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.
public class Solution {
public int ArrangeCoins(int n) {
return (int) ((Math.Sqrt(1 + 8.0 * n) - 1) / 2);
}
}Time Complexity: O(1)
Space Complexity: O(1)
Complexity Analysis Uniform cost model is used as Cost Model and n is the input number. b in this case would be 2.
Time Complexity:
Best Case O(1) : With respect to the input, the algorithm will always perform basic mathematical operation that run in constant time. Average Case O(1) : With respect to the input, the algorithm will always perform basic mathematical operation that run in constant time. Worst Case O(1) : With respect to the input, the algorithm will always perform basic mathematical operation that run in constant time. Auxiliary Space:
Worst Case O(1) : No extra space is used. Algorithm Approach: Mathematics
The problem is basically asking the maximum length of consecutive number that has the running sum lesser or equal to n. In other word, find x that satisfy the following condition:
1 + 2 + 3 + 4 + 5 + 6 + 7 + ... + x <= n sum_{i=1}^x i <= n Running sum can be simplified,
(x * ( x + 1)) / 2 <= n Using quadratic formula, x is evaluated to be,
x = 1 / 2 * (-sqrt(8 * n + 1)-1) (Inapplicable) or x = 1 / 2 * (sqrt(8 * n + 1)-1) Negative root is ignored and positive root is used instead. Note that 8.0 * n is very important because it will cause Java to implicitly autoboxed the intermediate result into double data type. The code will not work if it is simply 8 * n. Alternatively, an explicit casting can be done 8 * (long) n).
