def findMin(arr):
 
    # Stores the length of the maximum size subarray
    max_range = 0
 
    # Traverse the array and consider each element as a starting point of a subarray
    for i in range(len(arr)):
 
        '''
            Subarray invariant: max < 2×min
        '''
 
        # To keep track of the minimum and the maximum elements in a subarray
        min_arr = max_arr = arr[i]
 
        # Find the maximum size subarray `arr[i…j]` that satisfies
        # the above invariant
        for j in range(i, len(arr)):
 
            # Find the minimum and the maximum elements in the current subarray.
            # We must do this in constant time.
            min_arr = min(min_arr, arr[j])
            max_arr = max(max_arr, arr[j])
 
            # Discard the subarray if the invariant is violated
            if 2 * min_arr <= max_arr:
                break
 
            # Update `max_range` when a bigger subarray is found
            max_range = max(max_range, j - i + 1)
 
    # Return array size - length of the maximum size subarray
    return len(arr) - max_range
 
 
if __name__ == '__main__':
 
    arr = [4, 6, 1, 7, 5, 9, 2]
    print("The minimum number of removals is", findMin(arr))

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