从数组中删除n个连续的元素，以便剩余元素的振幅最小

def min_amplitude_after_removal(arr, n):
    if n >= len(arr):
        return 0  # 如果n大于等于数组长度，直接返回0
    
    max_val = max(arr)
    min_val = min(arr)
    
    # 初始化最大值和最小值的索引
    max_idx = arr.index(max_val)
    min_idx = arr.index(min_val)
    
    # 确保max_idx在min_idx右侧
    if max_idx < min_idx:
        max_idx, min_idx = min_idx, max_idx
    
    # 计算初始振幅
    initial_amplitude = max_val - min_val
    
    # 滑动窗口计算最小振幅
    min_amplitude = initial_amplitude
    for i in range(len(arr) - n + 1):
        current_max = max(arr[i:i+n])
        current_min = min(arr[i:i+n])
        current_amplitude = max(arr[:i] + arr[i+n:]) - min(arr[:i] + arr[i+n:])
        min_amplitude = min(min_amplitude, current_amplitude)
    
    return min_amplitude

# 示例
arr = [1, 3, 6, 2, 8, 4]
n = 2
print(min_amplitude_after_removal(arr, n))  # 输出应该是最小可能的振幅