Sunday, November 29, 2015

Facebook interview - Suffix Array Order - 我的博客 - ITeye技术网站


Facebook interview - Suffix Array Order - 我的博客 - ITeye技术网站
首先定义了suffix string 或者说suffix arrary
如果有个数组是 int[] text = {10, 20, 30, 25}
那么 suffix[0] = {10, 20, 30, 25}
           suffix[1] = {20, 30, 25}
           suffix[2] = {30, 25}
           suffix[3] = {25}
如果对这些数组进行lexical order 的排序,我们可以得到
suffix[0] < suffix[1] < suffix[3] < suffix[2]
问题是:
    input: int[] text
    output: int[] suffix_array_order
e.g.
input: int[] text = {10, 20, 30, 25}
output: int[] suffix_array_order = {0, 1, 3, 2}

Solution:
注意: Arrays.sort(T[], Comparator)只能用于排序对象数组,不能用于排序int, long之类的原始值数组!
  1. public Integer[] suffixArrayOrder(int[] A) {  
  2.     int n = A.length;  
  3.     Integer[] order = new Integer[n];  
  4.     for(int i=0; i<n; i++) {  
  5.         order[i] = i;  
  6.     }  
  7.       
  8.     Arrays.sort(order, new Comparator<Integer>(){  
  9.         @Override  
  10.         public int compare(Integer o1, Integer o2) {  
  11.             int res = A[o1] - A[o2];  
  12.             while(res == 0 && o1 < n && o2 < n) {  
  13.                 res = A[o1++] - A[o2++];  
  14.             }  
  15.             return res;  
  16.         }  
  17.     });  
  18.       
  19.     System.out.println(Arrays.toString(order));  
  20.     return order;  
  21. }

第二题: input:  int[] text, int[] subText
              output: boolean isExist;
检查text数组中有没有一个subarray 是subText。要求时间小于O(N^2), N == text.length;
这里假设我们有了第一题的 suffix_array_order.
(做法是binary search)

http://www.geeksforgeeks.org/suffix-array-set-1-introduction/

就是相当于把所有的substring 排序,然后binary search
You don't need to create an LIst<List> for all the suffix arrays since that will take extra space and time to create. What I did is to create an wrapper class:
Class Index{
    public int index;
    public int[] array;
}
Just pass the reference of whole input array to the wrapper class object.
I answered O(n log n), n is the length of the array at the interview and passed.

After that, I think if we consider the worse case of comparing two string, the worst case time should be O(n^2 log n).
  1. public class Index {
  2.     int[] suff;
  3.     int idx;
  4.     Index(int[] _suff, int _idx) {
  5.         suff = _suff;
  6.         idx = _idx;
  7.     } 
  8. }

  9. //Sort
  10. int[] suffixOrder(int[] text, int n) {
  11.     int[] res = new int[n];
  12.     Index[] indices = new Index[n];
  13.     for (int i=0; i < n; i++) {
  14.         int[] suffix = Arrays.copyOfRange(text, i, text.length);
  15.         indices[i] = new Index(suffix, i);
  16.     }

  17.     Arrays.sort(indices, new Comparator<Index>() {
  18.         public int compare(Index i1, Index i2) {
  19.             int p1 = 0;
  20.             int p2 = 0;
  21.             while (p1 < i1.suff.length && p2 < i2.suff.length) {
  22.                 if(i1.suff[p1] != i2.suff[p2]) {
  23.                     return i1.suff[p1] - i2.suff[p2];
  24.                 } else {
  25.                     p1 ++;
  26.                     p2 ++;
  27.                 }
  28.             }
  29.             if (p2 == i2.suff.length) {
  30.                 return 1;
  31.             }
  32.             return -1;
  33.         }
  34.     });

  35.     for (int i = 0; i < n; i++) {
  36.         res[i] = indices[i].idx;
  37.     }
  38.     return res;
  39. }

  40. // Search
  41. boolean search(int[] text, int n, int[] sub_text, int m) {
  42.     int[] suffix_order = suffixOrder(text, n);
  43.     Comparator<Index> textCompare = new Comparator<Index>(){
  44.         @Override
  45.         public int compare(Index i1, Index i2){
  46.             while (a < text.length && b < sub_text.length) {
  47.                 if (text[a] != sub_text[b]) {
  48.                     return text[a] - sub_text[b];
  49.                 } else {
  50.                     a ++;
  51.                     b ++;.
  52.                 }
  53.             }
  54.             if(a == text.length) return -1;
  55.             return 0;.
  56.         }
  57.     };

  58.     int lo = 0;
  59.     int hi = n - 1;
  60.     while (lo <= hi) {
  61.     int mid = (lo + hi) / 2;
  62.     int[] suffix = Arrays.copyOfRange(text, suffix_order[mid], text.length);
  63.     int cmp = textCompare(suffix, sub_text); 
  64.     if (cmp == 0) {
  65.         return true;
  66.     } else if(cmp < 0) {
  67.         lo = mid + 1;
  68.     } else {}
  69.         hi = mid - 1;
  70.     }
  71.     return false;
  72. }



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