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

Solution 1: Suffix Sum

For each character in the string $s$, we need to calculate its final shift amount, which is the sum of $\textit{shifts}[i]$, $\textit{shifts}[i + 1]$, $\textit{shifts}[i + 2]$, and so on. We can use the concept of suffix sum, traversing $\textit{shifts}$ from back to front, calculating the final shift amount for each character, and then taking modulo $26$ to get the final character.

The time complexity is $O(n)$, where $n$ is the length of the string $s$. Ignoring the space consumption of the answer, the space complexity is $O(1)$.

Python3JavaC++Go

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)

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;
    }
};

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)
}

848. Shifting Letters

Description

Solutions

Solution 1: Suffix Sum

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