Saturday, November 28, 2015

largest K such that A^K is a subsequence of B - Zenefits


http://www.1point3acres.com/bbs/thread-145290-1-1.html
// Find the largest K such that A^K is a subsequence of B, where A, B are two strings. |A| << |B|.

// Power of a string:
// Let A = xyxz 鏉ユ簮涓€浜�.涓夊垎鍦拌鍧�. 
// Then,
// A^1 = A = xyxz
// A^2 = xxyyxxzz
// A^3 = xxxyyyxxxzzz
. 1point 3acres 璁哄潧
// Example:-google 1point3acres
// A = xyxz
// B = xabyzxz

// B xaybyxzxxxzz

// A xxyyxxzz
The key of the problem is to find out the largest possible k, that the A^k is the subsequence of B. 

How to find out the maximum K. The easiest way is k = lenB / lenA. But that could result to a false larger K. 

A better approach is to calculate the frequency of each character in A and B. Then calculate the MIN(B[i] / A[i]).
    public int findK(String A, String B) {
        int len1 = A.length(), len2 = B.length();
        if (len1 > len2) {
            return 0;
        }
         
        if (len1 <= 0) {
          return 0;
        }
         
        // remove space weird characters, capital to lower case etc if needed
        A = preprocess(A);
             
        // find signature for each A and B and get the max possible value for k
        int[] sigA = new int[256];
        int[] sigB = new int[256];
        for (int i = 0; i < len1; i++) {
            sigA[A.charAt(i)]++;
        }
        for (int i = 0; i < len2; i++) {
            sigB[B.charAt(i)]++;
        }
         
        int kMax = Integer.MAX_VALUE;
        for (int i = 0; i < len1; i++) {
            if (sigA[A.charAt(i)] != 0) {
                kMax = Math.min(sigB[A.charAt(i)] / sigA[A.charAt(i)], kMax);
            }
        }
         
        int lo = 0, hi = kMax;
        int ret = 0;
        while (lo <= hi) {
            int med = lo + (hi - lo) / 2;
            if (isSubSeq(expand(A, med), B)){
                ret = med;
                lo = med + 1;
            } else {
                hi = med - 1;
            }
        }
        return ret;
    }
   
    String expand(String A, int k){
        StringBuilder sbuf = new StringBuilder();
        for (int i = 0; i < A.length(); i++) {
            char c = A.charAt(i);
            for (int j = 0; j < k; j++) {
                sbuf.append(c);
            }
        }
        return sbuf.toString();
    }
         
    boolean isSubSeq(String A, String B) {
        int len1 = A.length(), len2 = B.length();
        int a = 0, b = 0;
        while (a < len1 && b < len2){
            if (A.charAt(a) == B.charAt(b)){
                a++;
                b++;
            } else {
                b++;
            }
        }
        return a == len1;
    }
         
