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Maximum height when coins are arranged in a triangle
• Difficulty Level : Basic
• Last Updated : 31 Mar, 2021

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
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,

## CPP

 `//  C++ program to find maximum height of arranged``// coin triangle``#include ``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;``}`

## Java

 `// Java program to find maximum height``// of arranged coin triangle``class` `GFG``{``    ` `    ``/* Returns the square root of n.``    ``Note that the function */``    ``static` `float` `squareRoot(``float` `n)``    ``{``        ` `        ``/* We are using n itself as``        ``initial approximation.This``        ``can definitely be improved */``        ``float` `x = n;``        ``float` `y = ``1``;``        ` `        ``// e decides the accuracy level``        ``float` `e = ``0``.000001f;``        ``while` `(x - y > e)``        ``{``            ``x = (x + y) / ``2``;``            ``y = n / x;``        ``}``        ` `        ``return` `x;``    ``}``    ` `    ``// Method to find maximum height``    ``// of arrangement of coins``    ``static` `int` `findMaximumHeight(``int` `N)``    ``{``        ` `        ``// calculating portion inside``        ``// the square root``        ``int` `n = ``1` `+ ``8``*N;``        ``int` `maxH = (``int``)(-``1` `+ squareRoot(n)) / ``2``;``        ` `        ``return` `maxH;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `N = ``12``;``        ` `        ``System.out.print(findMaximumHeight(N));``    ``}``}` `// This code is contributed by Anant Agarwal.`

## 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`

## C#

 `// C# program to find maximum height``// of arranged coin triangle``using` `System;` `class` `GFG``{``    ``/* Returns the square root of n.``    ``Note that the function */``    ``static` `float` `squareRoot(``float` `n)``    ``{``        ``/* We are using n itself as``        ``initial approximation.This``        ``can definitely be improved */``        ``float` `x = n;``        ``float` `y = 1;` `        ``// e decides the accuracy level``        ``float` `e = 0.000001f;``        ``while` `(x - y > e)``        ``{``            ``x = (x + y) / 2;``            ``y = n / x;``        ``}``        ``return` `x;``    ``}``    ` `    ``static` `int` `findMaximumHeight(``int` `N)``    ``{` `        ``// calculating portion inside``        ``// the square root``        ``int` `n = 1 + 8*N;``        ``int` `maxH = (``int``)(-1 + squareRoot(n)) / 2;` `        ``return` `maxH;``    ``}` `    ``/* program to test above function */``    ``public` `static` `void` `Main()``    ``{``        ``int` `N = 12;``        ``Console.Write(findMaximumHeight(N));``    ``}``}` `// This code is contributed by _omg`

## PHP

 ` ``\$e``)``    ``{``        ``\$x` `= (``\$x` `+ ``\$y``) / 2;``        ``\$y` `= ``\$n``/``\$x``;``    ``}``    ``return` `\$x``;``}` `// Method to find maximum height of``// arrangement of coins``function` `findMaximumHeight( ``\$N``)``{``    ` `    ``// calculating portion inside``    ``// the square root``    ``\$n` `= 1 + 8 * ``\$N``;``    ``\$maxH` `= (-1 + squareRoot(``\$n``)) / 2;``    ``return` `floor``(``\$maxH``);``}` `// Driver code to test above method``\$N` `= 12;``echo` `findMaximumHeight(``\$N``) ;` `// This code is contributed by anuj_67.``?>`

## Javascript

 ``

Output:

`4`

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