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TCS Codevita | Holes And Balls
  • Difficulty Level : Medium
  • Last Updated : 17 Nov, 2020

Given two arrays of H[] and B[] consisting of N and M integers respectively, denoting the diameter of holes and balls respectively. M number of balls are made to roll from A to B on a sloping surface with N holes, each having different depth as shown in the figure below:

The task is to find the eventual position of each ball in the order of the ball released considering the following:

  • A ball will fall into a hole if its diameter is less than or equal to the diameter of the hole.
  • A hole Hi will become full if i numbers of balls fall into it.
  • If a hole is full, then no more balls fall into it.
  • A ball will reach B from A, if and only if it is not falling into any one of the holes.
  • If a ball is in hole Pi, then its position is i. If a ball reached the bottom point B, then take its position as 0.


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
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

Approach: Follow the steps below to solve the problem:

  • Initialize an array position[] of size N to store the final position of each ball and an array depth[] of size N to store the capacity of each hole.
  • Iterate over the range [1,  N] using the variable i and set the initial depth[i] of the hole[i] to i+1.
  • Traverse the array ball[] using the variable i and do the following:
    • Iterate over the array hole[] using variable j in reverse order.
      • Check if the diameter of the hole is greater than or equal to that of the ball, i.e., hole[j] ≥ ball[i], and if that hole is not full, i.e., depth[j] > 0 then, place the ball in that hole by appending j + 1 in the position[] array and decrementing the depth of the hole by 1 and break out of the loop.
    • If the ball doesn’t fit in any hole (has reached at end of the slope), then append 0 in the position[] array.
  • After the above steps, print the value stored in the array position[] as the result.

Below is the implementation of the above approach:






# Python program for the above approach
# Function to find and print the final
# position of balls
def ballPositionFinder(diameter_of_holes,
    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):
    # 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):
                max_hole_limit_counter[-j] -= 1
            # If ball has reached at end B
            if j == len(diameter_of_holes):
    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,
    print(*output, sep =' ')



1 0 3 0 0 3 3 2 2 0 0

Time Complexity: O(N*M)
Auxiliary Space: O(N)

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