Week3 Level Easy Problems

Week2/Swarnima_Shishodia/Que1.cpp

@@ -0,0 +1,38 @@
+#include<iostream>
+#include<string.h>
+using namespace std;
+int main()
+{
+    /*variable s store's the string input by the user
+    p stores the modified string
+    w stores the final string that we need to print*/
+
+string s,p="",w="";
+cin>>s;
+//concatenated '.' at left side of string so that last word can be concatenated when . appears
+s="."+s;
+//q stores the length of the string
+//i is a loop variable
+int q=s.size()-1,i;
+//traversing the string from right side
+for(i=s.size()-1;i>=0;i--)
+{
+    //When . appears the this part will get executed
+    if(s[i]=='.')
+    {
+    int j=i;
+    //all the characters to the right of the . get stored till next . appears
+    while(j<=q)
+    {
+    p=p+s[j];
+    j=j+1;
+    }
+    q=i-1;// so that the size of remaining string is stored
+    }
+}
+// As the first character is a dot which I have concatenated so run a loop from 1 position
+for(int i=1;i<p.size();i++)
+    w=w+p[i];//concatenate the characters in w and print them
+cout<<w<<endl;
+return 0;
+}

Week2/Swarnima_Shishodia/Que2.cpp

@@ -0,0 +1,15 @@
+#include <bits/stdc++.h>
+using namespace std;
+void allpermutation(string str)
+{
+    sort(str.begin(),str.end());//Sort the string lexicographically
+     while(next_permutation(str.begin(),str.end()))// nextt_permutation function prints all the permutations of a string
+     cout << str << endl;
+}
+int main()
+{
+    string str;
+    cin>>str;//Input the string
+    allpermutation(str);//Call the function
+    return 0;
+}

Week2/Swarnima_Shishodia/Que2I.cpp

@@ -0,0 +1,36 @@
+#include <iostream>
+using namespace std;
+int main()
+{
+int t;
+	cin>>t;
+	while(t--)
+	{
+	    string st;
+	    cin>>st;
+	    int s=4,b=1,c=0;
+	    for(int i=0;i<st.size();i++)
+	    {
+	        if(st[i]=='W')
+            {
+	        s=s+b;
+	        b=1;
+            }
+	        else
+	        {
+	            s=s-b;
+	            b=b*2;
+	        if(s<b&&i!=st.size()-1)
+            {
+	            c=1;
+	            break;
+	        }
+	        }
+	    }
+	    if(c==1)
+	    cout<<-1<<endl;
+	    else
+	    cout<<s<<endl;
+	}
+	return 0;
+}

Week2/Swarnima_Shishodia/Que3.cpp

@@ -0,0 +1,37 @@
+#include<iostream>
+using namespace std;
+int main()
+{
+    string c,d;
+    cin>>c;
+    cin>>d;
+    //Make two arrays of length 26 mark all the position in both the arrays equal to 1
+    int a[26],b[26];
+    for(int i=0;i<26;i++)
+    {
+        a[i]=0;
+        b[i]=0;
+    }
+    int i,j=0;
+    //increase the array position by 1 when character appears, After loop gets over It will give each character numbers
+    //Suppose 'a' comes then its ASCII value is 97, 97-97=0 at 0th position 1 gets stored .
+    //In similar way 1st position represents 'b' as 98-97=1
+    for(i=0;i<c.size();i++)
+        a[int(c[i])-97]=a[int(c[i])-97]+1;
+
+    for(i=0;i<d.size();i++)
+        b[int(d[i])-97]=b[int(d[i])-97]+1;
+
+    //Check whether array a and b are equal if true then print yes a otherwise print no
+    for(i=0;i<26;i++)
+    {
+        if(a[i]==b[i])
+            j=j+1;
+    }
+    if(j==26)
+        cout<<"Yes\n";
+    else
+        cout<<"No\n";
+
+return 0;
+}

Week2/Swarnima_Shishodia/Que4.cpp

@@ -0,0 +1,27 @@
+#include<iostream>
+#include<algorithm>
+#include<string.h>
+using namespace std;
+int main()
+{
+    string s;
+    getline(cin,s);//to input a space separated string
+    int a[257];
+    for(int i=0;i<257;i++)
+        a[i]=0;
+
+/* Make a array for 257 size, whenever a character comes, its ASCII value is calculated and 1 is stored at the same position in the array.
+Print the character, and store 1 at that position when the same character  will come reappear
+then it will not enter into the if position as it already 1*/
+   for(int i=0;i<s.size();i++)
+    {
+      if(a[int(s[i])]==0)
+      {
+          cout<<s[i];
+          a[int(s[i])]=1;
+      }
+      if(s[i]==' ')
+        cout<<' ';
+    }
+    return 0;
+}

Week2/Swarnima_Shishodia/Que4I.cpp

@@ -0,0 +1,42 @@
+#include<bits/stdc++.h>
+using namespace std;
+int main()
+ {
+	int t;
+	cin>>t;
+	while(t--)
+        {
+	    int n;
+	    cin>>n;
+	    vector<int> v;
+	    //Store the digits of a number in vector
+	    while(n!=0)
+	    {
+	        v.push_back(n%10);
+	        n/=10;
+	    }
+	    //Sort the vector
+	    sort(v.begin(), v.end());
+	    int len=v.size();//Store the length of a vector
+	    bool found=false;
+	     while(next_permutation(v.begin(), v.end()))// Using next_permutation find all the permutation of a number
+	     {
+    	     int sum=0;
+    	     for(int j=0;j<len;j++)
+    	     {
+    	         sum=sum*10+v[j];
+    	     }//Form the number
+    	     if(sum%8 == 0)//If it is divisible by 8 then print found
+    	     {
+    	         found = true;
+    	         break;
+    	     }
+	     }
+	     //Print Yes if any permutation is divisible by 8 else print no
+	    if(found)
+	        cout<<"Yes\n";
+	    else
+	        cout<<"No\n";
+	}
+	return 0;
+}

