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848. Shifting Letters

Description

You are given a string s of lowercase English letters and an integer array shifts of the same length.

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.

Example 2:

Input: s = "aaa", shifts = [1,2,3]
Output: "gfd"

 

Constraints:

  • 1 <= s.length <= 105
  • s consists of lowercase English letters.
  • shifts.length == s.length
  • 0 <= shifts[i] <= 109

Solutions

  • class Solution {
        public String shiftingLetters(String s, int[] shifts) {
            char[] cs = s.toCharArray();
            int n = cs.length;
            long t = 0;
            for (int i = n - 1; i >= 0; --i) {
                t += shifts[i];
                int j = (int) ((cs[i] - 'a' + t) % 26);
                cs[i] = (char) ('a' + j);
            }
            return String.valueOf(cs);
        }
    }
    
  • class Solution {
    public:
        string shiftingLetters(string s, vector<int>& shifts) {
            long long t = 0;
            int n = s.size();
            for (int i = n - 1; ~i; --i) {
                t += shifts[i];
                int j = (s[i] - 'a' + t) % 26;
                s[i] = 'a' + j;
            }
            return s;
        }
    };
    
  • class Solution:
        def shiftingLetters(self, s: str, shifts: List[int]) -> str:
            n, t = len(s), 0
            s = list(s)
            for i in range(n - 1, -1, -1):
                t += shifts[i]
                j = (ord(s[i]) - ord('a') + t) % 26
                s[i] = ascii_lowercase[j]
            return ''.join(s)
    
    
  • func shiftingLetters(s string, shifts []int) string {
    	t := 0
    	n := len(s)
    	cs := []byte(s)
    	for i := n - 1; i >= 0; i-- {
    		t += shifts[i]
    		j := (int(cs[i]-'a') + t) % 26
    		cs[i] = byte('a' + j)
    	}
    	return string(cs)
    }
    

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