TCS Codevita |孔和球

给定H []和B []的两个数组，分别由N和M个整数组成，分别表示孔和球的直径。使M个球在具有N个孔的倾斜表面上从A滚动到B ，每个孔的深度不同，如下图所示：

该任务是考虑以下因素，以释放的球的顺序找到每个球的最终位置：

• 如果球的直径小于或等于孔的直径，则球将掉入孔中。
• 如果i个球落入一个孔H _i中，它将充满。
• 如果孔已满，则不再有球落入其中。
• 当且仅当球未掉入任何一个孔中时，球才会从A到达B。
• 如果球在孔P _i中，则其位置为i 。如果球到达最低点B，则将其位置设为0 。

例子：

Input: H[] = {21, 3, 6}, B[] = {20, 15, 5, 7, 10, 4, 2, 1, 3, 6, 8}
Output: 1 0 3 0 0 3 3 2 2 0 0
Explanation:
Ball of diameter 20 will fall into the hole H₁ and the hole H₁ will become full.
Balls with diameter 15, 7 and 10 will reach bottom, since the hole H₁ is full and diameters of holes H₂ and H₃ are less than the diameters of the balls.
Balls with diameters 5, 4 and 2 will fall into the hole H₃.
Ball with diameter 1 will fall into the hole H₂ since the hole H₃ is already full.
Ball with diameter 3 will fall into hole H₂.
Balls with diameters 6, and 8 will reach the bottom point B.
The position of ball 20 is 1 because it is in hole H₁.
Positions of ball 15, 7, 10, 3, 6, and 8 are 0 because they reached the bottom point B.
Therefore, the balls with diameter 5, 4 and 2 are in the 3rd hole H₃, the ball with diameter 1 and 3 are in the 2nd hole H₂.

Input: H[] = {20, 15, 10, 5, 25}, B[] = {5, 10, 15, 20, 25, 30, 4, 9, 14, 19}
Output: 5 5 5 5 5 0 4 3 2 1

方法：请按照以下步骤解决问题：

• 初始化的大小为N的阵列位置[]存储每个球的最终位置和大小为N的一个阵列深度[]存储每个孔的容量。
• 使用变量i在范围[1，N]上进行迭代，并将hole [i]的初始深度[i]设置为i + 1 。
• 使用变量i遍历数组ball []并执行以下操作：
  • 以相反的顺序使用变量j遍历数组hole []。
    • 检查孔的直径是否大于或等于球的直径，即，孔[j]≥球[i] ，并且如果该孔未满，即深度[j]> 0 ，则将孔放置球在孔由位置[]阵列中追加J + 1和由1递减所述孔的深度和中断环路的进行。
  • 如果球不适合任何孔(已到达坡度的末端)，则将0附加到position []数组中。
• 完成上述步骤后，打印存储在数组position []中的值作为结果。

下面是上述方法的实现：

Python3

# Python program for the above approach
  
# Function to find and print the final
# position of balls
def ballPositionFinder(diameter_of_holes,
                       diameter_of_balls):
    
    max_hole_limit_counter = []
    position_value = []
      
    # Stores the positions of balls
    ball_positions = []
  
    # Determine the maximum balls a hole
    # can store and note the position
    # of holes in position_value
    for i in range(1, len(diameter_of_holes)+1):
        max_hole_limit_counter.append(i)
        position_value.append(i)
  
    # Iterate over all possible holes
    # for every ball released
    for i in range(0, len(diameter_of_balls)):
        for j in range(1, len(diameter_of_holes)+1):
  
            # Place ball in hole if it fits
            # in and if hole is not full
            if (diameter_of_holes[-j] >= diameter_of_balls[i]) and (
                    max_hole_limit_counter[-j] != 0):
                
                ball_positions.append(position_value[-j])
                max_hole_limit_counter[-j] -= 1
                break
  
            # If ball has reached at end B
            if j == len(diameter_of_holes):
                ball_positions.append(0)
                break
  
    return ball_positions
  
  
# Driver Code
if __name__ == "__main__":
    
    diameter_of_holes = [21, 3, 6]
      
    diameter_of_balls = [20, 15, 5, 7, 10, 4,
                         2, 1, 3, 6, 8]
  
    # Function Call
    output = ballPositionFinder(diameter_of_holes,
                                diameter_of_balls)
    print(*output, sep =' ')

输出：

1 0 3 0 0 3 3 2 2 0 0

时间复杂度： O(N * M)
辅助空间： O(N)