Week3/Que1Array.cpp

@@ -0,0 +1,30 @@
+#include <iostream>
+using namespace std;
+int main() {
+	int t,n,m,i,j,num1,num2;
+	cin>>t;
+	for(i=0;i<t;i++)
+	{
+	    cin>>n;
+	    int a[n];
+	    num1=0;
+	    num2=0;
+	    for(j=0;j<n;j++)//to input first array
+	    cin>>a[j];
+
+	    cin>>m;
+	    int b[m];
+	    for(j=0;j<m;j++)//to input second array
+	    cin>>b[j];
+
+	    for(j=0;j<n;j++)
+	    num1=a[j]+num1*10;//to form number from 1st array take digit and multiply the number with 10 everytime
+
+
+	    for(j=0;j<m;j++)
+	    num2=b[j]+num2*10;//to form number from 2nd array take each digit and multiply the number by 10 and add it to number
+
+	    cout<<(num1+num2)<<endl;
+	}
+	return 0;
+}

Week3/Que1MediumArray.cpp

@@ -0,0 +1,30 @@
+#include <iostream>
+#include<climits>
+//To find the contiguous subarray with maximum sum
+//Decalre two variables max_so_far and max_ending_here
+//max_so_far stores the maximum sum till now
+//max_ending_here considers the place where max_so_far get decreased
+using namespace std;
+int main()
+{
+    int t,n;
+    cin>>t;
+    for(int i=0;i<t;i++)
+    {
+        cin>>n;
+        long long a[n];
+        for(int j=0;j<n;j++)
+        cin>>a[j];
+    int max_so_far=INT_MIN,max_ending_here=0;
+    for (int i=0;i<n;i++)
+    {
+        max_ending_here = max_ending_here+a[i];
+        if (max_so_far < max_ending_here)// If the current sum is greater then make max_so_far equal to curent sum value
+            max_so_far = max_ending_here;
+
+        if (max_ending_here < 0)
+            max_ending_here = 0;
+    }
+    cout<<max_so_far<<endl;
+    }
+}

Week3/Que2Array.cpp

@@ -0,0 +1,56 @@
+#include <iostream>
+using namespace std;
+//1. Reverse from 0 to d-1
+//2. Reverse from d to n-1
+//3. Reverse from 0 to n-1
+//4. Print the output
+int main() {
+	int t,i,j,n,d;
+	cin>>t;
+	for(i=0;i<t;i++)
+	{
+	    cin>>n;
+	    cin>>d;
+	    int a[n];
+	    for(j=0;j<n;j++)
+	    cin>>a[j];
+	    j=0;
+	    int k=d-1;
+	    // Reverse from 0 to d-1
+	    while(j<=k)
+	    {
+	        int temp=a[j];
+	        a[j]=a[k];
+	        a[k]=temp;
+	        j=j+1;
+	        k=k-1;
+	    }
+	    j=d;
+	    k=n-1;
+	    //Reverse from d to n-1
+	    while(j<=k)
+	    {
+	        int temp=a[j];
+	        a[j]=a[k];
+	        a[k]=temp;
+	        j=j+1;
+	        k=k-1;
+	    }
+	    j=0;
+	    k=n-1;
+	    //Reverse the whole array
+	    while(j<=k)
+	    {
+	       int temp=a[j];
+	        a[j]=a[k];
+	        a[k]=temp;
+	        j=j+1;
+	        k=k-1;
+	    }
+	    for(j=0;j<n;j++)
+	    cout<<a[j]<<" ";
+	    cout<<endl;
+
+	}
+	return 0;
+}

Week3/Que3Array.cpp

@@ -0,0 +1,16 @@
+#include <iostream>
+using namespace std;
+//Logic - Traverse the string from right to left and print the final output
+int main() {
+	int t,i,j;
+	cin>>t;
+	for(i=0;i<t;i++)
+	{
+	    string s;
+	    cin>>s;
+	    for(j=s.length()-1;j>=0;j--)
+	    cout<<s[j];
+	    cout<<endl;
+	}
+	return 0;
+}

Week3/Que4Array.cpp

@@ -0,0 +1,44 @@
+// Use sliding window technique
+// Keep on adding the elements if sum of elements is smaller than um we are looking for
+// Subtract the leftmost element if sum of elements is greater than sum we are looking for
+// If equal print it, Maintain the variable for left variable
+#include<iostream>
+using namespace std;
+int main()
+{
+    int t,s,i,j,p,q;
+    long long int sum=0,n;
+    cin>>t;
+    for(i=0;i<t;i++)
+    {
+        cin>>n;
+        cin>>s;
+       long long int a[n];
+        for(j=0;j<n;j++)
+        cin>>a[j];
+        p=0;
+        q=0;
+        sum=0;
+        while(q<n)
+        {
+            while(sum<s)
+            {
+            sum=sum+a[q];
+                        q=q+1;
+            }
+            while(sum>s)
+            {
+            sum=sum-a[p];
+            p=p+1;
+            }
+            if(sum==s)
+            break;
+        }
+
+        if(sum==s)
+        cout<<p+1<<" "<<q<<endl;
+        else
+        cout<<"-1"<<endl;
+    }
+    return 0;
+}