Minimum Window Substring

 Given two strings s and t of lengths m and n respectively, 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 "".


The testcases will be generated such that the answer is unique.




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

Output: "BANC"

Explanation: The minimum window substring "BANC" includes 'A', 'B', and 'C' from string t.


Input: s = "a", t = "a"

Output: "a"

Explanation: The entire string s is the minimum window.

LeetCode:-


import java.util.*;

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

        Map<Character, Integer> need = new HashMap<>();
        Map<Character, Integer> window = new HashMap<>();

        // Count characters in t
        for (char c : t.toCharArray()) {
            need.put(c, need.getOrDefault(c, 0) + 1);
        }

        int left = 0, right = 0;
        int valid = 0;
        int start = 0, len = Integer.MAX_VALUE;

        while (right < s.length()) {
            char c = s.charAt(right);
            right++;

            // Update window
            if (need.containsKey(c)) {
                window.put(c, window.getOrDefault(c, 0) + 1);
                if (window.get(c).equals(need.get(c))) {
                    valid++;
                }
            }

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

                char d = s.charAt(left);
                left++;

                if (need.containsKey(d)) {
                    if (window.get(d).equals(need.get(d))) {
                        valid--;
                    }
                    window.put(d, window.get(d) - 1);
                }
            }
        }

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


Code :

public class codefile{
    static String solve(String s,String t){
            if (s.length() < t.length()) return "";

        // Step 1: Count characters in t
        Map<Character, Integer> need = new HashMap<>();
        Map<Character, Integer> window = new HashMap<>();

        for (char c : t.toCharArray()) {
            need.put(c, need.getOrDefault(c, 0) + 1);
        }

        int left = 0, right = 0;
        int valid = 0;  // counts how many characters satisfy the required frequency
        int minLen = Integer.MAX_VALUE;
        int start = 0;  // to track the starting index of the best window

        while (right < s.length()) {
            char c = s.charAt(right);
            right++;

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

                char d = s.charAt(left);
                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);
    }
}

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