395. Longest Substring with At Least K Repeating Characters
2016-11-09 15:18
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Find the length of the longest substring T of a given string (consists of lowercase letters only) such that every character in T appears no
less than k times.
Example 1:
Example 2:
摘自
https://discuss.leetcode.com/topic/57372/java-3ms-divide-and-conquer-recursion-solution
First, let us see how does this code work.
The idea is to keep finding the character which does not satisfy the requirement. (repeat less than k times).
And divide the problem into 2 parts, to find the longest substring on the lhs and rhs of the breaking point.
For divide and conquer, it needs O(log(n)) calls of helper function to check the entire string and in the helper function, it costs O(n) to find such breaking point. In total, this algorithm is O(nlog(n)).
public class Solution {
public int longestSubstring(String s, int k)
{
return findlen(s, 0, s.length(), k);
}
public int findlen(String s ,int start,int end,int k)
{
if(end<start||end-start<k)
return 0;
int[] count=new int[26];
for(int i=start;i<end;i++)
count[s.charAt(i)-'a']++;
for(int i=0;i<26;i++)
{
if(count[i]==0)
continue;
else if(count[i]<k)
{
for(int j=start;j<end;j++)
if(s.charAt(j)-'a'==i)
{
int leftlen=findlen(s, start, j, k);
int rightlen=findlen(s, j+1, end, k);
return Math.max(leftlen, rightlen);
}
}
}
return end-start;
}
}
less than k times.
Example 1:
Input: s = "aaabb", k = 3 Output: 3 The longest substring is "aaa", as 'a' is repeated 3 times.
Example 2:
Input: s = "ababbc", k = 2 Output: 5 The longest substring is "ababb", as 'a' is repeated 2 times and 'b' is repeated 3 times.
摘自
https://discuss.leetcode.com/topic/57372/java-3ms-divide-and-conquer-recursion-solution
First, let us see how does this code work.
The idea is to keep finding the character which does not satisfy the requirement. (repeat less than k times).
And divide the problem into 2 parts, to find the longest substring on the lhs and rhs of the breaking point.
For divide and conquer, it needs O(log(n)) calls of helper function to check the entire string and in the helper function, it costs O(n) to find such breaking point. In total, this algorithm is O(nlog(n)).
public class Solution {
public int longestSubstring(String s, int k)
{
return findlen(s, 0, s.length(), k);
}
public int findlen(String s ,int start,int end,int k)
{
if(end<start||end-start<k)
return 0;
int[] count=new int[26];
for(int i=start;i<end;i++)
count[s.charAt(i)-'a']++;
for(int i=0;i<26;i++)
{
if(count[i]==0)
continue;
else if(count[i]<k)
{
for(int j=start;j<end;j++)
if(s.charAt(j)-'a'==i)
{
int leftlen=findlen(s, start, j, k);
int rightlen=findlen(s, j+1, end, k);
return Math.max(leftlen, rightlen);
}
}
}
return end-start;
}
}
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