diff --git a/11-05-2020/Ques1.JPG b/11-05-2020/Ques1.JPG
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diff --git a/11-05-2020/Ques1.cpp b/11-05-2020/Ques1.cpp
new file mode 100644
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+++ b/11-05-2020/Ques1.cpp
@@ -0,0 +1,31 @@
+/*
+Question 1:
+Single Number
+Given an array of integers, every element appears twice except for one. Find that single
+one. Note: Your algorithm should have a linear runtime complexity. Could you implement it without
+using extra memory?
+*/
+#include <iostream>
+#include <bits/stdc++.h>
+using namespace std;
+
+int main()
+{   vector<int> a;
+    int n,p,e=0;
+    cout<<"Enter the no. Of Element In array:"<<endl;
+    cin>>n;
+    cout<<"Enter an Array"<<endl;
+    for(int i =0;i<n;i++)
+    {
+        cout<<"Enter Element:";
+        cin>>p;
+        a.push_back(p);
+    }
+
+    for (int i = 0;i<n;i++)
+    {
+        e = e^a[i];
+    }
+    cout<<"The Single One:"<<e<<endl;
+}
+
diff --git a/11-05-2020/Ques2.JPG b/11-05-2020/Ques2.JPG
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diff --git a/11-05-2020/Ques2.cpp b/11-05-2020/Ques2.cpp
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+++ b/11-05-2020/Ques2.cpp
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+/*
+         Square Root of Integer
+Given an integar A. Compute and return the square root of A.
+If A is not a perfect square, return floor(sqrt(A)).
+
+*/
+#include <iostream>
+#include<math.h>
+using namespace std;
+
+int main()
+{
+int n;
+cout<<"Enter the number:";
+cin>>n;
+float h = n;
+float l = 1;
+float mid;
+
+for(int i = 1;i<n;i++)
+{
+  mid = (l+h)/2;
+    if(mid*mid > n)
+    {
+       h = mid-1;
+    }
+    else
+    {if(mid*mid < n)
+      {
+        l = mid+1;
+      }
+    }
+}
+cout<<"Using B search:"<<mid<<endl;
+
+cout<<"Using sqrt fun:"<<sqrt(n);
+}
+
diff --git a/11-05-2020/Ques3.JPG b/11-05-2020/Ques3.JPG
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diff --git a/11-05-2020/Ques3.cpp b/11-05-2020/Ques3.cpp
new file mode 100644
index 0000000..b7d3d73
--- /dev/null
+++ b/11-05-2020/Ques3.cpp
@@ -0,0 +1,22 @@
+/*Question 3:
+Maximum height of the staircase
+Given an integer A representing the square blocks. The height of each square block is 1.
+The task is to create a staircase of max height using these blocks. The first stair would
+ require only one block, the second stair would require two blocks
+and so on. Find and return the maximum height of the staircase
+
+*/
+#include<iostream>
+#include<bits/stdc++.h>
+#include<math.h>
+using namespace std;
+int main(){
+    int n;
+    cout<<"Enter the no. of Blocks:"<<endl;
+    cin>>n;
+    int x;
+    x=  (-1+sqrt(1+8*n))/2;
+    cout<<"The Height is "<<x<<endl;
+
+}
+
diff --git a/11-05-2020/Ques4.JPG b/11-05-2020/Ques4.JPG
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diff --git a/11-05-2020/Ques4.cpp b/11-05-2020/Ques4.cpp
new file mode 100644
index 0000000..d166007
--- /dev/null
+++ b/11-05-2020/Ques4.cpp
@@ -0,0 +1,53 @@
+/*
+
+Count of paths in a grid
+Given an integer A, find and return the number of paths in a grid of size (A x A) that
+starts from (1, 1) and reaches (A, A) without crossing the major diagonal. Since the result can be large, return the result modulo (10^9 + 7)
+*/
+
+#include<iostream>
+#include <bits/stdc++.h>
+using namespace std;
+
+int main()
+{
+    int r;
+    cout<<"Enter A:"<<endl;
+    cin>>r;
+    vector<vector<int>> a(r ,vector<int>(r,0));
+    cout<<"Array:"<<endl;
+     for(int i =0;i<a.size();i++)
+    {   for(int j =0;j<a[i].size();j++)
+    {
+         cout<<"  "<<a[i][j];
+    }
+    cout<<endl;
+    }
+    for(int i =1;i<a.size();i++){
+        a[i][0]=1;
+    }
+
+    for(int i =1;i<a.size();i++)
+    {
+        for(int j =1;j<a.size();j++)
+    {
+        if (i<j)
+        {
+        continue;
+        }
+        else{
+                a[i][j] = a[i-1][j] + a[i][j-1];
+        }
+    }
+    }
+    cout<<"Array:"<<endl;
+  for(int i =0;i<a.size();i++)
+    {   for(int j =0;j<a[i].size();j++)
+    {
+         cout<<"  "<<a[i][j];
+    }
+cout<<endl;
+}
+ cout<<"\n\nTotal no. of path"<<a[a.size()-1][a.size()-1]<<endl;
+
+}
diff --git a/11-05-2020/Ques5.JPG b/11-05-2020/Ques5.JPG
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diff --git a/11-05-2020/Ques5.cpp b/11-05-2020/Ques5.cpp
new file mode 100644
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+++ b/11-05-2020/Ques5.cpp
@@ -0,0 +1,40 @@
+/*
+Given an integer array A of N integers, find the pair of integers in the array which have
+minimum XOR value. Report the minimum XOR value. You have an array A[] with N elements. We have 2 types of operation available on this
+array :
+. We can split a element B into 2 elements C and D such that B = C + D.
