Formatted question description: https://leetcode.ca/all/1427.html

# 1427. Perform String Shifts

Easy

## Description

You are given a string s containing lowercase English letters, and a matrix shift, where shift[i] = [direction, amount]:

• direction can be 0 (for left shift) or 1 (for right shift).
• amount is the amount by which string s 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;
}
}