# Remove minimum elements from array so that max <= 2 * min

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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++ implementation of the approach #include 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 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; }

 # 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

Output:
1

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Improved By : mohit kumar 29

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