LeetCode 2381. Shifting Letters II Solution in Java, C++, Python & More | Explanation + Code

Description

You are given a string s of lowercase English letters and a 2D integer array shifts where shifts[i] = [starti, endi, directioni]. For every i, shift the characters in s from the index starti to the index endi (inclusive) forward if directioni = 1, or shift the characters backward if directioni = 0.

Shifting a character forward means replacing it with the next letter in the alphabet (wrapping around so that 'z' becomes 'a'). Similarly, shifting a character backward means replacing it with the previous letter in the alphabet (wrapping around so that 'a' becomes 'z').

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

 

Example 1:

Input: s = "abc", shifts = [[0,1,0],[1,2,1],[0,2,1]]
Output: "ace"
Explanation: Firstly, shift the characters from index 0 to index 1 backward. Now s = "zac".
Secondly, shift the characters from index 1 to index 2 forward. Now s = "zbd".
Finally, shift the characters from index 0 to index 2 forward. Now s = "ace".

Example 2:

Input: s = "dztz", shifts = [[0,0,0],[1,1,1]]
Output: "catz"
Explanation: Firstly, shift the characters from index 0 to index 0 backward. Now s = "cztz".
Finally, shift the characters from index 1 to index 1 forward. Now s = "catz".

 

Constraints:

• 1 <= s.length, shifts.length <= 5 * 104
• shifts[i].length == 3
• 0 <= starti <= endi < s.length
• 0 <= directioni <= 1
• s consists of lowercase English letters.

Solutions

Solution 1

PythonJavaC++GoTypeScript

class Solution: def shiftingLetters(self, s: str, shifts: List[List[int]]) -> str: n = len(s) d = [0] * (n + 1) for i, j, v in shifts: if v == 0: v = -1 d[i] += v d[j + 1] -= v for i in range(1, n + 1): d[i] += d[i - 1] return ''.join( chr(ord('a') + (ord(s[i]) - ord('a') + d[i] + 26) % 26) for i in range(n) )(code-box)

class Solution { public String shiftingLetters(String s, int[][] shifts) { int n = s.length(); int[] d = new int[n + 1]; for (int[] e : shifts) { if (e[2] == 0) { e[2]--; } d[e[0]] += e[2]; d[e[1] + 1] -= e[2]; } for (int i = 1; i <= n; ++i) { d[i] += d[i - 1]; } StringBuilder ans = new StringBuilder(); for (int i = 0; i < n; ++i) { int j = (s.charAt(i) - 'a' + d[i] % 26 + 26) % 26; ans.append((char) ('a' + j)); } return ans.toString(); } }(code-box)

class Solution { public: string shiftingLetters(string s, vector<vector<int>>& shifts) { int n = s.size(); vector<int> d(n + 1); for (auto& e : shifts) { if (e[2] == 0) { e[2]--; } d[e[0]] += e[2]; d[e[1] + 1] -= e[2]; } for (int i = 1; i <= n; ++i) { d[i] += d[i - 1]; } string ans; for (int i = 0; i < n; ++i) { int j = (s[i] - 'a' + d[i] % 26 + 26) % 26; ans += ('a' + j); } return ans; } };(code-box)

func shiftingLetters(s string, shifts [][]int) string { n := len(s) d := make([]int, n+1) for _, e := range shifts { if e[2] == 0 { e[2]-- } d[e[0]] += e[2] d[e[1]+1] -= e[2] } for i := 1; i <= n; i++ { d[i] += d[i-1] } ans := []byte{} for i, c := range s { j := (int(c-'a') + d[i]%26 + 26) % 26 ans = append(ans, byte('a'+j)) } return string(ans) }(code-box)

function shiftingLetters(s: string, shifts: number[][]): string { const n: number = s.length; const d: number[] = new Array(n + 1).fill(0); for (let [i, j, v] of shifts) { if (v === 0) { v--; } d[i] += v; d[j + 1] -= v; } for (let i = 1; i <= n; ++i) { d[i] += d[i - 1]; } let ans: string = ''; for (let i = 0; i < n; ++i) { const j = (s.charCodeAt(i) - 'a'.charCodeAt(0) + (d[i] % 26) + 26) % 26; ans += String.fromCharCode('a'.charCodeAt(0) + j); } return ans; }(code-box)