Leetcode 2381 Shifting Letters II Solution in Java | Hindi Coding Community

 

You are given a string s of lowercase English letters and a 2D integer array shifts where shifts[i] = [starti, endi, directioni]. For every i, shift the characters in s from the index starti to the index endi (inclusive) forward if directioni = 1, or shift the characters backward if directioni = 0.

Shifting a character forward means replacing it with the next letter in the alphabet (wrapping around so that 'z' becomes 'a'). Similarly, shifting a character backward means replacing it with the previous letter in the alphabet (wrapping around so that 'a' becomes 'z').

Return the final string after all such shifts to s are applied.

 

Example 1:

Input: s = "abc", shifts = [[0,1,0],[1,2,1],[0,2,1]]
Output: "ace"
Explanation: Firstly, shift the characters from index 0 to index 1 backward. Now s = "zac".
Secondly, shift the characters from index 1 to index 2 forward. Now s = "zbd".
Finally, shift the characters from index 0 to index 2 forward. Now s = "ace".

Example 2:

Input: s = "dztz", shifts = [[0,0,0],[1,1,1]]
Output: "catz"
Explanation: Firstly, shift the characters from index 0 to index 0 backward. Now s = "cztz".
Finally, shift the characters from index 1 to index 1 forward. Now s = "catz".

 

Constraints:

• 1 <= s.length, shifts.length <= 5 * 104
• shifts[i].length == 3
• 0 <= starti <= endi < s.length
• 0 <= directioni <= 1
• s consists of lowercase English letters.

Java Code :

public String shiftingLetters(String s, int[][] shifts) {

char[] ch = s.toCharArray();

int[] count = new int[s.length()+1];

for(int[] shift : shifts){

int value = shift[2] == 1 ? 1 : -1;

count[shift[0]] += value;

count[shift[1] + 1] -= value;

}

int sum = 0;

for(int i = 0; i < count.length - 1; i++){

sum += count[i];

int newChar = ((ch[i] - 'a') + sum) % 26;

if(newChar < 0) newChar+= 26;

ch[i] = (char)('a' + newChar);

}

return String.valueOf(ch);

}