# Perform String Shifts

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).&#x20;
* `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`

```java
class Solution {
    public String stringShift(String s, int[][] shift) {
        int rotateCount = 0;
        
        for (int[] value : shift) {
            if (value[0] == 0) {
                rotateCount += value[1];
            } else {
                rotateCount -= value[1];
            }
        }
        rotateCount = rotateCount % s.length();
        
        if (rotateCount > 0) {
            s = leftShift(s, rotateCount);
        } else if (rotateCount < 0) {
            s = rightShift(s, -(rotateCount));
        }
        return s;
    }
    
    private static String leftShift(String s, int num) {
        return s.substring(num) + s.substring(0, num);
    }
    
    private static String rightShift(String s, int num) {
        return s.substring(s.length() - num) + s.substring(0, s.length() - num);
    }
}
```