    String preprocess(String A){
        return A.toLowerCase();
    }
  1. public int findK(String A, String B){
  2.             int len1 = A.length(), len2 = B.length();
  3.             if (len1 > len2) return 0;
  4.             if (len1 <= 0) return 0;
  5.             // remove space weird characters, capital to lower case etc if needed.
  6.             A = preprocess(A);
  7.             // find signature for each A and B and get the max possible value for k. 鐗涗汉浜戦泦,涓€浜╀笁鍒嗗湴
  8.             int[] sigA = new int[256];
  9.             int[] sigB = new int[256];. Waral 鍗氬鏈夋洿澶氭枃绔�,
  10.             for (int i = 0; i < len1; i++) sigA[A.charAt(i)]++;
  11.             for (int i = 0; i < len2; i++) sigB[B.charAt(i)]++;
  12.             int kMax = Integer.MAX_VALUE;. 鐣欏鐢宠璁哄潧-涓€浜╀笁鍒嗗湴
  13.             for (int i = 0; i < len1; i++){
  14.                 if (sigA[A.charAt(i)] != 0) kMax = Math.min(sigB[A.charAt(i)] / sigA[A.charAt(i)], kMax);
  15.             }
  16.             int lo = 0, hi = kMax;. 鍥磋鎴戜滑@1point 3 acres
  17.             int ret = 0;
  18.             while (lo <= hi){
  19.                 int med = lo + (hi - lo) / 2;
  20.                 if (isSubSeq(expand(A, med), B)){
  21.                     ret = med;
  22.                     lo = med + 1; 鏉ユ簮涓€浜�.涓夊垎鍦拌鍧�. 
  23.                 } else {
  24.                     hi = med - 1;
  25.                 }
  26.             }
  27.             return ret;
  28.         }
  29.         
  30.         String expand(String A, int k){
  31.             StringBuilder sbuf = new StringBuilder();
  32.             for (int i = 0; i < A.length(); i++){
  33.                 char c = A.charAt(i);
  34.                 for (int j = 0; j < k; j++) sbuf.append(c);
  35.             }. more info on 1point3acres.com
  36.             return sbuf.toString();
  37.         }
  38.         
  39.         boolean isSubSeq(String A, String B){
  40.             int len1 = A.length(), len2 = B.length();
  41.             int a = 0, b = 0;
  42.             while (a < len1 && b < len2){
  43.                 if (A.charAt(a) == B.charAt(b)){
  44.                     a++;
  45.                     b++;
  46.                 } else {
  47.                     b++;. 鐗涗汉浜戦泦,涓€浜╀笁鍒嗗湴
  48.                 }
  49.             }
  50.             return a == len1;. 鐗涗汉浜戦泦,涓€浜╀笁鍒嗗湴
  51.         }
  52.         
  53.         String preprocess(String A){
  54.             return A.toLowerCase();
  55.         }
http://www.1point3acres.com/bbs/forum.php?mod=viewthread&tid=143470&highlight=zenefit
一开始思路就是k从1一个个试,不行就返回。找subsequence用dp。但是面试官说复杂度太高,就想到用双指针遍历。然后问了时间复杂度。 写好后他还要improve.想到用二分查找k.因为最小是1,最大是B.length()/A.length();

subsequence是不需要连续的。如果x^(K+1)是y的subsequence,x^(K)一定是y的subsequence
  1.         public int findLargestK(String s1, String s2){.鏈枃鍘熷垱鑷�1point3acres璁哄潧
  2.                 if(s1.length() < 1 || s1.length() > s2.length())
  3.                         return -1;
  4.                 int k = 1;
  5.                 int left = 1, right = s2.length() / s1.length();
  6.                 int lastK = -1;
  7.                 while(left <= right){
  8.                         k = left + (right - left)/2;
  9.                         String akString = generateKth(s1, k);
  10.                         if(isSub(akString, s2)){
  11.                                 lastK = k;
  12.                                 left = k + 1;
  13.                         }else{. visit 1point3acres.com for more.
  14.                                 right = k - 1;
  15.                         }
  16.                 }
  17.                 return lastK;
  18.         }
  19.         
  20.         private String generateKth(String s, int k){
  21.                 StringBuilder temp = new StringBuilder();
  22.                 for(int i = 0; i < s.length(); i++){
  23.                         for(int j = 0; j < k; j++){. Waral 鍗氬鏈夋洿澶氭枃绔�,
  24.                                 temp.append(s.charAt(i));
  25.                         }
  26.                 }
  27.                 return temp.toString();
  28.         }. more info on 1point3acres.com
  29.         
  30.         private boolean isSub(String s1, String s2){
  31.                 int i = 0;
  32.                 int j = 0;
  33.                 while(i < s1.length() && j < s2.length()){
  34.                         if(s1.charAt(i) == s2.charAt(j)){. 1point 3acres 璁哄潧
  35.                                 i++;
  36.                         }
  37.                         j++;-google 1point3acres
  38.                 }
  39.                 if(i == s1.length())
  40.                         return true;
  41.                 return false;
  42.         }


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