You have a total of n coins that you want to form in a staircase shape, where every k-th row must have exactly k coins.
Given n, find the total number of full staircase rows that can be formed.
n is a non-negative integer and fits within the range of a 32-bit signed integer.
Example 1:
n = 5 The coins can form the following rows: ¤ ¤ ¤ ¤ ¤ Because the 3rd row is incomplete, we return 2.
Example 2:
n = 8 The coins can form the following rows: ¤ ¤ ¤ ¤ ¤ ¤ ¤ ¤ Because the 4th row is incomplete, we return 3.
C++: Time: O(n), Space: O(1)
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;
}
};
C++: Time: O(logn), Space: O(1)
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;
}
};
C++: Time: O(logn), Space: O(1) . n = (1 + x) * x / 2, x = (-1 + sqrt(8 * n + 1)) / 2
class Solution {
public:
int arrangeCoins(int n) {
return (int)((-1 + sqrt(1 + 8 * (long)n)) / 2); # sqrt is O(logn) time.
}
};
Python: O(logn), Space: O(1)
class Solution2(object):
def arrangeCoins(self, n):
"""
:type n: int
:rtype: int
"""
left, right = 1, n
while left <= right:
mid = left + (right - left) / 2
if 2 * n < mid * (mid+1):
right = mid - 1
else:
left = mid + 1
return left - 1
Python: O(logn), Space: O(1)
class Solution(object):
def arrangeCoins(self, n):
"""
:type n: int
:rtype: int
"""
return int((math.sqrt(8*n+1)-1) / 2) # sqrt is O(logn) time.