Thursday, June 16, 2016

LeetCode 358 - Rearrange String k Distance Apart


http://blog.csdn.net/jmspan/article/details/51678257
Given a non-empty string str and an integer k, rearrange the string such that the same characters are at least distance k from each other.
All input strings are given in lowercase letters. If it is not possible to rearrange the string, return an empty string "".
Example 1:
str = "aabbcc", k = 3

Result: "abcabc"

The same letters are at least distance 3 from each other.
Example 2:
str = "aaabc", k = 3 

Answer: ""

It is not possible to rearrange the string.
Example 3:
str = "aaadbbcc", k = 2

Answer: "abacabcd"

Another possible answer is: "abcabcda"

The same letters are at least distance 2 from each other.

X. https://discuss.leetcode.com/topic/49022/greedy-solution-beats-95
for example: "aaabbcc", k = 3
  1. Count the statistics of letters, sort them in terms of frequency in a descending way.
so it has the result: a - 3, b - 2, c - 2.
  1. Suppose the rewrite string length is len, divide the len into bins of size k, so in total
    you have
bin number of nBin = (len - 1) / k + 1,
with last bin size:
lastBinSize = len % k.
in the example, nBin = 3, lastBinSize = 1;
  1. Fill the same letter in different bins:
after filling 'a' ---> result = a##a##a
after filling 'b' ---> result = ab#ab#a
after filling 'c' ---> result = abcabca

http://dartmooryao.blogspot.com/2016/06/leetcode-358-rearrange-string-k.html
    public String rearrangeString(String str, int k) {
        if(k<=1){ return str; }
        int[] count = new int[26];
        for(int i=0; i<str.length(); i++){
            count[str.charAt(i)-'a']++;
        }
        PriorityQueue<int[]> pq = new PriorityQueue<>((a,b)->b[0]-a[0]);
        for(int i=0; i<count.length; i++){ pq.add(new int[]{ count[i], i}); }
       
        char[] result = new char[str.length()];
        int idx = 0;
        int start = 0;
        while(!pq.isEmpty()){
            int[] num = pq.remove();
            for(int i=0; i<num[0]; i++){
                result[idx] = (char)(num[1]+'a');
                if(idx>0 && result[idx-1]==result[idx]){ return ""; }
                idx+=k;
                if(idx>=str.length()){ idx=++start; }
            }
        }
        return new String(result);
    }
https://leetcode.com/discuss/108232/java_solution_in_12_ms-o-n-time-and-space
方法:根据出现频率将字母从大到小排列,以k为间隔进行重排。
https://leetcode.com/discuss/108174/c-unordered_map-priority_queue-solution-using-cache
http://www.cnblogs.com/grandyang/p/5586009.html
这道题给了我们一个字符串str,和一个整数k,让我们对字符串str重新排序,使得其中相同的字符之间的距离不小于k,这道题的难度标为Hard,看来不是省油的灯。的确,这道题的解法用到了哈希表,堆,和贪婪算法。这道题我最开始想的算法没有通过OJ的大集合超时了,下面的方法是参考网上大神的解法,发现十分的巧妙。我们需要一个哈希表来建立字符和其出现次数之间的映射,然后需要一个堆来保存这每一堆映射,按照出现次数来排序。然后如果堆不为空我们就开始循环,我们找出k和str长度之间的较小值,然后从0遍历到这个较小值,对于每个遍历到的值,如果此时堆为空了,说明此位置没法填入字符了,返回空字符串,否则我们从堆顶取出一对映射,然后把字母加入结果res中,此时映射的个数减1,如果减1后的个数仍大于0,则我们将此映射加入临时集合v中,同时str的个数len减1,遍历完一次,我们把临时集合中的映射对由加入堆
    string rearrangeString(string str, int k) {
        if (k == 0) return str;
        string res;
        int len = (int)str.size();
        unordered_map<char, int> m;
        priority_queue<pair<int, char>> q;
        for (auto a : str) ++m[a];
        for (auto it = m.begin(); it != m.end(); ++it) {
            q.push({it->second, it->first});
        }
        while (!q.empty()) {
            vector<pair<int, int>> v;
            int cnt = min(k, len);
            for (int i = 0; i < cnt; ++i) {
                if (q.empty()) return "";
                auto t = q.top(); q.pop();
                res.push_back(t.second);
                if (--t.first > 0) v.push_back(t);
                --len;
            }
            for (auto a : v) q.push(a);
        }
        return res;
    }
https://discuss.leetcode.com/topic/48125/java_solution_in_12_ms-o-n-time-and-space
    public String rearrangeString(String str, int k) {
        if(k < 2) return str;
      int[] times = new int[26];
      for(int i = 0; i < str.length(); i++){
          ++times[str.charAt(i) - 'a'];
      }
      SortedSet<int[]> set = new TreeSet<int[]>(new Comparator<int[]>(){
          @Override
          public int compare(int[] a, int[] b){
              return a[0] == b[0] ? Integer.compare(a[1], b[1]) : Integer.compare(b[0], a[0]);
          }
      });
      for(int i = 0; i < 26; i++){
          if(times[i] != 0){
            set.add(new int[]{times[i], i});
          }
      }
      int cycles = 0;
      int cur = cycles;
      Iterator<int[]> iter = set.iterator();
      char[] res = new char[str.length()];
      while(iter.hasNext()){
          int[] e = iter.next();
          for(int i = 0; i < e[0]; i++){
              res[cur] = (char)('a'+e[1]);
              if(cur > 0 && res[cur] == res[cur-1])
                return "";
              cur += k;
              if(cur >= str.length()){
                  cur = ++cycles;
              }
          }
      }
      return new String(res);
    }
http://blog.csdn.net/jmspan/article/details/51678257
方法:根据出现频率将字母从大到小排列,以k为间隔进行重排。
  1.     public String rearrangeString(String str, int k) {  
  2.         if (k <= 0return str;  
  3.         int[] f = new int[26];  
  4.         char[] sa = str.toCharArray();  
  5.         for(char c: sa) f[c-'a'] ++;  
  6.         int r = sa.length / k;  
  7.         int m = sa.length % k;  
  8.         int c = 0;  
  9.         for(int g: f) {  
  10.             if (g-r>1return "";  
  11.             if (g-r==1) c ++;  
  12.         }  
  13.         if (c>m) return "";  
  14.         Integer[] pos = new Integer[26];  
  15.         for(int i=0; i<pos.length; i++) pos[i] = i;  
  16.         Arrays.sort(pos, new Comparator<Integer>() {  
  17.            @Override  
  18.            public int compare(Integer i1, Integer i2) {  
  19.                return f[pos[i2]] - f[pos[i1]];  
  20.            }  
  21.         });  
  22.         char[] result = new char[sa.length];  
  23.         for(int i=0, j=0, p=0; i<sa.length; i++) {  
  24.             result[j] = (char)(pos[p]+'a');  
  25.             if (-- f[pos[p]] == 0) p ++;  
  26.             j += k;  
  27.             if (j >= sa.length) {  
  28.                 j %= k;  
  29.                 j ++;  
  30.             }  
  31.         }  
  32.         return new String(result);  
  33.     } 

