Sunday, October 30, 2016

LeetCode 441 - Arranging Coins


http://bookshadow.com/weblog/2016/10/30/leetcode-arranging-coins/
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.
https://discuss.leetcode.com/topic/65631/c-three-solutions-o-n-o-logn-o-1
The author means O(k), O(1), O(log(k)), where k is the number of staircases. In fact k is of order O(log(n)),

    int way1(int n) {
        int level = 1;
        for (long sum = 0; sum <= n; level++) 
            sum += level;
        return max(level - 2, 0);    
    }
    
    int way2(int n) {
        return sqrt(2 * (long)n + 1 / 4.0) - 1 / 2.0;
    }
    
    int arrangeCoins(int n) {
        long low = 1, high = n;
        while (low < high) {
            long mid = low + (high - low + 1) / 2;
            if ((mid + 1) * mid / 2.0 <= n) low = mid;
            else high = mid - 1;
        }
        return high;
    }

解法I 解一元二次方程(初等数学):
x ^ 2 + x = 2 * n
解得:
x = sqrt(2 * n + 1/4) - 1/2
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+...+xn
i=1xin
Running sum can be simplified,
x(x+1)2n
Using quadratic formula, x is evaluated to be,
x=12(-8n+1-1) (Inapplicable) or x=12(8n+1-1)

http://blog.csdn.net/mebiuw/article/details/52981579
public int arrangeCoins(int n) { /* 数学推导 (1+k)*k/2 = n k+k*k = 2*n k*k + k + 0.25 = 2*n + 0.25 (k + 0.5) ^ 2 = 2*n +0.25 k + 0.5 = sqrt(2*n + 0.25) k = sqrt(2*n + 0.25) - 0.5 */ return (int) (Math.sqrt(2*(long)n+0.25) - 0.5); }
https://discuss.leetcode.com/topic/65575/java-o-1-solution-math-problem
http://blog.csdn.net/cloudox_/article/details/53005388
The idea is about quadratic equation, the formula to get the sum of arithmetic progression is
sum = (x + 1) * x / 2
so for this problem, if we know the the sum, then we can know the x = (-1 + sqrt(8 * n + 1)) / 2
(1 + X) * X = 2n
4X * X + 4 * X = 8n
(2X + 1)(2X + 1) - 1 = 8n
X = (Math.sqrt(8 * n + 1) - 1)/ 2
I just had to add the inequalities to completely convince me:
If x is the answer, then n coins do fill x rows but don't fill x+1 rows. So we have x(x+1)/2 ≤ n < (x+1)((x+1)+1)/2. Which, using the other formula, turns into x ≤ (sqrt(8n+1) - 1) / 2 < x+1. So we indeed get the right answer by rounding down (sqrt(8n+1) - 1) / 2.
    public int arrangeCoins(int n) {
        return (int)((-1 + Math.sqrt(1 + 8 * (long)n)) / 2);
    }
https://discuss.leetcode.com/topic/65593/java-clean-code-with-explanations-and-running-time-2-solutions
解法II 二分枚举答案(Binary Search):
等差数列前m项和:m * (m + 1) / 2
在上下界l, r = [0, n]范围内二分枚举答案
X.  https://ratchapong.com/algorithm-practice/leetcode/arranging-coins
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+...+xn
i=1xin
Running sum can be simplified,
x(x+1)2n
Binary search is used in this case to slowly narrow down the x that will satisfy the equation. Note that 0.5 * mid * mid + 0.5 * mid does not have overflow issue as the intermediate result is implicitly autoboxed to double data type.

    public int arrangeCoins(int n) {
        int start = 0;
        int end = n;
        int mid = 0;
        while (start <= end) {
            mid = (start + end) >>> 1;
            if ((0.5 * mid * mid + 0.5 * mid) <= n) {
                start = mid + 1;
            } else {
                end = mid - 1;
            }
        }
        return start - 1;
    }

O(k)
http://www.cnblogs.com/grandyang/p/6026066.html
这道题给了我们n个硬币,让我们按一定规律排列,第一行放1个,第二行放2个,以此类推,问我们有多少行能放满。通过分析题目中的例子可以得知最后一行只有两种情况,放满和没放满。由于是按等差数列排放的,我们可以快速计算出前i行的硬币总数。我们先来看一种O(n)的方法,非常简单粗暴,就是从第一行开始,一行一行的从n中减去,如果此时剩余的硬币没法满足下一行需要的硬币数了,我们之间返回当前行数即可
    int arrangeCoins(int n) {
        int cur = 1, rem = n - 1;
        while (rem >= cur + 1) {
            ++cur;
            rem -= cur;
        }
        return n == 0 ? 0 : cur;
    }


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