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 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;
}
}
```