diff --git a/Backtracking/binary_string.py b/Backtracking/binary_string.py
new file mode 100644
index 0000000..acbee29
--- /dev/null
+++ b/Backtracking/binary_string.py
@@ -0,0 +1,22 @@
+# To print all the binary string for a given 'n' value
+
+def binary(n,bi):
+
+    if (n == -1):
+        print ''.join(bi)
+        return
+    else:
+        bi[n] = str(0)
+        binary(n-1, bi)
+        bi[n] = str(1)
+        binary(n-1, bi)
+
+
+n = input()
+bi = []
+
+# Defining a list of size n with dummy values
+for i in range(n):
+    bi.append(-1)
+
+binary(n-1, bi)
diff --git a/Backtracking/coin_denom.py b/Backtracking/coin_denom.py
new file mode 100644
index 0000000..3efd32d
--- /dev/null
+++ b/Backtracking/coin_denom.py
@@ -0,0 +1,34 @@
+"""
+Coin denomination problem - Topdown Memoization Approach (Dynamic Programming)
+Example: Given the coins with denomination [1,2,3]. Assume that there are infinite number of coins. Find the minimum number of coins to get the value 'x'.
+"""
+
+# maxint returns maximum integer value supported python 2.7
+from sys import maxint
+
+
+def coin_denom(coins,dictionary,remain):
+
+    if remain == 0:
+        return 0
+    elif remain < 0:
+        return -1
+
+    if remain in dictionary:
+        return dictionary[remain]
+
+    # In Python 2.7, maximum integer value is 2147483647
+    min_val = maxint
+
+    for i in coins:
+        temp = 1 + coin_denom(coins,dictionary,remain-i)
+        if temp < min_val and temp != 0:
+            min_val = temp
+
+    if remain not in dictionary:
+        dictionary[remain] = min_val
+
+    return min_val
+
+                                ### Testcases ###
+#print coin_denom([1,2,3,4], {}, 7)
diff --git a/Backtracking/n_queen.py b/Backtracking/n_queen.py
new file mode 100644
index 0000000..63029c3
--- /dev/null
+++ b/Backtracking/n_queen.py
@@ -0,0 +1,63 @@
+"""
+N-Queen problem
+Time Complexiy is exponential (it is based on the size of the table)
+Space Complexiy - O(n)
+"""
+
+class N_Queen(object):
+
+    def possibility(self,row,column,q_row,q_column):
+        left_diagonal = q_row + q_column
+        right_diagonal = q_row - q_column
+
+        if q_row == row:
+            return False
+        elif q_column == column:
+            return False
+        elif row + column == left_diagonal:
+            return False
+        elif row - column == right_diagonal:
+            return False
+        else:
+            return True
+
+
+    def queen(self,n,row,previous_queen,result):
+        if row == n:
+            print result
+            return True
+
+        # Iterating the columns
+        for j in range(n):
+            valid = True
+            for q in range(len(previous_queen)):
+                if(not(self.possibility(row,j,previous_queen[q][0],previous_queen[q][1]))):
+                    valid = False
+                    break
+
+            if(valid):
+                previous_queen.append([row,j])
+                result[row][j] = 'Queen'
+                if(self.queen(n,row+1,previous_queen,result)):
+                    return True
+                else:
+                    previous_queen.remove([row,j])
+                    result[row][j] = 0
+
+        return False
+
+
+    def solveNQueens(self, n):
+
+        result = []
+        for i in range(n):
+            result.append([])
+            for j in range(n):
+                # Void is zero - 0 
+                result[i].append(0)
+        previous_queen = []
+        self.queen(n,0,previous_queen,result)
+
+
+obj = N_Queen()
+obj.solveNQueens(4)
diff --git a/Backtracking/string_permutations.py b/Backtracking/string_permutations.py
new file mode 100644
index 0000000..aca29ba
--- /dev/null
+++ b/Backtracking/string_permutations.py
@@ -0,0 +1,19 @@
+"""
+Print all the string permutations using backtracking method.
+"""
+def permutations(string,index,size):
+
+	if index == size:
+		print ''.join(string)
+		return
+
+	for i in range(index,size):
+		string[i], string[index] = string[index], string[i]
+		permutations(string,index+1,size)
+		string[i], string[index] = string[index], string[i]
+
+
+string = raw_input()
+string = list(string)
+size = len(string)
+print permutations(string,0,size)
diff --git a/Bit manipulation/brianKerninghan.py b/Bit manipulation/brianKerninghan.py
new file mode 100644
index 0000000..bdfbb23
--- /dev/null
+++ b/Bit manipulation/brianKerninghan.py	
@@ -0,0 +1,17 @@
+"""
+Brian Kerninghan - Count the number of set bits in an integer
+Time Complexity - O(log n)
+Key idea: n & (n-1)
+"""
+
+def brianKerninghan(n):
+    count = 0
+    while (n):
+        n = n & (n-1)
+        count += 1
+    return count
+
+
+#                               ### Testcases ###
+# print(brianKerninghan(7))
+# print(brianKerninghan(1024))
diff --git a/Bit manipulation/checkBitSet.py b/Bit manipulation/checkBitSet.py
new file mode 100644
index 0000000..96b22d2
--- /dev/null
+++ b/Bit manipulation/checkBitSet.py	
@@ -0,0 +1,15 @@
+"""
+To check whether the m'th bit in an integer is set or not
+"""
+
+def checkBitSet(n, m):
+        aux = 1 << (m-1)
+        if (n & m):
+            return True
+        else:
+            return False
+
+
+#                                   ### Testcases ###
+# print(checkBitSet(5,2))
+# print(checkBitSet(5,3))
diff --git a/Bit manipulation/checkTwoPower.py b/Bit manipulation/checkTwoPower.py
new file mode 100644
index 0000000..008e0c4
--- /dev/null
+++ b/Bit manipulation/checkTwoPower.py	
@@ -0,0 +1,15 @@
+"""
+To check whether a number is divisible by powers of 2
+"""
+
+def checkTwoPower(n):
+    if (not(n & n -1)):
+        return True
+    else:
+        return False
+
+
+#                               ### Testcases ###
+# print(checkTwoPower(8))
+# print(checkTwoPower(5))
+# print(checkTwoPower(1))