[JAVA] Readable solution with HashMap and Deque

• class Solution {
public String minWindow(String s, String t) {

// By default the window is the entire string...
int bestMin = 0;
int bestMax = s.length() -1;

// This holds the amount of each character we need to see
Map<Character, Integer> charCount = new HashMap<>();

for (int i = 0; i < t.length(); i++) {
char c = t.charAt(i);
Integer count = charCount.getOrDefault(c, 0);
charCount.put(c, count+1);
}

// This holds characters and their last seen indexes in the string
Map<Character, Deque<Integer>> lastSeen = new HashMap<>();
// This holds characters that we have seen the minimum amount of
Set<Character> seenMin = new HashSet<>();

for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);

// check if this is a character we are looking for
if (charCount.containsKey(c)) {

Deque<Integer> positions = lastSeen.getOrDefault(c, new LinkedList<>());

// If we have stored the maximum positions, remove the earliest and add the newest to the end
if (positions.size() == charCount.get(c)) {
positions.removeFirst();
}

if (positions.size() == charCount.get(c)) {
}

lastSeen.put(c, positions);

// If we have seen all our characters the minimum amount of times
// we can try and calculate the minimum total length that holds all those characters
if (seenMin.size() == charCount.size()) {
int min = Integer.MAX_VALUE;
int max = Integer.MIN_VALUE;

// Iterate over the positions for each character and grab their min/max positions
for (Deque<Integer> values : lastSeen.values()) {
min = Math.min(min, values.getFirst());
max = Math.max(max, values.getLast());
}

// If we have a smaller length than our current best length, save it.
if (max - min < bestMax - bestMin) {
bestMax = max;
bestMin = min;
}
}
}
}

// If we have seen all of our characters the minimum amount of times we know we will have a valid solution
if (seenMin.size() == charCount.size()) {
return s.substring(bestMin, bestMax+1);
} else {
return "";
}
}
}

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