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

  1. direction can be 0 (for left shift) or 1 (for right shift). 
  2. amount is the amount by which string s is to be shifted.
  3. A left shift by 1 means remove the first character of s and append it to the end.
  4. 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. 1 <= s.length <= 100
  2. s only contains lower case English letters.
  3. 1 <= shift.length <= 100
  4. shift[i].length == 2
  5. 0 <= shift[i][0] <= 1
  6. 0 <= shift[i][1] <= 100

Solution – Easy to understand

C# Program

public class Solution {
    public string StringShift(string s, int[][] shift) {
        int sum = 0;
        foreach(int[] arr in shift)
        {
            if(arr[0] == 0){
                sum = sum - arr[1];
            }
            else{
                sum = sum + arr[1];
            }
        }
        
        int len = s.Length;
        
        //To order length below string length
        if(sum > len || -sum > len){
            sum = sum % len;
        }
        
        if(sum > 0){
        //Performing right reversal 
            int d = len - sum;
            string result = s.Substring(d) + s.Substring(0,d);
            return result;
        }
        else if(sum < 0){
        //Performing left reversal
            sum = Math.Abs(sum);
            string result = s.Substring(sum) + s.Substring(0,sum);
            return result;
        }
        return s;
    }
}

Code Walk Through

Time Complexity: O(n)

Space Complexity: O(1)

One thought on “Perform String Shifts

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