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.


Input : N = 7
Output : 3
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
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.

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,

//  C++ program to find maximum height of arranged
// coin triangle
#include <bits/stdc++.h>
using namespace std;

/* Returns the square root of n. Note that the function */
float squareRoot(float n)
    /* We are using n itself as initial approximation
      This can definitely be improved */
    float x = n;
    float y = 1;

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

//  Driver code to test above method
int main()
    int N = 12;
    cout << findMaximumHeight(N) << endl;
    return 0;



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