_start == -``1``) { ` `                    ``longest_start = start; ` `                    ``longest_end = end; ` `                ``} ` `            ``} ` `        ``} ` ` `  `        ``// If not even a single element follow the property,  ` `        ``// then return n  ` `        ``if` `(longest_start == -``1``) { ` `            ``return` `n; ` `        ``} ` ` `  `        ``// Return the number of elements to be removed  ` `        ``return` `(n - (longest_end - longest_start + ``1``)); ` `    ``} ` ` `  `// Driver program to test above functions  ` `    ``public` `static` `void` `main(String[] args) { ` `        ``int` `arr[] = {``4``, ``5``, ``100``, ``9``, ``10``, ``11``, ``12``, ``15``, ``200``}; ` `        ``int` `n = arr.length; ` `        ``System.out.println(minRemovalsDP(arr, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by PrinciRaj1992  `

 `# A O(n*n) solution to find the minimum of ` `# elements to be removed ` `import` `sys; ` ` `  `# Returns the minimum number of removals  ` `# from either end in arr[l..h] so that  ` `# 2*min becomes greater than max. ` `def` `minRemovalsDP(arr, n): ` ` `  `    ``# Initialize starting and ending indexes  ` `    ``# of the maximum sized subarray  ` `    ``# with property 2*min > max ` `    ``longest_start ``=` `-``1``; ` `    ``longest_end ``=` `0``; ` ` `  `    ``# Choose different elements as starting point ` `    ``for` `start ``in` `range``(n): ` `         `  `        ``# Initialize min and max  ` `        ``# for the current start ` `        ``min` `=` `sys.maxsize; ` `        ``max` `=` `-``sys.maxsize; ` ` `  `        ``# Choose different ending points for current start ` `        ``for` `end ``in` `range``(start,n): ` `            ``# Update min and max if necessary ` `            ``val ``=` `arr[end]; ` `            ``if` `(val < ``min``): ` `                ``min` `=` `val; ` `            ``if` `(val > ``max``): ` `                ``max` `=` `val; ` ` `  `            ``# If the property is violated, then no ` `            ``# point to continue for a bigger array ` `            ``if` `(``2` `*` `min` `<``=` `max``): ` `                ``break``; ` ` `  `            ``# Update longest_start and longest_end if needed ` `            ``if` `(end ``-` `start > longest_end ``-` `longest_start ``or` `longest_start ``=``=` `-``1``): ` `                ``longest_start ``=` `start; ` `                ``longest_end ``=` `end; ` ` `  `    ``# If not even a single element follow the property, ` `    ``# then return n ` `    ``if` `(longest_start ``=``=` `-``1``): ` `        ``return` `n; ` ` `  `    ``# Return the number of elements to be removed ` `    ``return` `(n ``-` `(longest_end ``-` `longest_start ``+` `1``)); ` ` `  `# Driver Code ` `arr ``=` `[``4``, ``5``, ``100``, ``9``, ``10``, ``11``, ``12``, ``15``, ``200``]; ` `n ``=` `len``(arr); ` `print``(minRemovalsDP(arr, n)); ` ` `  `# This code is contributed by mits `

 `// A O(n*n) solution to find the minimum of elements to  ` `// be removed  ` ` `  `using` `System;  ` `public` `class` `GFG { ` `  `  `// Returns the minimum number of removals from either end  ` `// in arr[l..h] so that 2*min becomes greater than max.  ` `    ``static` `int` `minRemovalsDP(``int` `[]arr, ``int` `n) { ` `        ``// Initialize starting and ending indexes of the maximum  ` `        ``// sized subarray with property 2*min > max  ` `        ``int` `longest_start = -1, longest_end = 0; ` `  `  `        ``// Choose different elements as starting point  ` `        ``for` `(``int` `start = 0; start < n; start++) { ` `            ``// Initialize min and max for the current start  ` `            ``int` `min = ``int``.MaxValue, max = ``int``.MinValue; ` `  `  `            ``// Choose different ending points for current start  ` `            ``for` `(``int` `end = start; end < n; end++) { ` `                ``// Update min and max if necessary  ` `                ``int` `val = arr[end]; ` `                ``if` `(val < min) { ` `                    ``min = val; ` `                ``} ` `                ``if` `(val > max) { ` `                    ``max = val; ` `                ``} ` `  `  `                ``// If the property is violated, then no  ` `                ``// point to continue for a bigger array  ` `                ``if` `(2 * min <= max) { ` `                    ``break``; ` `                ``} ` `  `  `                ``// Update longest_start and longest_end if needed  ` `                ``if` `(end - start > longest_end - longest_start ` `                        ``|| longest_start == -1) { ` `                    ``longest_start = start; ` `                    ``longest_end = end; ` `                ``} ` `            ``} ` `        ``} ` `  `  `        ``// If not even a single element follow the property,  ` `        ``// then return n  ` `        ``if` `(longest_start == -1) { ` `            ``return` `n; ` `        ``} ` `  `  `        ``// Return the number of elements to be removed  ` `        ``return` `(n - (longest_end - longest_start + 1)); ` `    ``} ` `  `  `// Driver program to test above functions  ` `    ``public` `static` `void` `Main() { ` `        ``int` `[]arr = {4, 5, 100, 9, 10, 11, 12, 15, 200}; ` `        ``int` `n = arr.Length; ` `        ``Console.WriteLine(minRemovalsDP(arr, n)); ` `    ``} ` `} ` `  `  `// This code is contributed by Rajput-Ji `

 ` max ` `    ``\$longest_start` `= -1; ` `    ``\$longest_end` `= 0; ` ` `  `    ``// Choose different elements as starting point ` `    ``for` `(``\$start` `= 0; ``\$start` `< ``\$n``; ``\$start``++) ` `    ``{ ` `         `  `        ``// Initialize min and max  ` `        ``// for the current start ` `        ``\$min` `= PHP_INT_MAX; ` `        ``\$max` `= PHP_INT_MIN; ` ` `  `        ``// Choose different ending points for current start ` `        ``for` `(``\$end` `= ``\$start``; ``\$end` `< ``\$n``; ``\$end` `++) ` `        ``{ ` `            ``// Update min and max if necessary ` `            ``\$val` `= ``\$arr``[``\$end``]; ` `            ``if` `(``\$val` `< ``\$min``) ``\$min` `= ``\$val``; ` `            ``if` `(``\$val` `> ``\$max``) ``\$max` `= ``\$val``; ` ` `  `            ``// If the property is violated, then no ` `            ``// point to continue for a bigger array ` `            ``if` `(2 * ``\$min` `<= ``\$max``) ``break``; ` ` `  `            ``// Update longest_start and longest_end if needed ` `            ``if` `(``\$end` `- ``\$start` `> ``\$longest_end` `- ``\$longest_start` `|| ` `                ``\$longest_start` `== -1) ` `            ``{ ` `                ``\$longest_start` `= ``\$start``; ` `                ``\$longest_end` `= ``\$end``; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// If not even a single element follow the property, ` `    ``// then return n ` `    ``if` `(``\$longest_start` `== -1) ``return` `\$n``; ` ` `  `    ``// Return the number of elements to be removed ` `    ``return` `(``\$n` `- (``\$longest_end` `- ``\$longest_start` `+ 1)); ` `} ` ` `  `// Driver Code ` `\$arr` `= ``array``(4, 5, 100, 9, 10, 11, 12, 15, 200); ` `\$n` `= sizeof(``\$arr``); ` `echo` `minRemovalsDP(``\$arr``, ``\$n``); ` ` `  `// This code is contributed by jit_t ` `?> `

Output:
`4`