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:

$\begin{aligned} k*(k + 1) / 2 &<= n\\ k^2 + k &<= 2*n\\ k^2 + k - 2*n &<=0 \end{aligned}$We can use Sridharacarya’s formula to solve this equation:

$k = \frac{-1 + \sqrt{1 + 8n}}{2}$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: $O(1)$