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).


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++ 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++)
    // 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);
        // 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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