http://www.cnblogs.com/grandyang/p/5351347.html
https://discuss.leetcode.com/topic/48827/java-o-nlogk-using-treemap-to-keep-last-occurrence-interview-follow-up-question/3
The interviewer may say that the string is given as a steam. In this situation, we can't maintain a "left pointer" as the classical O(n) hashmap solution.
http://www.1point3acres.com/bbs/thread-209202-1-1.html
第一轮，Longest Substring with At Most K Distinct Characters变形题，输入是stream而不是string，所以无法知道输入的长度，而且内存不够，无法存入整个stream，所以维持sliding window来更新最大长度的算法无法解决该问题。最后是通过一个map纪录每个字符上次出现的index来实现。提供了函数streamreader.reader()来返回下一个字符
第一轮，Longest Substring with At Most K Distinct Characters变形题: 假设API streamreader.read()返回'\0' 若stream已读完（具体的要求不影响实现）。因为内存有限制，那么这个题还有一个细节就是所给的k大小能不能认为是有界的，一般若假设是26或256个字符的话那么用一个unordered_map<char, int> lastInx记录当前最后k个distinct char出现的最后index 就可以了。但是这样实际会造成O(Nk)时间复杂度，因为在前进substring front的时候还会涉及判断front index是不是某个char的最后index。我是再定义一个reverse mapping: unordered_map<int, char> chars来记录每个最后index所对应的char. 这样用O(2k)的空间就可以实现O(N)的时间。
当然假设内存虽然有限制，但至少是O(k)的 （即使不是O(N)的）。
关键看你的sliding window在advance front index的时候需要记录什么。
在无内存限制的时候，advance front index需要判断删除char s[front_index]后是否减少了某种字符的出现，若是这样实现必然要存储substring从front_index一直到cur_index的字符，这需要O(N)空间，也就是这道题不允许的。所以按照Leetcode原题只记录char frequency是不行的。.鏈枃鍘熷垱鑷�1point3acres璁哄潧
实际上每次front_index++时并不需要知道s[front_index]本身是什么，只需要知道该字符是否是最后出现的instance。这个改进应该就是这个题的考点吧。
当然，前提是内存虽然没有O(N)，但至少有O(k). 注意O(k)可以远远小于O(charset).
Unicode allows for 17 planes, each of 65,536 possible characters (or 'code points'). This gives a total of 1,114,112 possible characters. At present, only about 10% of this space has been allocated.
Given a string, find the length of the longest substring T that contains at most k distinct characters.
For example, Given s =
“eceba” and k = 2,
T is "ece" which its length is 3.
public int lengthOfLongestSubstringKDistinct(String s, int k) {
int[] count = new int[256];
int num = 0, i = 0, res = 0;
for (int j = 0; j < s.length(); j++) {
if (count[s.charAt(j)]++ == 0) num++;
if (num > k) {
while (--count[s.charAt(i++)] > 0);//??
num--;
}
res = Math.max(res, j - i + 1);
}
return res;
}
http://buttercola.blogspot.com/2016/06/leetcode-340-longest-substring-with-at.html
The problem is very similar to the Leetcode question 3 (Longest Substring Without Repeating Characters).
The key idea of the solution is to use two pointers to maintain a "sliding window". We also use a HashMap to store all the characters in the window. Each time we move forward the "start" pointer and increase the size of the sliding window until the size of the window is greater than K. Then we can easily calculate the size of the window which contains at most K distinct characters. The next step is to shrink the window by moving forward the "end" pointer. In order to get the maximum window size, we must move the minimum steps of the end pointer. So each step we move the end pointer, we need to update the map and remove the character out of the sliding window. The stop condition is that when the window size is again equal to the K, which means the window contains K distinct characters. That's the minimum steps we need to move forward the end pointer.
The only trick here is we need to check at the last that if the start pointer is out of boundary, we still need to check if the largest window size. That's a common trick for major string w/ two pointer problems.
https://leetcode.com/discuss/95698/generic-solution-in-java-that-can-be-used-for-unicode
The problem is very similar to the Leetcode question 3 (Longest Substring Without Repeating Characters).
The key idea of the solution is to use two pointers to maintain a "sliding window". We also use a HashMap to store all the characters in the window. Each time we move forward the "start" pointer and increase the size of the sliding window until the size of the window is greater than K. Then we can easily calculate the size of the window which contains at most K distinct characters. The next step is to shrink the window by moving forward the "end" pointer. In order to get the maximum window size, we must move the minimum steps of the end pointer. So each step we move the end pointer, we need to update the map and remove the character out of the sliding window. The stop condition is that when the window size is again equal to the K, which means the window contains K distinct characters. That's the minimum steps we need to move forward the end pointer.
The only trick here is we need to check at the last that if the start pointer is out of boundary, we still need to check if the largest window size. That's a common trick for major string w/ two pointer problems.
https://leetcode.com/discuss/95698/generic-solution-in-java-that-can-be-used-for-unicode
This problem can be solved using two pointers. The important part is
while (map.size() > k), we move left pointer to make sure the map size is less or equal to k. This can be easily extended to any number of unique characters.
