diff --git a/Data Structure & Algorithm/Pair_with_given_sum.c b/Data Structure & Algorithm/Pair_with_given_sum.c
new file mode 100644
index 0000000..58bc06a
--- /dev/null
+++ b/Data Structure & Algorithm/Pair_with_given_sum.c	
@@ -0,0 +1,84 @@
+//Method 1:
+
+#include <stdio.h>
+
+// Naive method to find a pair in an array with given sum
+void findPair(int arr[], int n, int sum)
+{
+	// consider each element except last element
+	for (int i = 0; i < n - 1; i++)
+	{
+		// start from i'th element till last element
+		for (int j = i + 1; j < n; j++)
+		{
+			// if desired sum is found, print it and return
+			if (arr[i] + arr[j] == sum)
+			{
+				printf("Pair found at index %d and %d", i, j);
+				return;
+			}
+		}
+	}
+ 
+	// No pair with given sum exists in the array
+	printf("Pair not found");
+}
+ 
+// Find pair with given sum in the array
+int main()
+{
+	int arr[] = { 8, 7, 2, 5, 3, 1 };
+	int sum = 10;
+ 
+	int n = sizeof(arr)/sizeof(arr[0]);
+ 
+	findPair(arr, n, sum);
+ 
+	return 0;
+}
+
+
+//Method 2: O(n) Solution using Hashing
+
+#include <iostream>
+#include <unordered_map>
+
+// Function to find a pair in an array with given sum using Hashing
+void findPair(int arr[], int n, int sum)
+{
+	// create an empty map
+	std::unordered_map<int, int> map;
+
+	// do for each element
+	for (int i = 0; i < n; i++)
+	{
+		// check if pair (arr[i], sum-arr[i]) exists
+
+		// if difference is seen before, print the pair
+		if (map.find(sum - arr[i]) != map.end())
+		{
+			std::cout << "Pair found at index " << map[sum - arr[i]] << 
+						 " and " << i;
+			return;
+		}
+
+		// store index of current element in the map
+		map[arr[i]] = i;
+	}
+
+	// we reach here if pair is not found
+	std::cout << "Pair not found";
+}
+
+// Find pair with given sum in the array
+int main()
+{
+	int arr[] = { 8, 7, 2, 5, 3, 1};
+	int sum = 10;
+
+	int n = sizeof(arr)/sizeof(arr[0]);
+
+	findPair(arr, n, sum);
+
+	return 0;
+}