Sliding Window Problems

See:

Window Sliding Technique

This technique shows how a nested for loop in few problems can be converted to single for loop and hence reducing the time complexity.

Template

public class Solution {
public List<Integer> slidingWindowTemplateByHarryChaoyangHe(String s, String t) {
//init a collection or int value to save the result according the question.
List<Integer> result = new LinkedList<>();
if(t.length()> s.length()) return result;

//create a hashmap to save the Characters of the target substring.
//(K, V) = (Character, Frequence of the Characters)
Map<Character, Integer> map = new HashMap<>();
for(char c : t.toCharArray()){
map.put(c, map.getOrDefault(c, 0) + 1);
}
//maintain a counter to check whether match the target string.
int counter = map.size();//must be the map size, NOT the string size because the char may be duplicate.

//Two Pointers: begin - left pointer of the window; end - right pointer of the window
int begin = 0, end = 0;

//the length of the substring which match the target string.
int len = Integer.MAX_VALUE;

//loop at the begining of the source string
while(end < s.length()){

char c = s.charAt(end);//get a character

if( map.containsKey(c) ){
map.put(c, map.get(c)-1);// plus or minus one
if(map.get(c) == 0) counter--;//modify the counter according the requirement(different condition).
}
end++;

//increase begin pointer to make it invalid/valid again
while(counter == 0 /* counter condition. different question may have different condition */){

char tempc = s.charAt(begin);//***be careful here: choose the char at begin pointer, NOT the end pointer
if(map.containsKey(tempc)){
map.put(tempc, map.get(tempc) + 1);//plus or minus one
if(map.get(tempc) > 0) counter++;//modify the counter according the requirement(different condition).
}

/* save / update(min/max) the result if find a target*/
// result collections or result int value

begin++;
}
}
return result;
}
}