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+# LeetCode 76.Minimum Window Substring
+
+## Problem Description
+
+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.**
+
+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?
+
+### Link
+
+https://leetcode.com/problems/minimum-window-substring/
diff --git a/Hard/76.Minimum Window Substring/Sol1.java b/Hard/76.Minimum Window Substring/Sol1.java
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+import java.util.HashMap;
+
+class Solution {
+    public String minWindow(String s, String t) {
+        if (s == null || t == null || s.length() < t.length()) {
+            return "";
+        }
+        
+     
+        HashMap<Character, Integer> tMap = new HashMap<>();
+        for (char c : t.toCharArray()) {
+            tMap.put(c, tMap.getOrDefault(c, 0) + 1);
+        }
+        
+        int left = 0; 
+        int minLen = Integer.MAX_VALUE; 
+        int minLeft = 0; 
+        int count = tMap.size(); 
+
+        for (int right = 0; right < s.length(); right++) {
+            char rightChar = s.charAt(right);
+
+            if (tMap.containsKey(rightChar)) {
+                tMap.put(rightChar, tMap.get(rightChar) - 1);
+                if (tMap.get(rightChar) == 0) {
+                    count--;
+                }
+            }
+
+  
+            while (count == 0) {
+                if (right - left + 1 < minLen) {
+                    minLen = right - left + 1;
+                    minLeft = left;
+                }
+
+                char leftChar = s.charAt(left);
+
+                if (tMap.containsKey(leftChar)) {
+                    tMap.put(leftChar, tMap.get(leftChar) + 1);
+                    if (tMap.get(leftChar) > 0) {
+                        count++;
+                    }
+                }
+
+                left++;
+            }
+        }
+
+        return minLen == Integer.MAX_VALUE ? "" : s.substring(minLeft, minLeft + minLen);
+    }
+}