(Medium) Shifting Letters - LeetCode

Description:

We have a string S of lowercase letters, and an integer array shifts.

Call the shift of a letter, the next letter in the alphabet, (wrapping around so that 'z' becomes 'a'). 

For example, shift('a') = 'b', shift('t') = 'u', and shift('z') = 'a'.

Now for each shifts[i] = x, we want to shift the first i+1 letters of S, x times.

Return the final string after all such shifts to S are applied.

Example 1:

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.

Note:

1. 1 <= S.length = shifts.length <= 20000
2. 0 <= shifts[i] <= 10 ^ 9

Accepted

15,306

Submissions

36,610

 

Solution:

class Solution {
    public String shiftingLetters(String S, int[] shifts) {
      if(S==null||S.length()==0){
          return null;
      }
    if(shifts==null||shifts.length==0){
        return S;
    }
        String res= "";
       
       for(int i = 0; i< shifts.length; i++){
           
           long times = GetTimes(shifts, i);
           
           res = res + Shift(S.charAt(i),times);
       } 
         String sub = S.substring(shifts.length, S.length());
        
        // System.out.println(Shift('m',505870226%26));
        
        return res;
    }
    //97 to 122
    static long GetTimes(int[] arr, int k ){
        long sum = 0;
        for (int i = k; i<arr.length; i++){
            sum = sum + arr[i];
        }
        return sum ; 
    }
    static Character Shift(char a,long k ){
        char tmp = a; 
        k = k%26;
        while (k >0){
         
          if(tmp =='z'){
              tmp = 'a';
          }
            else{
                
               tmp=  (char) ((int)tmp+1);
            }
            
            
            k=k-1;
            
            
        }
        return tmp;
         
    }
}