LeetCode //C - 848. Shifting Letters

848. Shifting Letters

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:

• $1<=s.length<=10^5$
• s consists of lowercase English letters.
• shifts.length == s.length
• $0<=shifts[i]<=10^9$

From: LeetCode
Link: 848. Shifting Letters

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Solution:

Ideas:

• Each shifts[i] means: shift the first i+1 letters of s by shifts[i].

• Instead of applying each shift step by step (which would be O(n²)), we compute the total shift contribution for each character from the end.

• Observation:

  • The last character (s[n-1]) is shifted by shifts[n-1].
  • The second-to-last (s[n-2]) is shifted by shifts[n-2] + shifts[n-1].
  • In general, character s[i] is shifted by the sum of all shifts from i to n-1.

• Use modulo 26 (% 26) because shifting by multiples of 26 returns to the same character.

Code:

char* shiftingLetters(char* s, int* shifts, int shiftsSize) {long long total = 0;  // use long long because shifts[i] can be up to 1e9int n = shiftsSize;// Traverse from end to startfor (int i = n - 1; i >= 0; i--) {total = (total + shifts[i]) % 26;  // keep mod 26 to prevent overflows[i] = ((s[i] - 'a' + total) % 26) + 'a';}return s;
}