+. We can merge 2 elements P and Q as one element R such that R = P^Q i.e XOR of P
+and Q. You have to determine whether it is possible to make array A[] containing only 1
+element 0 after several splits and/or merge?
+*/
+
+#include <iostream>
+#include <bits/stdc++.h>
+using namespace std;
+
+int main()
+{   vector<int> a;
+    int n,p;
+    cout<<"Enter the no. Of Element In array:"<<endl;
+    cin>>n;
+    cout<<"Enter an Array"<<endl;
+    for(int i =0;i<n;i++)
+    {
+        cout<<"Enter Element:";
+        cin>>p;
+        a.push_back(p);
+    }
+
+
+    int i,s,m=INT_MAX;
+    sort(a.begin(),a.end());
+    for(int i =0;i<a.size();i++)
+    {
+        s=a[i]^a[i-1];
+        if(m>s)
+        m=s;
+    }
+    cout<<"Min X-or Value:"<<m<<endl;
+
+}
+
diff --git a/12-05-2020/Q1_gcd.JPG b/12-05-2020/Q1_gcd.JPG
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diff --git a/12-05-2020/Q1_gcd.cpp b/12-05-2020/Q1_gcd.cpp
new file mode 100644
index 0000000..acc1232
--- /dev/null
+++ b/12-05-2020/Q1_gcd.cpp
@@ -0,0 +1,33 @@
+ //The logic behind it is GCD of consecutive numbers is always 1 but if numbers are equal return number
+ //   if(A==B)
+  //      return A;
+ //   else
+ //       return "1";
+//GCD OF A AND B
+#include<iostream>
+using namespace std;
+int gcd(int a,int b)
+{
+    int q,r;
+    q=a/b;
+    r=a%b;
+    if (r==0){
+        return b;
+    }
+    else{
+    gcd(b,r);
+    }
+}
+int main(){
+int x,y,a,b;
+cout<<"Enter the Numbers For GCD:";
+cin>>x>>y;
+cout<<endl;
+a=(x>y?x:y);
+b=((x+y)-a);
+cout<<"Greatest divisor that divides both "<<x<<" and "<<y<< " is " <<(b==1?1:gcd(a,b));
+return 0;
+}
+
+
+
diff --git a/12-05-2020/Q2_ABC.JPG b/12-05-2020/Q2_ABC.JPG
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diff --git a/12-05-2020/Q2_ABC.cpp b/12-05-2020/Q2_ABC.cpp
new file mode 100644
index 0000000..54fe653
--- /dev/null
+++ b/12-05-2020/Q2_ABC.cpp
@@ -0,0 +1,36 @@
+/*
+
+	Scooby has 3 three integers A, B and C.
+ 
+Scooby calls a positive integer special if it is divisible by B and it is divisible by C. You need to tell number of special integers less than or equal to A.
