Algorithm-Day12-Minimum-Window-Substring-lc-76

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šŸ§© Problem Description

Given two strings s and t, return the minimum window substring of s such that every character in t (including duplicates) is included in the window.

If there is no such substring, return the empty string "".

Example:

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Input: s = "ADOBECODEBANC", t = "ABC"
Output: "BANC"

šŸ’” Brute Force? Not Efficient

Try all substrings of s and check if they contain all characters of t.

  • Time: O(nĀ³)
  • āŒ Way too slow

šŸ’” Optimal Approach: Sliding Window + Hash Map

šŸ§  Key Ideas

  • Use a sliding window defined by two pointers: left and right
  • Expand the window to the right until all required characters are included
  • Then try to shrink the window from the left to find the minimum
  • Track character counts using hash maps

šŸ”¢ Java Code

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class Solution {
    public String minWindow(String s, String t) {
        if (s.length() < t.length()) return "";

        Map<Character, Integer> need = new HashMap<>();
        for (char c : t.toCharArray()) {
            need.put(c, need.getOrDefault(c, 0) + 1);
        }

        Map<Character, Integer> window = new HashMap<>();
        int left = 0, right = 0;
        int valid = 0;
        int minLen = Integer.MAX_VALUE;
        int start = 0;

        while (right < s.length()) {
            char c = s.charAt(right++);
            if (need.containsKey(c)) {
                window.put(c, window.getOrDefault(c, 0) + 1);
                if (window.get(c).intValue() == need.get(c).intValue()) {
                    valid++;
                }
            }

            // Try to shrink the window
            while (valid == need.size()) {
                if (right - left < minLen) {
                    minLen = right - left;
                    start = left;
                }

                char d = s.charAt(left++);
                if (need.containsKey(d)) {
                    if (window.get(d).intValue() == need.get(d).intValue()) {
                        valid--;
                    }
                    window.put(d, window.get(d) - 1);
                }
            }
        }

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

ā± Time and Space Complexity

  • Time Complexity: O(n + m)
    Where n is length of s, m is length of t
  • Space Complexity: O(m)
    For the hash maps tracking counts

āœļø Summary

  • A classic and foundational sliding window problem
  • Mastering this helps with problems like:
    • lc-567 (Permutation in String)
    • lc-438 (Find All Anagrams)
    • lc-239 (Sliding Window Maximum)

Key challenge: tracking when the window becomes valid and shrinking it optimally.