Formatted question description: https://leetcode.ca/all/1427.html
1427. Perform String Shifts
Level
Easy
Description
You are given a string s
containing lowercase English letters, and a matrix shift
, where shift[i] = [direction, amount]
:
direction
can be0
(for left shift) or1
(for right shift).amount
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
0 <= shift[i][0] <= 1
0 <= shift[i][1] <= 100
Solution
Calculate the accumulated amount of shifts. If a right shift is performed, the amount is counted as a negative value. After all operations, the total amount of left shifts is calculated. If the amount is x
where 0 <= x < s.length()
, then the final string is s.substring(x) + s.substring(0, x)
. Return the final string.
class Solution {
public String stringShift(String s, int[][] shift) {
int totalShift = 0;
for (int[] curShift : shift) {
int direction = curShift[0], amount = curShift[1];
if (direction == 1)
amount = -amount;
totalShift += amount;
}
int length = s.length();
totalShift %= length;
if (totalShift < 0)
totalShift += length;
String shifted = s.substring(totalShift) + s.substring(0, totalShift);
return shifted;
}
}