Remove minimum elements from either side such that 2*min becomes more than max (original) (raw)

Last Updated : 23 Jul, 2025

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

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)

Find min and max in arr[l..h]
If 2*min > max, then return 0.
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++ `

// 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 2min 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, 2min > 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[0]); cout << minRemovals(arr, 0, n-1); return 0; }

Java

// Java implementation of above approach import java.io.; import java.util.;

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

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#

// 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 2min 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, 2min > 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

PHP

l,l, l,h) { mn=mn = mn=arr[$l]; for ($i = l+1;l + 1; l+1;i <= h;h; h;i++) if ($mn > arr[arr[arr[i]) mn=mn = mn=arr[$i]; return $mn; } // A utility function to find // maximum in arr[l..h] function max_1(&$arr, l,l, l,h) { mx=mx = mx=arr[$l]; for ($i = l+1;l + 1; l+1;i <= h;h; h;i++) if ($mx < arr[arr[arr[i]) mx=mx = mx=arr[$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,l, 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 >= $h) return 0; // 1) Find minimum and maximum in arr[l..h] mn=min1(mn = min_1(mn=min1(arr, l,l, l,h); mx=max1(mx = max_1(mx=max1(arr, l,l, l,h); // If the property is followed, // no removals needed if (2 * mn>mn > mn>mx) return 0; // Otherwise remove a character from left // end and recur, then remove a character // from right end and recur, take the // minimum of two is returned return min(minRemovals($arr, l+1,l + 1, l+1,h), minRemovals($arr, l,l, l,h - 1)) + 1; } // Driver Code $arr = array(4, 5, 100, 9, 10, 11, 12, 15, 200); n=sizeof(n = sizeof(n=sizeof(arr); echo minRemovals($arr, 0, $n - 1); // This code is contributed // by ChitraNayal ?>

JavaScript

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).
Auxiliary Space: O(1)

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++ `

// 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 ,
// from diagonal elements to table[0][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 int main() { int arr[] = {20, 4, 1, 3}; int n = sizeof(arr)/sizeof(arr[0]); cout << minRemovalsDP(arr, n); return 0; }

Java

// Java program of above approach import java.util.; import java.io.;

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 https://www.geeksforgeeks.org/dsa/minimum-insertions-to-form-a-palindrome-dp-28/),
    // from diagonal elements to table[0][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

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[0][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#

// 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 https://www.geeksforgeeks.org/dsa/minimum-insertions-to-form-a-palindrome-dp-28/),
    // from diagonal elements to table[0][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

PHP

l,l, l,h) { mn=mn = mn=arr[$l]; for ($i = l+1;l + 1; l+1;i <= h;h; h;i++) if ($mn > arr[arr[arr[i]) mn=mn = mn=arr[$i]; return $mn; } // A utility function to find // maximum in arr[l..h] function max1($arr, l,l, l,h) { mx=mx = mx=arr[$l]; for ($i = l+1;l + 1; l+1;i <= h;h; h;i++) if ($mx < arr[arr[arr[i]) mx=mx = mx=arr[$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=arrayfill(0,table = array_fill(0, table=arrayfill(0,n, array_fill(0, $n, 0)); // Fill table using above recursive formula. // Note that the table is filled in diagonal fashion // (similar to https://www.geeksforgeeks.org/dsa/minimum-insertions-to-form-a-palindrome-dp-28/), from diagonal elements // to table[0][n-1] which is the result. for ($gap = 0; gap<gap < gap<n; ++$gap) { for ($i = 0, j=j = j=gap; j<j < j<n; ++$i, ++$j) { mn=min1(mn = min1(mn=min1(arr, i,i, i,j); mx=max1(mx = max1(mx=max1(arr, i,i, i,j); table[table[table[i][$j] = (2 * mn>mn > mn>mx) ? 0 : min($table[$i][$j - 1] + 1, table[table[table[i + 1][$j] + 1); } } return table[0][table[0][table[0][n - 1]; } // Driver Code $arr = array(20, 4, 1, 3); n=count(n = count(n=count(arr); echo minRemovalsDP($arr, $n); // This code is contributed by mits ?>

JavaScript

Time Complexity: O(n3) where n is the number of elements in arr[].
Auxiliary Space: O(n^2)

Further Optimizations:

The above code can be optimized in many ways.

We can avoid calculation of min() and/or max() when min and/or max is/are not changed by removing corner elements.
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.

C++ `

// 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 2min becomes greater than max. int minRemovalsDP(int arr[], int n) { // Initialize starting and ending indexes of the maximum // sized subarray with property 2min > 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_MAX, max = INT_MIN;

    // 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 int main() { int arr[] = {4, 5, 100, 9, 10, 11, 12, 15, 200}; int n = sizeof(arr)/sizeof(arr[0]); cout << minRemovalsDP(arr, n); return 0; }

Java

// A O(nn) solution to find the minimum of elements to // be removed import java.util.; import java.io.*;

class GFG {

// Returns the minimum number of removals from either end // in arr[l..h] so that 2min becomes greater than max. static int minRemovalsDP(int arr[], int n) { // Initialize starting and ending indexes of the maximum // sized subarray with property 2min > 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

Python3

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

C#

// 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 2min becomes greater than max. static int minRemovalsDP(int []arr, int n) { // Initialize starting and ending indexes of the maximum // sized subarray with property 2min > 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

PHP

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

JavaScript

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

Remove minimum elements from either side such that 2*min becomes more than max (original) (raw)

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

Python3 implementation of above approach

A utility function to find

minimum in arr[l..h]

A utility function to find

maximum in arr[l..h]

Returns the minimum number of

removals from either end in

arr[l..h] so that 2*min becomes

greater than max.

Driver Code

This code is contributed

by sahilshelangia

Python3 program of above approach

A utility function to find

minimum in arr[l..h]

A utility function to find

maximum in arr[l..h]

Returns the minimum number of removals

from either end in arr[l..h] so that

2*min becomes greater than max.

Driver Code

This code is contributed by mits

A O(n*n) solution to find the minimum of

elements to be removed

Returns the minimum number of removals

from either end in arr[l..h] so that

2*min becomes greater than max.

Driver Code

This code is contributed by mits