## DEV Community

Ayan Banerjee

Posted on • Originally published at notescs.gitlab.io on

# Maximum Height When Coins are Arranged in Staircase Fashion - A GoDaddy Interview Question

Given a total of `n` coins find the total number of full staircase rows that can be built.

Staircase rows are those where `i`-th row has `i` coins.

For example, given n = 6, return 3 as you can form staircase like this:

``````*
* *
* * *
``````

Given n = 9, return 3.

``````*
* *
* * *
* * *
``````

Note that, the 4th row is invalid as it doesn't have 4 coins.

## Approach - 1: Binary Search

To build a staircase till k-th row, we need:

1 + 2 + 3 + ... + k = k*(k + 1) / 2 coins.

So we need to find maximum k such that, k*(k + 1) / 2 <= n.

Since n is an increasing function of k, we can use binary search to solve this problem.

We initialize `low` and `high` as `0` and `n` respectively. In each step, we calculate
the value of coins required using the formula n = k*(k + 1) / 2 where `k` is the
middle element between `low` and `high`. If the required coins are greater than
`n` the value of `high` is updated to `k - 1` and if its less than `n`, the value
of `low` is updated to `k + 1`. Since we reduce the search space by half at each
iteration, the time complexity is `O(logN)`, where N is the number of coins.

C++ code:

``````#include <iostream>
using namespace std;

int arrangeCoins(int n) {
long low = 0, high = n;
while (low <= high) {
long k = low + (high - low) / 2;
long cur = k * (k + 1) / 2;

if (cur == n) return (int)k;

if (n < cur) {
high = k - 1;
} else {
low = k + 1;
}
}
return (int)high;
}

int main() {
cout << 6 << " " << arrangeCoins(6) << endl;
cout << 9 << " " << arrangeCoins(9) << endl;
}
``````

Python code:

``````def arrangeCoins(n):
low = 0
high = n
while low <= high:
k = low + (high - low) // 2
cur = k * (k + 1) // 2

if cur == n: return k

if n < cur:
high = k - 1
else:
low = k + 1
return high

if __name__ == '__main__':
print(6, arrangeCoins(6))  # n = 6, prints 3
print(9, arrangeCoins(9))  # n = 9, prints 3
``````

Time Complexity: `O(logN)` due to binary search

Space Complexity: `O(1)`

## Approach - 2: Math

We have formulated the equation:

``````k*(k + 1) / 2 <= n
k^2 + k <= 2*n
k^2 + k - 2*n <=0
``````

We can use Sridharacarya's formula to solve this equation:

``````k = (-1 + sqrt(1 + 8n))/2
``````

C++ code:

``````#include <iostream>
#include <cmath>
using namespace std;

int arrangeCoins(int n) {
int(-1 + sqrt(1 + (long)8 * n)) / 2;
}

int main() {
cout << 6 << " " << arrangeCoins(6) << endl;
cout << 9 << " " << arrangeCoins(9) << endl;
}
``````

Python code:

``````def arrangeCoins(n):
return int((-1 + ((1 + 8 * n) ** 0.5)) / 2)

if __name__ == '__main__':
print(6, arrangeCoins(6))  # n = 6, prints 3
print(9, arrangeCoins(9))  # n = 9, prints 3
``````

Time Complexity: `O(1)`

Space Complexity: `O(1)` 