LeetCode-in-Java

3992. Rearrange String to Avoid Character Pair

Easy

You are given a string s and two distinct lowercase English letters x and y.

Rearrange the characters of s to construct a new string t such that:

• t is a permutation of s.
• Every occurrence of y appears before every occurrence of x in t.

Return any valid string t.

Example 1:

Input: s = “aabc”, x = “a”, y = “c”

Output: “cbaa”

Explanation:

The string "cbaa" is a permutation of "aabc", and every occurrence of 'c' appears before every occurrence of 'a'.

Example 2:

Input: s = “dcab”, x = “d”, y = “b”

Output: “cabd”

Explanation:

The string "cabd" is a permutation of "dcab", and every occurrence of 'b' appears before every occurrence of 'd'.

Example 3:

Input: s = “axe”, x = “o”, y = “x”

Output: “axe”

Explanation:

The string "axe" is already valid. Since 'o' does not occur in the string, the required condition is automatically satisfied.

Constraints:

• 1 <= s.length <= 100
• s consists of lowercase English letters.
• x and y are lowercase English letters.
• x != y

Solution

@SuppressWarnings("java:S1172")
public class Solution {
    public String rearrangeString(String s, char x, char y) {
        char[] arr = s.toCharArray();
        int i = 0;
        for (int j = 0; j < arr.length; j++) {
            if (arr[j] == y) {
                char temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
                i++;
            }
        }
        return new String(arr);
    }
}

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