+
+
+*/
+
+#include<iostream>
+using namespace std;
+int gcd(int a,int b)
+{
+    int q,r;
+    q=a/b;
+    r=a%b;
+    if (r==0){
+        return b;
+    }
+    else{
+    gcd(b,r);
+    }
+}
+
+int main(){
+
+    int a,b,c;
+    cout<<"Enter A:";
+    cin>>a;
+    cout<<"\nEnter B:";
+    cin>>b;
+    cout<<"\nEnter C:";
+    cin>>c;
+    int lcm=(b*c)/gcd(b,c);
+    cout<<"Total nos between range 1 to "<<a<<" that are divisible by "<<b<<" and "<<c<<" are:"<<a/lcm;
+}
diff --git a/13-05-2020/Q1_3.JPG b/13-05-2020/Q1_3.JPG
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diff --git a/13-05-2020/Q1_3.cpp b/13-05-2020/Q1_3.cpp
new file mode 100644
index 0000000..f5787fe
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+++ b/13-05-2020/Q1_3.cpp
@@ -0,0 +1,29 @@
+/*
+You have an array lets say a vector or a static array
+  that arrays contains some numbers.You have to remove minimum number of
+   elements from array such that sum of adjacent elements is always even
+
+*/
+#include<bits/stdc++.h>
+using namespace std;
+int main(){
+    int n;
+    cout<<"How May Array elements:"<<endl;
+    cin>>n;
+    vector<int> A(n);
+    cout<<"Enter an Array"<<endl;
+    for(int i =0;i<A.size();i++)
+    {
+        cout<<"Enter Element:";
+        cin>>A[i];
+        cout<<endl;
+    }
+    int i,o=0,e=0;
+    for(i=0;i<A.size();++i){
+        if(A[i]%2==0)
+            ++e;
+        else
+            ++o;
+    }
+    cout<<"Minimum number of elements to be removed is "<<min(o,e)<<endl;
+}
diff --git a/13-05-2020/Q2_3.JPG b/13-05-2020/Q2_3.JPG
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diff --git a/13-05-2020/Q2_3.py b/13-05-2020/Q2_3.py
new file mode 100644
index 0000000..bce9795
--- /dev/null
+++ b/13-05-2020/Q2_3.py
@@ -0,0 +1,32 @@
+#For an given array,count the number of subarrays with even number only.
+
+def even(l):
+    f = False
+    for i in l:
+        if i % 2 == 0:
+            f = True
+        else:
+            f = False
+            break
+    return f
+
+n = int(input("Enter No. of elements:"))
+l = []
+print("Enter Array:")
+for i in range(n):
+    l.append(int(input("Enter Elements:")))
+
+print(l)
+sub_arr = []
+for i in range(0, n):
+    for j in range(i, n):
+        temp = []
+        for k in range(i, j + 1):
+            temp += [l[k]]
+        sub_arr += [temp]
+
+sub_arr_even = []
+sub_arr_even += list(filter(even, sub_arr))
+
+print("Sub Arrays:", sub_arr_even)
+print("Total:", len(sub_arr_even))
diff --git a/13-05-2020/Q3_3.JPG b/13-05-2020/Q3_3.JPG
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diff --git a/13-05-2020/Q3_3.cpp b/13-05-2020/Q3_3.cpp
new file mode 100644
index 0000000..aa3138a
--- /dev/null
+++ b/13-05-2020/Q3_3.cpp
@@ -0,0 +1,30 @@
+//GCD OF A AND B + LCM
+
+#include<iostream>
+using namespace std;
+int gcd(int a,int b)
+{
+    int q,r;
+    q=a/b;
+    r=a%b;
+    if (r==0){
+        return b;
+    }
+    else{
+    gcd(b,r);
+    }
+}
+int main(){
+int x,y,a,b;
+cout<<"Enter the Numbers For GCD:";
+cin>>x>>y;
+cout<<endl;
+a=(x>y?x:y);
+b=((x+y)-a);
+int gd = (b==1?1:gcd(a,b));
+cout<<"Greatest divisor that divides both "<<x<<" and "<<y<< " is " <<gd;
+cout<<"\nLCM IS:"<<(a*b)/gd;
+return 0;
+}
+
+
diff --git a/13-05-2020/Q4_3.JPG b/13-05-2020/Q4_3.JPG
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diff --git a/13-05-2020/Q4_3.cpp b/13-05-2020/Q4_3.cpp
new file mode 100644
index 0000000..5575681
--- /dev/null
+++ b/13-05-2020/Q4_3.cpp
@@ -0,0 +1,50 @@
+/*
+Print all the magic numbers from 1 to 1000
+( a magic number is a number whose sum of digits is always 1 provided sum is less than 10 otherwise we again have to compute sum of digits of the sum itself)
+For example :
+N=10     (1+0=1) Magic Number
+
+N=897  sum=(8+9+7=24
+          Again 2+4=6 which is not equal to 1) so output is not magic number
+
+*/
+/*bool isMagic(int n)
+{
+    int sum = 0;
+
+
+    while (n > 0 || sum > 9)
+    {
+        if (n == 0)
+        {
+            n = sum;
+            sum = 0;
+        }
+        sum += n % 10;
+        n /= 10;
+    }
+
+    return (sum == 1);
+}
+*/
+
+#include<bits/stdc++.h>
+using namespace std;
+int main(){
+    int n = 1000;
+    int i = 1;
+    cout<<"All Magic Numbers Between 1 to 1000:"<<endl;
+    for(i;i<= n;i++)
+    {
+        if (i % 9 == 1){
+            cout<<i<<"\t";
+        }
+       /* cout<<"\t";
+       if(isMagic(i))
+            cout<<i;      // Bruteforce
+        cout<<"\n";*/
+    }
+
+return 0;
+}
+