Given an array **arr**, the task is to remove minimum number of elements such that after their removal, **max(arr) <= 2 * min(arr)**.

**Examples:**

Input:arr[] = {4, 5, 3, 8, 3}

Output:1

Remove 8 from the array.

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

Output:1

Remove 1 from the array.

**Approach:** Let us fix each value as the minimum value say **x** and find number of terms that are in range **[x, 2*x]**. This can be done using prefix-sums, we can use map (implements self balancing BST) instead of array as the values can be large. The remaining terms which are not in range **[x, 2*x]** will have to be removed. So, across all values of **x**, we choose the one which maximises the number of terms in range **[x, 2*x]**.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to return the minimum removals from ` `// arr such that max(arr) <= 2 * min(arr) ` `int` `minimumRemovals(` `int` `n, ` `int` `a[]) ` `{ ` ` ` `// Count occurrence of each element ` ` ` `map<` `int` `, ` `int` `> ct; ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `ct[a[i]]++; ` ` ` ` ` `// Take prefix sum ` ` ` `int` `sm = 0; ` ` ` `for` `(` `auto` `mn : ct) { ` ` ` `sm += mn.second; ` ` ` `ct[mn.first] = sm; ` ` ` `} ` ` ` ` ` `int` `mx = 0, prev = 0; ` ` ` `for` `(` `auto` `mn : ct) { ` ` ` ` ` `// Chosen minimum ` ` ` `int` `x = mn.first; ` ` ` `int` `y = 2 * x; ` ` ` `auto` `itr = ct.upper_bound(y); ` ` ` `itr--; ` ` ` ` ` `// Number of elements that are in ` ` ` `// range [x, 2x] ` ` ` `int` `cr = (itr->second) - prev; ` ` ` `mx = max(mx, cr); ` ` ` `prev = mn.second; ` ` ` `} ` ` ` ` ` `// Minimum elements to be removed ` ` ` `return` `n - mx; ` `} ` ` ` `// Driver Program to test above function ` `int` `main() ` `{ ` ` ` `int` `arr[] = { 4, 5, 3, 8, 3 }; ` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]); ` ` ` `cout << minimumRemovals(n, arr); ` ` ` `return` `0; ` `} ` |

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## Python3

`# Python3 implementation of the approach ` `from` `bisect ` `import` `bisect_left as upper_bound ` ` ` `# Function to return the minimum removals from ` `# arr such that max(arr) <= 2 * min(arr) ` `def` `minimumRemovals(n, a): ` ` ` ` ` `# Count occurrence of each element ` ` ` `ct ` `=` `dict` `() ` ` ` `for` `i ` `in` `a: ` ` ` `ct[i] ` `=` `ct.get(i, ` `0` `) ` `+` `1` ` ` ` ` `# Take prefix sum ` ` ` `sm ` `=` `0` ` ` `for` `mn ` `in` `ct: ` ` ` `sm ` `+` `=` `ct[mn] ` ` ` `ct[mn] ` `=` `sm ` ` ` ` ` `mx ` `=` `0` ` ` `prev ` `=` `0` `; ` ` ` `for` `mn ` `in` `ct: ` ` ` ` ` `# Chosen minimum ` ` ` `x ` `=` `mn ` ` ` `y ` `=` `2` `*` `x ` ` ` `itr ` `=` `upper_bound(` `list` `(ct), y) ` ` ` ` ` `# Number of elements that are in ` ` ` `# range [x, 2x] ` ` ` `cr ` `=` `ct[itr] ` `-` `prev ` ` ` `mx ` `=` `max` `(mx, cr) ` ` ` `prev ` `=` `ct[mn] ` ` ` ` ` `# Minimum elements to be removed ` ` ` `return` `n ` `-` `mx ` ` ` `# Driver Code ` `arr ` `=` `[` `4` `, ` `5` `, ` `3` `, ` `8` `, ` `3` `] ` `n ` `=` `len` `(arr) ` `print` `(minimumRemovals(n, arr)) ` ` ` `# This code is contributed by Mohit Kumar ` |

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**Output:**

1

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