n = 5

The coins can form the following rows:
¤
¤ ¤
¤ ¤

Because the 3rd row is incomplete, we return 2.

n = 8

The coins can form the following rows:
¤
¤ ¤
¤ ¤ ¤
¤ ¤

Because the 4th row is incomplete, we return 3.

class Solution {
public:
    int arrangeCoins(int n) {
        int cur = 1, rem = n - 1;
        while (rem >= cur + 1) {
            ++cur;
            rem -= cur;
        }
        return n == 0 ? 0 : cur;
    }
};

class Solution {
public:
    int arrangeCoins(int n) {
        if (n <= 1) return n;
        long low = 1, high = n;
        while (low < high) {
            long mid = low + (high - low) / 2;
            if (mid * (mid + 1) / 2 <= n) low = mid + 1;
            else high = mid;
        }
        return low - 1;
    }
};

class Solution {
public:
    int arrangeCoins(int n) {
        return (int)((-1 + sqrt(1 + 8 * (long)n)) / 2);
    }
};