Merge pull request #228 from RAUNAK-PANDEY/Raunak

Added 6 programs in Bit Manipulation directory #221

Author: smv1999

Bit Manipulation/Bit_Difference.cpp

@@ -0,0 +1,51 @@
+ 
+/*
+You are given two numbers A and B. The task is to count the number of bits needed to be flipped to convert A to B.
+
+Example 1:
+
+Input: A = 10, B = 20
+Output: 4
+Explanation:
+A  = 01010
+B  = 10100
+As we can see, the bits of A that need 
+to be flipped are 01010. If we flip 
+these bits, we get 10100, which is B.*/
+//Initial Template for C++
+
+#include<bits/stdc++.h>
+using namespace std;
+ 
+// Function to find number of bits to be flip
+// to convert A to B
+int countBitsFlip(int a, int b){
+    
+    // Your logic here
+    int ans=a^b;
+    int c=0;
+    while(ans>0)
+    {
+        ans &=(ans-1);
+        c++;
+    }
+     return c;   
+    
+}
+
+// { Driver Code Starts.
+
+// Driver Code
+int main()
+{
+	int t;
+	cin>>t;// input the testcases
+	while(t--) //while testcases exist
+	{
+		int a,b;
+		cin>>a>>b; //input a and b
+
+		cout<<countBitsFlip(a, b)<<endl;
+	}
+	return 0;
+}  // } Driver Code Ends

Bit Manipulation/Count_total_set_bits.cpp

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+
+/*
+You are given a number N. Find the total count of set bits for all numbers from 1 to N(both inclusive).
+
+Example 1:
+
+Input: N = 4
+Output: 5
+Explanation:
+For numbers from 1 to 4.
+For 1: 0 0 1 = 1 set bits
+For 2: 0 1 0 = 1 set bits
+For 3: 0 1 1 = 2 set bits
+For 4: 1 0 0 = 1 set bits
+Therefore, the total set bits is 5.*/
+
+#include<bits/stdc++.h>
+using namespace std;
+ 
+// Function to count set bits in the given number x
+// n: input to count the number of set bits
+int largestpowerof2inrange(int n)
+{
+    int x=0;
+    while((1<<x) <=n)
+    {
+        x++;
+    }
+    return x-1;
+}
+int countSetBits(int n)
+{
+    // Your logic 
+    if(n==0) return 0;
+    int x=largestpowerof2inrange(n);
+    int first= x*(1<<(x-1));
+    int second= n-(1<<x)+1;
+    int third=n-(1<<x);
+    int ans=first+second+countSetBits(third);
+    return ans;
+}
+ 
+// Driver code
+int main()
+{
+	 int t;
+	 cin>>t;// input testcases
+	 while(t--) //while testcases exist
+	 {
+	       int n;
+	       cin>>n; //input n
+	       
+	       cout << countSetBits(n) << endl;// print the answer
+	  }
+	  return 0;
+}
+  // } Driver Code Ends

Bit Manipulation/Find_position_of_set_bit.cpp

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+
+/*
+Given a number N having only one ‘1’ and all other ’0’s in its binary representation, find position of the only set bit. If there are 0 or more than 1 set bit the answer should be -1. Position of  set bit '1' should be counted starting with 1 from LSB side in binary representation of the number.
+
+Example 1:
+
+Input:
+N = 2
+Output:
+2
+Explanation:
+2 is represented as "10" in Binary.
+As we see there's only one set bit
+and it's in Position 2 and thus the
+Output 2.*/
+#include <bits/stdc++.h>
+using namespace std;
+ 
+
+class Solution {
+  public:
+  
+  int isPowerOfTwo(unsigned N) 
+    { 
+        return N && (!(N & (N - 1))); 
+    } 
+    int findPosition(int N) {
+        // code here
+        if (!isPowerOfTwo(N)) 
+            return -1; 
+  
+        unsigned i = 1, pos = 1; 
+      
+        
+        while (!(i & N)) { 
+            
+            i = i << 1; 
+      
+            // increment position 
+            ++pos; 
+        } 
+      
+        return pos; 
+        
+        }
+};
+
+// { Driver Code Starts.
+int main() {
+    int t;
+    cin >> t;
+    while (t--) {
+        int N;
+
+        cin>>N;
+
+        Solution ob;
+        cout << ob.findPosition(N) << endl;
+    }
+    return 0;
+}  // } Driver Code Ends

