Thursday, July 27, 2017

LeetCode 483 - Smallest Good Base


https://leetcode.com/problems/smallest-good-base/
For an integer n, we call k>=2 a good base of n, if all digits of n base k are 1.
Now given a string representing n, you should return the smallest good base of n in string format. 
Example 1:
Input: "13"
Output: "3"
Explanation: 13 base 3 is 111.
Example 2:
Input: "4681"
Output: "8"
Explanation: 4681 base 8 is 11111.
Example 3:
Input: "1000000000000000000"
Output: "999999999999999999"
Explanation: 1000000000000000000 base 999999999999999999 is 11.
Note:
  1. The range of n is [3, 10^18].
  2. The string representing n is always valid and will not have leading zeros.
X.
http://blog.csdn.net/hanzheng992/article/details/54879692
Naive Solution是通过遍历base来搜索整个解空间的, 除此之外, 我们也可以通过遍历转换后1的位数来遍历搜索整个解空间, 这样搜索的范围会小很多.
我们假设在goodBase进制下, 最终得到的数是11...1, 其中有k个1. 那么k的取值范围就是[2, log(n, 2)]. 然后我们用二分查找的方式来判断是否存在这样一个整数base, 使得n通过进制转换后得到一个由k1组成的数.
def smallestGoodBase(self, n): """ :type n: str :rtype: str """ def getAnsofBase(length, base): """ Convert 11...11 (base `base`) to base 10 """ ans = 1 for i in xrange(length-1): ans = ans * base + 1 return ans def findLengthBase(length, n): """ Check whether there exist a base such that n in base `base` == 111...111 (length's 1s) """ start, end = 0, n/2 while start <= end: mid = (start + end) / 2 target = getAnsofBase(length, mid) if target == n: return mid elif target < n: start = mid + 1 else: end = mid - 1 return -1 num = int(n) thisLen = int(math.log(num,2)) + 1 while thisLen > 2: retVal = findLengthBase(thisLen, num) if retVal != -1: return str(retVal) thisLen -= 1 return str(num - 1)
https://discuss.leetcode.com/topic/82130/java-solution-with-hand-writing-explain
https://discuss.leetcode.com/topic/76406/java-c-binary-search-solutions-with-detailed-explanation
The java solution is submitted by lixx2100 to contest.
  1. n is equal to x^(k-1) + x^(k-2) + ... + x + 1, where k is from 2 to, say, 66
  2. for each k from 2 to 66, we get "x", and the minimum of all "x"s is the answer
  3. To get "x", we use binary search approach with left = 2, and right = Long.MAX_VALUE
  4. to compare whether n is equal to x^(k-1) + x^(k-2) + ... + x + 1, we don't need to calculate x^(k-1) + x^(k-2) + ... + x + 1 from x.
    if n = x^(k-1) + x^(k-2) + ... + x + 1, then n * (x - 1) = x^k -1.
    in the source code, cb is x^k -1 and wb is n * (x - 1).
public String smallestGoodBase(String nn) {
  long n = Long.parseLong(nn);
  long res = 0;
  for(int k = 60; k >= 2; k--){
    long s = 2, e = n;
    while(s < e){
        long m = s + (e - s) / 2;   
        
        BigInteger left = BigInteger.valueOf(m);
        left = left.pow(k).subtract(BigInteger.ONE);
        BigInteger right = BigInteger.valueOf(n).multiply(BigInteger.valueOf(m).subtract(BigInteger.ONE));
        int cmr = left.compareTo(right);
        if(cmr == 0){
            res =  m;
            break;
        } else if(cmr < 0){
            s = m + 1;
        } else {
            e = m;
        }
    }
    
    if(res != 0) break;
  }
  
  return "" + res;
}
https://discuss.leetcode.com/topic/76357/java-binary-search-solution-9-ms
    public String smallestGoodBase(String n) {
        long num = 0;
        for (char c : n.toCharArray()) num = num * 10 + c - '0';
        
        long x = 1;
        for (int p = 64; p >= 1; p--) {
            if ((x << p) < num) {
                long k = helper(num, p);
                if (k != -1) return String.valueOf(k);
            }
        }
        return String.valueOf(num - 1);
    }
    
