PROBLEM LINK:

Practice
Contest

Author:Sunny Aggarwal
Tester:Sergey Kulik
Editorialist:Sunny Aggarwal

DIFFICULTY:

Cakewalk

PREREQUISITES:

Implementaion, Mathematics, Brute Force.

PROBLEM STATEMENT:

You are given a number $N$ that represents the number of coins you have and you have to print the maximum possible height $H$ of triangle which can be made from these coins in such a way that $i^{th}$ row of the triangle contains $i$ coins.

QUICK EXPLANATION:

Maximum possible height of the triangle is the maximum possible value of $H$ such that $\frac{H\times(H+1)}{2} \le N$.

EXPLANATION:

It should be noted that $i^{th}$ row of triangle contains $i$ coins. Therefore, a traingle with $H$ height will contain $1$ coin in $1^{st}$ row, 2 coins in $2^{nd}$ row, ... , $H$ coins in $H^{th}$ row and total number of coins required to build a triangle of height $H$ will be equal to

$$\sum_{\substack{1 \le i \le H}}{i} \quad = \quad \frac{H \times (H+1)}{2}$$

As we are asked to find the maximum possible height of the triangle, we have to find the maximum value of $H$ such that $\frac{H \times (H+1)}{2} \le N$. Note that the value of $N$ ~= $10^9$ in worst case and therefore the maximum possible value of $H$ will be order of $\sqrt{N}$.

We can use following simple procedure to solve this problem.

int sum(int h) {
    return (h * (h+1) / 2);
}
void solve() {
    int n;
    cin >> n;
    int h = 1;
    while(sum(h) <= n) {
        h ++;
    }
    cout << h - 1 << "\n";
}

TIME COMPLEXITY:

$O(\sqrt{N})$

BONUS:

Try solving same problem with $1 \le N \le 10^{15}$.

AUTHOR'S AND TESTER'S SOLUTION:

Author's solution can be found here.
Tester's solution can be found here.