Zenefits 一道面试题 | 书脊


Zenefits 一道面试题 | 书脊
A, B两个String
//example
A = XYZ;
A^2 = XXYYZZ;
A^3 = XXXYYYZZZ;

B = XXadhflakjhelXXzzqqkkpoYYadadfhgakheZafhajkefhlZadhflkejhZfagjhfebhh

A^2 是B的subsequence, 所以
return k = 2;

A可能有重复的char, B可能有其他字符, 求k.

我用的run-length encode做的, 先找下a的count, 然后找下b的count, 这里的技巧是把在b但是不在a的字符存成负数. 这样encode的时候, 我们只encode a和b中共同的字符. 然后走一下数字的最小值就好了, 时间o(n).
http://www.mitbbs.com/article_t/JobHunting/33043355.html
Test if A^k is in B takes simple O(n), as follows:

var test = function(A, k, B) {
    var i = 0;
    var next = A[i];
    var quota = k;

    for(j = 0; j < B.length; j++) {
        if (B[j] == next) {
            quota--;
        }
        if (quota == 0) {
            i++;
            if (i >= A.length) {
                return true;
            } else {
                next = A[i];
                quota = k;
            }
        }
    }
    return false;
};

Then apply binary search to find the largest k, as follows:

var searchK = function(A, B) {
    var bot = 1;
    var top = Math.floor(B.length / A.length);

    while (bot < top) {
        var mid = (bot + top) / 2;
        if (test(A, mid, B)) {
            bot = mid;
        } else {
            top = mid;
        }
    }

    return test(A, top, B);
};

This take O(n*log(k)), overall.

建立一个一维dp table,大小为A的size;
开始dp表都是0,遇到A中一个元素,查看dp table的对应value和dp[i-1]的value。
如果dp[i -1] 》 dp[i] ,dp[i]++;
都完事了dp[n - 1]就是结果。

-- don't think this works
    public int findK(String a, String b) {
        int[] count_a = new int[256];
        int[] count_b = new int[256];
        for (int i = 0; i < a.length(); i++) {
            count_a[a.charAt(i)]++;
        }
        for (int i = 0; i < b.length(); i++) {
            if (count_a[b.charAt(i)] == 0)
                count_b[b.charAt(i)] --;
            else
                count_b[b.charAt(i)] ++;
        }
        StringBuffer sb_b = new StringBuffer();
        for (int i = 0; i < b.length(); i++) {
            if (count_b[b.charAt(i)] > 0)
                sb_b.append(b.charAt(i));
        }
        String t_b = encode(sb_b.toString());
        int min = Integer.MAX_VALUE;
        for (int i = 0; i <t_b.length(); i++) {
            if (t_b.charAt(i) <= '9' && t_b.charAt(i) >= '0')
                min = Math.min(min, t_b.charAt(i) - '0');
        }
        return min;
    }
    public String encode(String s) {
        StringBuffer sb = new StringBuffer();
        for (int i = 0; i < s.length(); i++) {
            int counter = 1;
            while (i+1 < s.length() && s.charAt(i) == s.charAt(i+1)) {
                counter++;
                i++;
            }
            sb.append(counter);
            sb.append(s.charAt(i));
        }
        return sb.toString();
    }
    public static void main(String[] args) {
        String a = "XYX";
        String b = "XXadhflakjhelXXzzqqkkpoYadadfhgakheZafhajkefhlZadhflkejhZfagjhfebhhXX";
        System.out.println(new Zenefits().findK(a,b));
    }
https://instant.1point3acres.com/thread/133828
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