    private long helper(long num, int p) {
        long l = 1, r = (long)(Math.pow(num, 1.0/p) + 1);
        while (l < r) {
            long mid = l + (r - l) / 2;
            long sum = 0, cur = 1;
            for (int i = 0; i <= p; i++) {
                sum += cur;
                cur *= mid;
            }
            if (sum == num) return mid;
            else if (sum > num) r = mid;
            else l = mid + 1;
        }
        return -1;
    }
http://bookshadow.com/weblog/2017/01/22/leetcode-smallest-good-base/
记k的最高次幂为m,从上界 int(log(n)) 向下界 1 递减枚举m

问题转化为计算1 + k + k^2 + ... + k^m = n的正整数解

由n > k^m得: k < n ** 1/m

由n < (k + 1)^m得: k > n ** 1/m - 1,此处使用了二项式定理

因此k可能的解为:int(n ** 1/m)

最后验证1 + k + k^2 + ... + k^m 是否等于 n
http://blog.csdn.net/guoyuhaoaaa/article/details/54782315
这道题刚开始看的时候毫无头绪,感觉唯一合适的思路就是穷举了,从base为2开始一直开始往前试(因为是找最小的基,因此要从小到大的开始试),试到的那个数如果满足条件那么我们就找到了答案。这样的方法显然很笨,绝对的超时。后来看了答案,才恍然大悟,有时候换个角度去思考问题,往往能得到不错的结果。 
题目中已经说了,要找到目标数aim的全‘1’表示的base。那么也就是说无论如何最后该aim的表示形式一定是全‘1’的,如果是最小的base那么也就意味着最长的‘1’串。我们可以固定一个‘1’串来找是否有合适的base可以满足,而在固定‘1’串找base的时候可以使用二分查找来节省时间。如果当前的‘1’串都不合适,那么我们就可以减少‘1’串的长度,知道找到为止,这时候‘1’串对应的base一定是最小的base。由于aim的范围是在[3, 10^18],那么‘1’串最长无非是base为2的时候(这里的放缩其实可以适当的放大范围,并不影响整体的复杂度)。

http://www.gegugu.com/2017/03/14/1356.html


X. http://blog.csdn.net/hanzheng992/article/details/54879692
根据题目意思, 很容易写出checkBase(base,n)这个函数, 对于一个给定的baes判断该base是否为给定数n的good base, 并且这个判断函数的时间复杂度为O(log N). 搜索的空间自然为0n - 1. 那么, 我们就能写出一个时间复杂度为O(N log N)算法.
然而题目给定n的范围为[3, 10^18], 即使是O(N log N)的方法我们也无法接受.
def smallestGoodBase(self, n): """ :type n: str :rtype: str """ def checkBase(base, n): """ Given a base, check whether it is a good base. Time complexity is O(log N) """ current = 1 while current < n: current = current * base + 1 return current == n thisNum = int(n) for i in xrange(2, thisNum): if checkBase(i, thisNum): return str(i) return str(thisNum - 1)

X.
http://blog.csdn.net/hanzheng992/article/details/54879692
假设base是我们最终需要求的good base, k为进制转换后1的个数, 那么, 我们可以得到如下等式:
base^(k-1) + base^(k-2) + … + base^1 + base^0 = N … [1]
base^k + base^(k-1) + … + base^2 + base^1 = N * base
因此, 我们可以得到:
base^k - base^0 = (base - 1) * N
N = (base^k - 1) / (base - 1) …. [2]
[1], 可以得:
base ^ (k-1) < N < (base+1) ^ (k-1) … 多项式展开可证右边的不等号
base < (k-1)-th root of N < base + 1 … [3]
根据[2][3], 我们可以通过遍历位数的方法得到最终答案
def smallestGoodBase(self, n): """ :type n: str :rtype: str """ num = int(n) thisLen = int(math.log(num,2)) + 1 while thisLen > 2: # from equation [3], we havve thisBase = int(num ** (1.0/(thisLen - 1))) # from equation [2], we have if num * (thisBase - 1) == thisBase ** thisLen - 1: return str(thisBase) thisLen -= 1 return str(num - 1)



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