LeetCode - Perform String Shifts - 30Days Challenge

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

As we know that one left shift and right shift will cancel each other, We can take advantage of it and count number of left and right shifts. Just shift Characters only difference of Mod(LShiftCnt-RShiftCnt)

Code


 class Solution {
    public String stringShift(String s, int[][] shift) {
       if (s.length()<2) return s;

		int rCnt = 0, lCnt = 0;

		for (int[] sArr : shift) {
			if (sArr[0] == 0)
				lCnt += sArr[1];
			else
				rCnt += sArr[1];
		}
        
		if (rCnt == lCnt)
			return s;

		if (rCnt > lCnt) {
			rCnt = rCnt - lCnt;
			rCnt = rCnt % s.length();
			return s.substring(s.length()-rCnt) + s.substring(0, s.length()-rCnt);
		} else {
			lCnt = lCnt - rCnt;
			lCnt = lCnt % s.length();
			return s.substring(lCnt) + s.substring(0,lCnt);
		}
    }
}