方法二:最多允许有str.length() % k个冗余的字符,所以可以不排序,O(N)时间复杂度。


https://reeestart.wordpress.com/201
  1.     public String rearrangeString(String str, int k) {  
  2.         if (str == null || str.length() <= 1 || k <= 0return str;  
  3.         char[] sa = str.toCharArray();  
  4.         int[] frequency = new int[26];  
  5.         for(char ch : sa) {  
  6.             frequency[ch - 'a'] ++;  
  7.         }  
  8.         int bucketSize = sa.length / k;  
  9.         int remainSize = sa.length % k;  
  10.         int[] remain = new int[remainSize];  
  11.         int count = 0;  
  12.         for(int i = 0; i < frequency.length; i++) {  
  13.             if (frequency[i] > bucketSize + 1return "";  
  14.             if (frequency[i] > bucketSize && count >= remainSize) return "";  
  15.             if (frequency[i] > bucketSize) remain[count++] = i;  
  16.         }  
  17.           
  18.         int offset = 0, j = 0;  
  19.         for(int i = 0; i < count; i++) {  
  20.             while (frequency[remain[i]] > 0) {  
  21.                 frequency[remain[i]] --;  
  22.                 sa[j] = (char)('a' + remain[i]);  
  23.                 j += k;  
  24.                 if (j >= sa.length) {  
  25.                     offset ++;  
  26.                     j = offset;  
  27.                 }  
  28.             }  
  29.         }  
  30.           
  31.         for(int i = 0; i < 26; i ++) {  
  32.             while (frequency[i] > 0) {  
  33.                 frequency[i] --;  
  34.                 sa[j] = (char)('a' + i);  
  35.                 j += k;  
  36.                 if (j >= sa.length) {  
  37.                     offset ++;  
  38.                     j = offset;  
  39.                 }  
  40.             }  
  41.         }  
  42.         return new String(sa);  
  43.     }  
6/06/23/rearrange-string-with-k-distance-apart/
第一个是统计frequency用hash table vs int[]的区别,事实证明array的确要比hash table快一点(80+ms vs 110+ms)。无论input string是不是只包含小写字母,都可以用array来替代hash table,只是size大点小点的关系。
之前onsite面试的时候就已经被面试官不止一次的指出这个问题,能用array的时候就别用hash table.
第二个是sort vs TreeSet的区别。这个真是震惊了,不比不知道,一比下一跳,TreeSet的运行时间最低居然只有8ms,比sorting快了将近10倍。但理论上时间复杂度都是O(nlogn),为什么会这样有点理解不能…
  public String rearrangeString(String str, int k) {
    if (str == null || str.isEmpty() || k <= 1) {
      return str;
    }
    int[] cnt = new int[26];
    for (char c : str.toCharArray()) {
      cnt[c - 'a']++;
    }
//    List<int[]> entryList = new ArrayList<>();
//    for (int i = 0; i < 26; i++) {
//      if (cnt[i] != 0) {
//        entryList.add(new int[] {i, cnt[i]});
//      }
//    }
//    Collections.sort(entryList, (a, b) -> (-(a[1] - b[1])));
    TreeSet<int[]> entryList = new TreeSet<>(new Comparator<int[]>() {
      public int compare(int[] a, int[] b) {
        if (a[1] == b[1]) {
          return a[0] - b[0];
        }
        return -(a[1] - b[1]);
      }
    });
    for (int i = 0; i < 26; i++) {
      if (cnt[i] != 0) {
        entryList.add(new int[] {i, cnt[i]});
      }
    }
    char[] ch = new char[str.length()];
    int i = 0;
    int start = 1;
    for (int[] entry : entryList) {
      for (int j = 0; j < entry[1]; j++) {
        ch[i] = (char) (entry[0] + 'a');
        if (i != 0 && ch[i] == ch[i - 1]) {
          return "";
        }
        i += k;
        if (i >= str.length()) {
          i = start;
          start++;
        }
      }
    }
    return new String(ch);
  }