A more generic solution as follows, can be solution for Unicode string:
public int lengthOfLongestSubstringKDistinct(String s, int k) {
Map<Character, Integer> map = new HashMap<>();
int left = 0;
int best = 0;
for(int i = 0; i < s.length(); i++) {
// character at the right pointer
char c = s.charAt(i);
map.put(c, map.getOrDefault(c, 0) + 1);
// make sure map size is valid, no need to check left pointer less than s.length()
while (map.size() > k) {
char leftChar = s.charAt(left);
if (map.containsKey(leftChar)) {
map.put(leftChar, map.get(leftChar) - 1);
if (map.get(leftChar) == 0) {
map.remove(leftChar);
}
}
left++;
}
best = Math.max(best, i - left + 1);
}
return best;
}
public static int lengthOfLongestSubstringKDistinct(String s, int k) {
int[] count = new int[256];
int num = 0, i = 0, res = 0;
for (int j = 0; j < s.length(); j++) {
if (count[s.charAt(j)]++ == 0) num++;
if (num > k) {
while (--count[s.charAt(i++)] > 0);
num--;
}
res = Math.max(res, j - i + 1);
}
return res;
}
public static int lengthOfLongestSubstringKDistinct(String s, int k) {
Map<Character, Integer> map = new HashMap<>();
int left = 0;
int res = 0;
for(int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
map.put(c, map.getOrDefault(c, 0) + 1);
while (map.size() > k) {
char leftChar = s.charAt(left);
left+=map.get(leftChar);
map.remove(leftChar);
}
res = Math.max(res, i - left + 1);
}
return res;
}
Follow uphttps://discuss.leetcode.com/topic/48827/java-o-nlogk-using-treemap-to-keep-last-occurrence-interview-follow-up-question/3
The interviewer may say that the string is given as a steam. In this situation, we can't maintain a "left pointer" as the classical O(n) hashmap solution.
Solving the problem with O(n) time is not enough, some interviewer may require this solution as a followup. Instead of recording each char's count, we keep track of char's last occurrence. If you consider k as constant, it is also a O(n) algorithm.
inWindow keeps track of each char in window and its last occurrence position
lastOccurrence is used to find the char in window with left most last occurrence. A better idea is to use a PriorityQueue, as it takes O(1) to getMin, However Java's PQ does not support O(logn) update a internal node, it takes O(n). TreeMap takes O(logn) to do both getMin and update.
Every time when the window is full of k distinct chars, we lookup TreeMap to find the one with leftmost last occurrence and set left bound j to be 1 + first to exclude the char to allow new char coming into window.
Every time when the window is full of k distinct chars, we lookup TreeMap to find the one with leftmost last occurrence and set left bound j to be 1 + first to exclude the char to allow new char coming into window.
public int lengthOfLongestSubstringKDistinct(String str, int k) {
if (str == null || str.isEmpty() || k == 0) {
return 0;
}
TreeMap<Integer, Character> lastOccurrence = new TreeMap<>();
Map<Character, Integer> inWindow = new HashMap<>();
int j = 0;
int max = 1;
for (int i = 0; i < str.length(); i++) {
char in = str.charAt(i);
while (inWindow.size() == k && !inWindow.containsKey(in)) {
int first = lastOccurrence.firstKey();
char out = lastOccurrence.get(first);
inWindow.remove(out);
lastOccurrence.remove(first);
j = first + 1;
}
//update or add in's position in both maps
if (inWindow.containsKey(in)) {
lastOccurrence.remove(inWindow.get(in));
}
inWindow.put(in, i);
lastOccurrence.put(i, in);
max = Math.max(max, i - j + 1);
}
return max;
}第一轮，Longest Substring with At Most K Distinct Characters变形题，输入是stream而不是string，所以无法知道输入的长度，而且内存不够，无法存入整个stream，所以维持sliding window来更新最大长度的算法无法解决该问题。最后是通过一个map纪录每个字符上次出现的index来实现。提供了函数streamreader.reader()来返回下一个字符
第一轮，Longest Substring with At Most K Distinct Characters变形题: 假设API streamreader.read()返回'\0' 若stream已读完（具体的要求不影响实现）。因为内存有限制，那么这个题还有一个细节就是所给的k大小能不能认为是有界的，一般若假设是26或256个字符的话那么用一个unordered_map<char, int> lastInx记录当前最后k个distinct char出现的最后index 就可以了。但是这样实际会造成O(Nk)时间复杂度，因为在前进substring front的时候还会涉及判断front index是不是某个char的最后index。我是再定义一个reverse mapping: unordered_map<int, char> chars来记录每个最后index所对应的char. 这样用O(2k)的空间就可以实现O(N)的时间。
当然假设内存虽然有限制，但至少是O(k)的 （即使不是O(N)的）。
关键看你的sliding window在advance front index的时候需要记录什么。
在无内存限制的时候，advance front index需要判断删除char s[front_index]后是否减少了某种字符的出现，若是这样实现必然要存储substring从front_index一直到cur_index的字符，这需要O(N)空间，也就是这道题不允许的。所以按照Leetcode原题只记录char frequency是不行的。.鏈枃鍘熷垱鑷�1point3acres璁哄潧
实际上每次front_index++时并不需要知道s[front_index]本身是什么，只需要知道该字符是否是最后出现的instance。这个改进应该就是这个题的考点吧。
当然，前提是内存虽然没有O(N)，但至少有O(k). 注意O(k)可以远远小于O(charset).
Unicode allows for 17 planes, each of 65,536 possible characters (or 'code points'). This gives a total of 1,114,112 possible characters. At present, only about 10% of this space has been allocated.
http://www.programcreek.com/2013/02/longest-substring-which-contains-2-unique-characters/
In this solution, a hashmap is used to track the unique elements in the map. When a third character is added to the map, the left pointer needs to move right.
public int lengthOfLongestSubstringTwoDistinct(String s) { int max=0; HashMap<Character,Integer> map = new HashMap<Character, Integer>(); int start=0; for(int i=0; i<s.length(); i++){ char c = s.charAt(i); if(map.containsKey(c)){ map.put(c, map.get(c)+1); }else{ map.put(c,1); } if(map.size()>2){ max = Math.max(max, i-start); while(map.size()>2){ char t = s.charAt(start); int count = map.get(t); if(count>1){ map.put(t, count-1); }else{ map.remove(t); } start++; } } } max = Math.max(max, s.length()-start); return max; } |
