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Formatted question description: https://leetcode.ca/all/2381.html

# 2381. Shifting Letters II

• Difficulty: Medium.
• Related Topics: Array, String, Prefix Sum.
• Similar Questions: The Skyline Problem, Range Sum Query - Mutable, Range Addition, Shifting Letters, Maximum Population Year, Describe the Painting.

## Problem

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.

## Solution (Java, C++, Python)

• class Solution {
public String shiftingLetters(String s, int[][] shifts) {
int[] diff = new int[s.length() + 1];
int l;
int r;
for (int[] shift : shifts) {
l = shift[0];
r = shift[1] + 1;
diff[l] += 26;
diff[r] += 26;
if (shift[2] == 0) {
diff[l]--;
diff[r]++;
} else {
diff[l]++;
diff[r]--;
}
diff[l] %= 26;
diff[r] %= 26;
}
StringBuilder sb = new StringBuilder();
int current = 0;
int val;
for (int i = 0; i < s.length(); ++i) {
current += diff[i];
val = s.charAt(i) - 'a';
val += current;
val %= 26;
sb.append((char) ('a' + val));
}
return sb.toString();
}
}

############

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();
}
}

• 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)
)

############

# 2381. Shifting Letters II
# https://leetcode.com/problems/shifting-letters-ii/

class Solution:
def shiftingLetters(self, s: str, shifts: List[List[int]]) -> str:
n = len(s)
A = [ord(x) - ord('a') for x in s]
B = [0] * (n + 1)

for start, end, direction in shifts:
c = 1 if direction == 1 else -1

B[start] += c
B[end + 1] -= c

for i in range(1, len(B)):
B[i] += B[i - 1]

for index, (a, b) in enumerate(zip(A, B)):
A[index] += b
A[index] %= 26

return "".join([chr(x + ord("a")) for x in A])


• 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;
}
};

• 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)
}


Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).