diff --git a/src/main/java/com/thealgorithms/slidingwindow/PermutationInString.java b/src/main/java/com/thealgorithms/slidingwindow/PermutationInString.java
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index 000000000000..4e1b7e56439b
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+++ b/src/main/java/com/thealgorithms/slidingwindow/PermutationInString.java
@@ -0,0 +1,63 @@
+package com.thealgorithms.slidingwindow;
+
+
+/**
+ * Given two strings s1 and s2, return true if s2 contains a permutation of s1, or false otherwise.
+ * In other words, return true if one of s1's permutations is the substring of s2.
+ * 
+ * Example:
+ * Input: s1 = "ab", s2 = "eidbaooo"
+ * Output: true
+ * Explanation: s2 contains one permutation of s1 ("ba"). 
+ * 
+ * Best Case Time Complexity: O(n)
+ * Worst Case Time Complexity: O(m)
+ * Space Complexity: O(1)
+ * 
+ * @author m-saurabh01
+ *
+ */
+
+class PermutationInString {  
+	
+    public boolean checkInclusion(String s1, String s2) {  
+        // Get the length of s1 and s2  
+        int n = s1.length();  
+        int m = s2.length();  
+
+        // Create frequency arrays for characters in s1 and the current window in s2  
+        int[] freq1 = new int[26]; // Frequency count for s1  
+        int[] freq2 = new int[26]; // Frequency count for current substring of s2  
+
+        // Fill the frequency array for s1  
+        for (char c : s1.toCharArray()) {  
+            freq1[c - 'a']++; 
+        }  
+
+        // Slide the window of size n across s2  
+        for (int i = 0; i < m; i++) {  
+            // Add current character to the frequency array of the window  
+            freq2[s2.charAt(i) - 'a']++;  
+
+            // Check if the window size exceeds s1's length  
+            if (i >= n) {  
+                // Remove the character that is no longer in the window  
+                freq2[s2.charAt(i - n) - 'a']--;  
+            }  
+
+            // If the frequency arrays match, return true  
+            if (i >= n - 1) { // Only check if we've processed enough characters  
+                boolean match = true; // Assume they match  
+                for (int j = 0; j < 26; j++) {  
+                    if (freq1[j] != freq2[j]) { // If there's a mismatch  
+                        match = false; // Set match to false  
+                        break; // No need to check further  
+                    }  
+                }  
+                if (match) return true; // If we found a match, return true  
+            }  
+        }  
+
+        return false; // No permutation of s1 found in s2  
+    }  
+}
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