diff --git a/Sliding Window Problems/duplicates_within_given_range_k_in_array.cpp b/Sliding Window Problems/duplicates_within_given_range_k_in_array.cpp
new file mode 100644
index 0000000..cea5497
--- /dev/null
+++ b/Sliding Window Problems/duplicates_within_given_range_k_in_array.cpp	
@@ -0,0 +1,73 @@
+/* 
+Problem:
+Given an array and a positive number k, 
+check whether the array contains any duplicate elements within range k.
+If k is more than size of the array, the solution should check for duplicates in the complete array
+
+Using Sliding Window Technique
+
+Example:
+
+A[] = {1, 2, 3, 4, 9, 3, 8, 1 }
+k = 4
+o/p:
+Duplicates found
+(element 3 is repeated at distance 3 which is <=k )
+
+A[] = {1, 2, 3, 9, 1 }
+k = 3
+o/p:
+No duplicates found
+(element 1 is repeated at distance 4 which is >k) 
+
+*/
+
+#include <iostream>
+#include <vector>
+#include <unordered_set>
+using namespace std;
+
+bool contains(unordered_set<int> const &set, int x) {
+	return set.find(x) != set.end();
+}
+
+bool hasDuplicates(vector<int> &vec, int k)
+{
+	// create an empty set to store elements within range k
+	unordered_set<int> window;
+
+	for (int i = 0; i < vec.size(); i++)
+	{
+		// if the current element already exists in the window,
+		// then it is repeated within range of k
+		if (contains(window, vec[i])) {
+			return true;
+		}
+
+		window.insert(vec[i]);
+
+		if (i >= k) {
+			window.erase(vec[i - k]);
+		}
+	}
+
+	// we reach here when no element is repeated within range k
+	return false;
+}
+
+int main()
+{
+	vector<int> vec = { 5, 6, 8, 2, 4, 6, 9 };
+	int k = 4;
+
+	if (hasDuplicates(vec, k)) {
+		cout << "Duplicates found";
+	}
+	else {
+		cout << "No Duplicates found";
+	}
+
+	return 0;
+}
+
+//Time complexity is O(n) and uses O(n) extra space
\ No newline at end of file
diff --git a/Sliding Window Problems/longest_substring_containing_distinct_characters.cpp b/Sliding Window Problems/longest_substring_containing_distinct_characters.cpp
new file mode 100644
index 0000000..d77ff33
--- /dev/null
+++ b/Sliding Window Problems/longest_substring_containing_distinct_characters.cpp	
@@ -0,0 +1,69 @@
+/* 
+Problem:
+Given a string, find the logest substring of given string 
+containing distinct characters.
+
+Using Sliding Window Technique
+
+Example:
+
+string: 'hacktoberfest'
+o/p is 'hacktoberf'
+
+string: 'abaa'
+o/p is 'ab'
+
+*/
+
+#include <iostream>
+#include <vector>
+using namespace std;
+
+#define CHAR_RANGE 128
+
+string longestSubstr(string str, int n)
+{
+	vector<bool> window(CHAR_RANGE);
+
+	int begin = 0, end = 0;
+
+	for (int low = 0, high = 0; high < n; high++)
+	{
+		if (window[str[high]])
+		{
+			// remove characters from the left of the window till
+			// we encounter current character
+			while (str[low] != str[high])
+				window[str[low++]] = false;
+
+			low++;		
+		}
+		else
+		{
+			// if current character is not present in the current
+			// window, include it
+			window[str[high]] = true;
+
+			if (end - begin < high - low)
+			{
+				begin = low;
+				end = high;
+			};
+		}
+	}
+
+	return str.substr(begin, end - begin + 1);
+}
+
+int main()
+{
+	string str;
+
+    cout << "Enter string : ";
+    cin >> str;
+
+	int n = str.length();
+	cout << longestSubstr(str, n);
+
+	return 0;
+}
\ No newline at end of file
diff --git a/Sliding Window Problems/longest_substring_containing_k_distinct_characters.cpp b/Sliding Window Problems/longest_substring_containing_k_distinct_characters.cpp
new file mode 100644
index 0000000..046b4b3
--- /dev/null
+++ b/Sliding Window Problems/longest_substring_containing_k_distinct_characters.cpp	
@@ -0,0 +1,75 @@
+/* 
+Problem:
+Given a string and a positive number k, find the logest substring of given string 
+containing k distinct characters.
+If k is more than the number of distict characters in the string, return the whole string.
