Python Program for Maximum height when coins are arranged in a triangle

We have N coins which need to arrange in form of a triangle, i.e. first row will have 1 coin, second row will have 2 coins and so on, we need to tell maximum height which we can achieve by using these N coins. Examples:

Input : N = 7
Output : 3
Explanation: Maximum height will be 3, putting 1, 2 and then 3 coins. It is not possible to use 1 coin left.

Input : N = 12
Output : 4
Explanation: Maximum height will be 4, putting 1, 2, 3 and 4 coins, it is not possible to make height as 5, because that will require 15 coins.

Brute-Force Approach:
A triangle takes coins in increasing order i.e., first line will have 1 coin, second line will have 2 coins and so on.
So we can subtract [1.2.3.,,,] till n is greater than the number, and increment our answer.

Below is the implementation of the above approach:

Python3

def triangle(input1):
    if input1 < 1:
        return 'Zero'
    else:
          # Counter for controlling the coins in a row
        a = 1
        height = 0
        while input1 > 0:
            if input1 - a >= 0:
                input1 = input1 - a
                a = a + 1
                height = height + 1
            else:
                break
        return height


if __name__ == "__main__":
    testinput1 = 22
    print(triangle(testinput1))

Output

6

Time complexity: O(sqrt(n))
Auxiliary space: O(1).

Efficient Approach: This problem can be solved by finding a relation between height of the triangle and number of coins. Let maximum height is H, then total sum of coin should be less than N, 

Sum of coins for height H <= N
            H*(H + 1)/2  <= N
        H*H + H – 2*N <= 0
Now by Quadratic formula 
(ignoring negative root)

Maximum H can be (-1 + √(1 + 8N)) / 2 

Now we just need to find the square root of (1 + 8N) for
which we can use Babylonian method of finding square root

Below code is implemented on above stated concept

Python3

# Python3 program to find
# maximum height of arranged
# coin triangle

# Returns the square root of n.
# Note that the function
def squareRoot(n):

    # We are using n itself as
        # initial approximation
    # This can definitely be improved
    x = n
    y = 1

    e = 0.000001  # e decides the accuracy level
    while (x - y > e):
        x = (x + y) / 2
        y = n/x

    return x


# Method to find maximum height
# of arrangement of coins
def findMaximumHeight(N):

    # calculating portion inside the square root
    n = 1 + 8*N
    maxH = (-1 + squareRoot(n)) / 2
    return int(maxH)


# Driver code to test above method
N = 12
print(findMaximumHeight(N))

# This code is contributed by
# Smitha Dinesh Semwal

Output

4

Time complexity: O(log N)
Auxiliary space: O(1)

Please refer complete article on Maximum height when coins are arranged in a triangle for more details!