1427. Perform String Shifts π
Description
You are given a string s
containing lowercase English letters, and a matrix shift
, where shift[i] = [directioni, amounti]
:
directioni
can be0
(for left shift) or1
(for right shift).amounti
is the amount by which strings
is to be shifted.- A left shift by 1 means remove the first character of
s
and append it to the end. - Similarly, a right shift by 1 means remove the last character of
s
and add it to the beginning.
Return the final string after all operations.
Example 1:
Input: s = "abc", shift = [[0,1],[1,2]] Output: "cab" Explanation: [0,1] means shift to left by 1. "abc" -> "bca" [1,2] means shift to right by 2. "bca" -> "cab"
Example 2:
Input: s = "abcdefg", shift = [[1,1],[1,1],[0,2],[1,3]] Output: "efgabcd" Explanation: [1,1] means shift to right by 1. "abcdefg" -> "gabcdef" [1,1] means shift to right by 1. "gabcdef" -> "fgabcde" [0,2] means shift to left by 2. "fgabcde" -> "abcdefg" [1,3] means shift to right by 3. "abcdefg" -> "efgabcd"
Constraints:
1 <= s.length <= 100
s
only contains lower case English letters.1 <= shift.length <= 100
shift[i].length == 2
directioni
is either0
or1
.0 <= amounti <= 100
Solutions
Solution 1: Simulation
We can denote the length of the string $s$ as $n$. Next, we traverse the array $shift$, accumulate to get the final offset $x$, then take $x$ modulo $n$, the final result is to move the first $n - x$ characters of $s$ to the end.
The time complexity is $O(n + m)$, where $n$ and $m$ are the lengths of the string $s$ and the array $shift$ respectively. The space complexity is $O(1)$.
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