+
+Using Sliding Window Technique
+
+Example:
+string: 'zootopia'
+for k = 2, o/p is 'ooto'
+for k = 3, o/p is 'zooto'
+
+*/
+
+#include <iostream>
+#include <string>
+#include <unordered_set>
+using namespace std;
+
+#define CHAR_RANGE 128
+
+string longestSubstr(string str, int k, int n)
+{
+	// for longest substring boundaries
+	int end = 0, begin = 0;
+
+	unordered_set<char> window;
+
+	// array to store frequency of characters present in
+	// current window
+	int freq[CHAR_RANGE] = { 0 };
+
+	for (int low = 0, high = 0; high < n; high++)
+	{
+		window.insert(str[high]);
+		freq[str[high]]++;
+
+		while (window.size() > k)
+		{
+			// if the frequency of leftmost character becomes 0 after
+			// removing it in the window, remove it from set as well
+			if (--freq[str[low]] == 0)
+				window.erase(str[low]);
+
+			low++;		
+		}
+
+		if (end - begin < high - low)
+		{
+			end = high;
+			begin = low;
+		}
+	}
+
+	return str.substr(begin, end - begin + 1);
+}
+
+int main()
+{
+	string str;
+	int k;
+
+    cout << "Enter string and value of k: ";
+    cin >> str >> k; 
+
+	int n = str.length();
+	cout << longestSubstr(str, k, n);
+
+	return 0;
+}
+
+/*
+Time Complexity: O(n)
+As it does 2 traversals of the given string
\ No newline at end of file
diff --git a/Sliding Window Problems/subarrays_of_array_having_distinct_elements.cpp b/Sliding Window Problems/subarrays_of_array_having_distinct_elements.cpp
new file mode 100644
index 0000000..4ccbaa5
--- /dev/null
+++ b/Sliding Window Problems/subarrays_of_array_having_distinct_elements.cpp	
@@ -0,0 +1,76 @@
+/* 
+Problem:
+Given an array of integers, 
+print all maximum size sub-arrays having all distinct elements in them
+
+Using Sliding Window Technique
+
+Example:
+
+A[] = {1, 2, 3, 3, 2, 4, 5}
+o/p:
+{1, 2, 3}
+{3, 2, 4, 5}
+*/
+
+#include <iostream>
+#include <unordered_map>
+using namespace std;
+
+void printSubArray(int A[], int i, int j, int n)
+{
+    //invalid input
+	if (i < 0 || i > j || j >= n) 
+		return;
+
+	for (int index = i; index < j; index++)
+		cout << A[index] << ", " ;
+
+	cout << A[j] << endl;
+}
+
+void calculate(int A[], int n)
+{
+	// Map to mark elements as visited in the current window
+	unordered_map<int, bool> visited;
+
+	int right = 0, left = 0;
+
+	while (right < n)
+	{
+		// keep increasing the window size if all elements in the
+		// current window are distinct
+		while (right < n && !visited[A[right]])
+		{
+			visited[A[right]] = true;
+			right++;
+		}
+
+		printSubArray(A, left, right - 1, n);
+
+		// As soon as duplicate is found (A[right]),
+		// terminate the above loop and reduce the window's size
+		// from its left to remove the duplicate
+		while (right < n && visited[A[right]])
+		{
+			visited[A[left]] = false;
+			left++;
+		}
+	}
+}
+
+int main()
+{
+    int n;
+    cout << "Enter the size of the array : ";
+    cin >> n;
+ 
+    int A[n];
+    cout << "Enter the elements: " << endl;
+    for(int i=0; i<n; i++)
+        cin >> A[i]; 
+
+	calculate(A, n);
+
+	return 0;
+}
\ No newline at end of file
diff --git a/Sliding Window Problems/substring_of_string_which_are_permutations_of_another_string.cpp b/Sliding Window Problems/substring_of_string_which_are_permutations_of_another_string.cpp
new file mode 100644
index 0000000..2f58fe7
--- /dev/null
+++ b/Sliding Window Problems/substring_of_string_which_are_permutations_of_another_string.cpp	
@@ -0,0 +1,82 @@
+/* 
+Problem:
+Find all substrings of a string that are permutations of another string.
+(Find Anagrams)
+
+Using Sliding Window Technique
+
+Example:
+
+string1: 'ABCCDBAC'
+string2: 'ABC'
+o/p:
+Anagram 'ABC' present at index 0
+Anagram 'BAC' present at index 5
+
+*/
+
+#include <iostream>
+#include <string>
+#include <unordered_multiset>
+using namespace std;
+
+void findAllAnagrams(string str1, string str2)
+{
+	int m, n;
+
+	// invalid input
+	if ((m = str2.length()) > (n = str1.length()))
+		return;
+
+	unordered_multiset<char> window;
+
+	unordered_multiset<char> set;
+
+	for (int i = 0; i < m; i++)
+		set.insert(str2[i]);
+
+	for (int i = 0; i < n; i++)
+	{
+		if (i < m)
+			window.insert(str1[i]);
+	
+		else
+		{
+            // If all characters in current window matches that of
+			// string2, we found an anagram
+			if (window == set)
+			{
+				cout << "Anagram " << str1.substr(i - m, m) <<
+					" present at index " << i - m << '\n';
+			}
+		
+			auto itr = window.find(X[i - m]);
+			if (itr != window.end())
+				window.erase(itr);
+
+			window.insert(str1[i]);
+		}
+	}
+
+	// if last m characters of string1 matches that of string2,
+	// we found an anagram
+	if (window == set)
+	{
+		cout << "Anagram " << X.substr(n - m, m) <<
+			" present at index " << n - m << '\n';
+	}
+}
+
+int main()
+{
+	string str1;
+	string str2;
+
+	findAllAnagrams(X, Y);
+
+	return 0;
+}
+
+//Time complexity: O((n-m)*m)
+// n -> length of first string 
+// m -> length of second string
\ No newline at end of file