76. Minimum Window Substring

Solution: sliding window O(∣S∣+∣T∣)

• t may include duplicates, so we need to count frequency of t

• Expand right pointer r to include characters

• Shrink left pointer l or left to minimize the window

• track # of UNIQUE char in cur window meets requirement

  • requirements: 1. present in t 2. frequency

• Track best (smallest) valid window at current step

class Solution {
    public String minWindow(String s, String t) {
        // frequency of each char in t
        Map<Character, Integer> tFreq = new HashMap<>();
        for (char c : t.toCharArray()) {
            tFreq.put(c, tFreq.getOrDefault(c, 0) + 1);
        }

        // to track the frequency of char in window
        Map<Character, Integer> windowFreq = new HashMap<>();

        // # of UNIQUE char in cur window meets requirement
        // requirements: 1. present in t 2. frequency
        int count = 0; 
        
        // for result
        int minLen = Integer.MAX_VALUE, minStart = -1, minEnd = -1;

        int left = 0, right = 0;
        while (left < s.length() && right < s.length()) {
            // Expand window via right ++
            char cur = s.charAt(right);
            windowFreq.put(cur, windowFreq.getOrDefault(cur, 0) + 1);
            // if uses ==, which is risky for boxed Integer objects (especially outside the [-128, 127] range).
            if (tFreq.containsKey(cur) && windowFreq.get(cur).intValue() == tFreq.get(cur).intValue()) {
                count++;
            }
            
            // Contract window size via left++
            while (count == tFreq.size()) {
                // best result currently
                if (right - left + 1 < minLen) {
                    minStart = left;
                    minEnd = right;
                    minLen = Math.min(minLen, right - left + 1);
                }
                char c = s.charAt(left);
                windowFreq.put(c, windowFreq.get(c) - 1);
                
                if (tFreq.containsKey(c)) {
                    if (windowFreq.get(c).intValue() < tFreq.get(c).intValue()) {
                        count--;
                    }
                }

                left++; // contract
            }
            

            right++; // expand 
        }

        return minLen == Integer.MAX_VALUE ? "" : s.substring(minStart, minEnd + 1);
    }
}