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Orientation (1) Google APAC (1) Graph But No Graph (1) Graph Transpose (1) Graph Traversal (1) Graph-Coloring (1) Graph-Longest Path (1) Gray Code (1) HOJ (1) Hanoi (1) Hard Algorithm (1) How Hash (1) How to Test (1) Improve It (1) In Place (1) Inorder-Reverse Inorder Traverse Simultaneously (1) Interpolation search (1) Interview (1) Interview - Easy (1) Interview - Facebook (1) Isomorphic (1) JDK8 (1) K Dimensional Tree (1) Knapsack - Fractional (1) Knapsack - ZeroOnePack (1) Knight (1) Kosaraju’s algorithm (1) Kruskal (1) Kruskal MST (1) Kth Element (1) Least Common Ancestor (1) LeetCode - Binary Tree (1) LeetCode - Coding (1) LeetCode - Detail (1) LeetCode - Related (1) LeetCode Diffcult (1) Linked List Reverse (1) Linkedin (1) Linkedin Interview (1) Local MinMax (1) Logic Pattern (1) Longest Common Subsequence (1) Longest Common Substring (1) Longest Prefix Suffix(LPS) (1) Manhattan Distance (1) Map && Reverse Map (1) Math - Induction (1) Math-Multiply (1) Math-Sum Of Digits (1) Matrix - O(N+M) (1) Matrix BFS (1) Matrix Graph (1) Matrix Search (1) Matrix+DP (1) Matrix-Rotate (1) Max Min So Far (1) Median (1) Memory-Efficient (1) MinHash (1) MinMax Heap (1) Monotone Queue (1) Monto Carlo (1) Multi-Reverse (1) Multiple DFS (1) Multiple Tasks (1) Next Successor (1) Offline Algorithm (1) PAT (1) Partition (1) Path Finding (1) Patience Sort (1) Persistent (1) Pigeon Hole Principle (1) Power Set (1) Pratical Algorithm (1) Probabilistic Data Structure (1) Proof (1) Python (1) Queue & Stack (1) RSA (1) Ranking (1) Rddles (1) ReHash (1) Realtime (1) Recurrence Relation (1) Recursive DFS (1) Recursive to Iterative (1) Red-Black Tree (1) Region (1) Regular Expression (1) Resources (1) Reverse Inorder Traversal (1) Robin (1) Selection (1) Self Balancing BST (1) Similarity (1) Sort && Binary Search (1) String Algorithm. Symbol Table (1) String DP (1) String Distance (1) SubMatrix (1) Subsequence (1) System of Difference Constraints(差分约束系统) (1) TSP (1) Ternary Search Tree (1) Test (1) Thread (1) TimSort (1) Top-Down (1) Tournament (1) Tournament Tree (1) Transform Tree in Place (1) Tree Diameter (1) Tree Rotate (1) Trie + DFS (1) Trie and Heap (1) Trie vs Hash (1) Trie vs HashMap (1) Triplet (1) Two Data Structures (1) Two Stacks (1) USACO - Classical (1) USACO - Problems (1) UyHiP (1) Valid Tree (1) Vector (1) Wiggle Sort (1) Wikipedia (1) Yahoo Interview (1) ZOJ (1) baozitraining (1) codevs (1) cos126 (1) javabeat (1) jum (1) namic Programming (1) sqrt(N) (1) 两次dijkstra (1) 九度 (1) 二进制枚举 (1) 夹逼法 (1) 归一化 (1) 折半枚举 (1) 枚举 (1) 状态压缩DP (1) 男人八题 (1) 英雄会 (1) 逆向思维 (1)

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