LeetCode-in-Java

76. Minimum Window Substring

Hard

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.

A substring is a contiguous sequence of characters within the string.

Example 1:

Input: s = “ADOBECODEBANC”, t = “ABC”

Output: “BANC”

Explanation: The minimum window substring “BANC” includes ‘A’, ‘B’, and ‘C’ from string t.

Example 2:

Input: s = “a”, t = “a”

Output: “a”

Explanation: The entire string s is the minimum window.

Example 3:

Input: s = “a”, t = “aa”

Output: “”

Explanation: Both ‘a’s from t must be included in the window. Since the largest window of s only has one ‘a’, return empty string.

Constraints:

• m == s.length
• n == t.length
• 1 <= m, n <= 105
• s and t consist of uppercase and lowercase English letters.

Follow up: Could you find an algorithm that runs in O(m + n) time?

To solve the “Minimum Window Substring” problem in Java with the Solution class, follow these steps:

1. Define a method minWindow in the Solution class that takes two strings s and t as input and returns the minimum window substring of s containing all characters from t.
2. Create two frequency maps: tFreqMap to store the frequency of characters in string t, and sFreqMap to store the frequency of characters in the current window of string s.
3. Initialize two pointers left and right to track the window boundaries. Initialize a variable minLength to store the minimum window length found so far.
4. Iterate over string s using the right pointer until the end of the string:
   • Update the frequency map sFreqMap for the character at index right.
   • Check if the current window contains all characters from t. If it does, move the left pointer to minimize the window while maintaining the condition.
   • Update the minLength if the current window length is smaller.
   • Move the right pointer to expand the window.
5. Return the minimum window substring found, or an empty string if no such substring exists.

Here’s the implementation of the minWindow method in Java:

import java.util.HashMap;
import java.util.Map;

class Solution {
    public String minWindow(String s, String t) {
        Map<Character, Integer> tFreqMap = new HashMap<>();
        Map<Character, Integer> sFreqMap = new HashMap<>();
        
        // Initialize tFreqMap with character frequencies from string t
        for (char ch : t.toCharArray()) {
            tFreqMap.put(ch, tFreqMap.getOrDefault(ch, 0) + 1);
        }
        
        int left = 0;
        int right = 0;
        int minLength = Integer.MAX_VALUE;
        int minStart = 0;
        
        while (right < s.length()) {
            char rightChar = s.charAt(right);
            sFreqMap.put(rightChar, sFreqMap.getOrDefault(rightChar, 0) + 1);
            right++;
            
            // Check if the current window contains all characters from t
            while (containsAllChars(sFreqMap, tFreqMap)) {
                // Update the minimum window length
                if (right - left < minLength) {
                    minLength = right - left;
                    minStart = left;
                }
                
                char leftChar = s.charAt(left);
                sFreqMap.put(leftChar, sFreqMap.get(leftChar) - 1);
                left++;
            }
        }
        
        return minLength == Integer.MAX_VALUE ? "" : s.substring(minStart, minStart + minLength);
    }
    
    // Helper method to check if sFreqMap contains all characters from tFreqMap
    private boolean containsAllChars(Map<Character, Integer> sFreqMap, Map<Character, Integer> tFreqMap) {
        for (Map.Entry<Character, Integer> entry : tFreqMap.entrySet()) {
            char ch = entry.getKey();
            int count = entry.getValue();
            if (sFreqMap.getOrDefault(ch, 0) < count) {
                return false;
            }
        }
        return true;
    }
}

This implementation finds the minimum window substring in O(m + n) time complexity, where m is the length of string s and n is the length of string t. It uses two frequency maps to keep track of character frequencies and adjusts the window boundaries to find the minimum window containing all characters from t.

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