Bit Manipulation/Non_Repeating_Numbers.cpp

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+
+/*
+Given an array A containing 2*N+2 positive numbers, out of which 2*N numbers exist in pairs whereas the other two number occur exactly once and are distinct. Find the other two numbers.
+
+
+Example 1:
+
+Input: 
+N = 2
+arr[] = {1, 2, 3, 2, 1, 4}
+Output:
+3 4 
+Explanation:
+3 and 4 occur exactly once.*/
+#include<bits/stdc++.h>
+using namespace std;
+ 
+
+class Solution
+{
+public:
+    vector<int> singleNumber(vector<int> nums) 
+    {
+        // Code here.
+        int res=0;
+        vector <int> vec,v;
+        for(int i=0;i<nums.size();i++)
+        {
+            res^=nums[i];
+            
+        }
+        int set_bit_no = res & ~(res-1);
+        int temp=res;
+            for(int i=0;i<nums.size();i++)
+            {
+                if((nums[i]& set_bit_no))
+                {
+                    vec.push_back(nums[i]);
+                }
+            }
+            for(int i=0;i<vec.size();i++)
+            {
+               temp^=vec[i]; 
+            }
+            v.push_back(temp);
+            res^=temp;
+            v.push_back(res);
+            sort(v.begin(),v.end());
+            return v;
+            
+    }
+};
+
+// { Driver Code Starts.
+int main(){
+    int T;
+    cin >> T;
+    while(T--)
+    {
+    	int n; 
+    	cin >> n;
+    	vector<int> v(2 * n + 2);
+    	for(int i = 0; i < 2 * n + 2; i++)
+    		cin >> v[i];
+    	Solution ob;
+    	vector<int > ans = ob.singleNumber(v);
+    	for(auto i: ans)
+    		cout << i << " ";
+    	cout << "\n";
+    }
+	return 0;
+}  // } Driver Code Ends

Bit Manipulation/Number_of_1_Bits.cpp

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+
+/* Given a positive integer N, print count of set bits in it. 
+
+Example 1:
+
+Input:
+N = 6
+Output:
+2
+Explanation:
+Binary representation is '110' 
+So the count of the set bit is 2.*/
+#include<bits/stdc++.h> 
+using namespace std; 
+
+  
+
+class Solution
+{
+public:
+    int setBits(int N)
+    {
+       int count =0;   
+        while (N) { 
+            N &=(N-1);
+            count ++;
+              
+        } 
+        return count; 
+     }
+};
+
+// { Driver Code Starts.
+int main()
+{
+    int t;
+    cin >> t;
+    while (t--)
+    {
+        int N;
+        cin >> N;
+        
+        Solution ob;
+        int cnt = ob.setBits(N);
+        cout << cnt << endl;
+    }
+    return 0;
+}
+  // } Driver Code Ends

Bit Manipulation/Power_of_2.cpp

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+ /*
+Given a non-negative integer N. The task is to check if N is a power of 2. More formally, check if N can be expressed as 2x for some x.
+ 
+
+Example 1:
+
+Input: N = 1
+Output: true
+Explanation:
+1 is equal to 2 raised to 0 (20 = 1).*/
+//Initial Template for C++
+
+#include<bits/stdc++.h>
+using namespace std;
+ 
+// Function to check power of two
+bool isPowerofTwo(long long n){
+    
+    // Your code here 
+    if(n==0)
+        return 0;
+        
+    else if(n>0)
+    {
+        if((n & n-1)==0)
+            return 1;
+    }
+    return 0;
+}
+ 
+// Driver code
+int main()
+{
+
+    int t;
+    cin>>t;//testcases
+
+    for(int i=0;i<t;i++)
+    {
+        long long n; //input a number n
+        cin>>n;
+
+         if(isPowerofTwo(n))//Now, if log2 produces an integer not decimal then we are sure raising 2 to this value
+             cout<<"YES"<<endl;
+         else
+            cout<<"NO"<<endl;
+
+    }
+
+    return 0;
+}
+  // } Driver Code Ends