Shifting Letters: —

Input: s = "abc", shifts = [3,5,9]
Output: "rpl"
Explanation: We start with "abc".
After shifting the first 1 letters of s by 3, we have "dbc".
After shifting the first 2 letters of s by 5, we have "igc".
After shifting the first 3 letters of s by 9, we have "rpl", the answer.

CODE: →

string shiftingLetters(string S, vector<int>& shifts) 
{
    for(int i = shifts.size() - 2 ;i >=0; i--)
        shifts[i] = (shifts[i] + shifts[i + 1]) % 26;
    for(int i = 0; i < S.size(); i++) 
        S[i] = ( ((S[i] - 'a') + shifts[i]) % 26 + 'a');    return S;
}

Detect Loop in linked list: —

Input:
N = 4
value[] = {3,2,0,-4}
x = 1
Output: true
Explanation: There is a cycle in the linked list, where the tail connects to the 2nd node (1-indexed).

Input:
N = 3
value[] = {1,3,4}
x = 2
Output: true
Explanation: In above test case N = 3. 
The linked list with nodes N = 3 is given. Then value of x=2 is given which means last node is connected with xth node of linked list. Therefore, there exists a loop.

CODE: →

bool detectLoop(Node* head)    
{        
    // your code here        
    unordered_set<Node*> s;        
    while (head != NULL)        
    {
        if (s.find(head) != s.end())       
            return true; 
          
        s.insert(head);           
        head = head->next;        
    }
    return false;    
}

bool detectLoop(Node* head)    
{        
    // your code here        
    Node* slow = head;    
    Node* fast = head;               while(fast != NULL && fast->next != NULL)   
   {        
       slow = slow->next;       
       fast = fast->next->next;                        if(slow == fast)           
           return true;        
   }       
   return false;    
}

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