LeetCode-in-Java

3722. Lexicographically Smallest String After Reverse

Medium

You are given a string s of length n consisting of lowercase English letters.

You must perform exactly one operation by choosing any integer k such that 1 <= k <= n and either:

• reverse the first k characters of s, or
• reverse the last k characters of s.

Return the lexicographically smallest string that can be obtained after exactly one such operation.

A string a is lexicographically smaller than a string b if, at the first position where they differ, a has a letter that appears earlier in the alphabet than the corresponding letter in b. If the first min(a.length, b.length) characters are the same, then the shorter string is considered lexicographically smaller.

Example 1:

Input: s = “dcab”

Output: “acdb”

Explanation:

• Choose k = 3, reverse the first 3 characters.
• Reverse "dca" to "acd", resulting string s = "acdb", which is the lexicographically smallest string achievable.

Example 2:

Input: s = “abba”

Output: “aabb”

Explanation:

• Choose k = 3, reverse the last 3 characters.
• Reverse "bba" to "abb", so the resulting string is "aabb", which is the lexicographically smallest string achievable.

Example 3:

Input: s = “zxy”

Output: “xzy”

Explanation:

• Choose k = 2, reverse the first 2 characters.
• Reverse "zx" to "xz", so the resulting string is "xzy", which is the lexicographically smallest string achievable.

Constraints:

• 1 <= n == s.length <= 1000
• s consists of lowercase English letters.

Solution

public class Solution {
    public String lexSmallest(String s) {
        int n = s.length();
        char[] arr = s.toCharArray();
        char[] best = arr.clone();
        // Check all reverse first k operations
        for (int k = 1; k <= n; k++) {
            if (isBetterReverseFirstK(arr, k, best)) {
                updateBestReverseFirstK(arr, k, best);
            }
        }
        // Check all reverse last k operations
        for (int k = 1; k <= n; k++) {
            if (isBetterReverseLastK(arr, k, best)) {
                updateBestReverseLastK(arr, k, best);
            }
        }
        return new String(best);
    }

    private boolean isBetterReverseFirstK(char[] arr, int k, char[] best) {
        for (int i = 0; i < arr.length; i++) {
            char currentChar = (i < k) ? arr[k - 1 - i] : arr[i];
            if (currentChar < best[i]) {
                return true;
            }
            if (currentChar > best[i]) {
                return false;
            }
        }
        return false;
    }

    private boolean isBetterReverseLastK(char[] arr, int k, char[] best) {
        int n = arr.length;
        for (int i = 0; i < n; i++) {
            char currentChar = (i < n - k) ? arr[i] : arr[n - 1 - (i - (n - k))];
            if (currentChar < best[i]) {
                return true;
            }
            if (currentChar > best[i]) {
                return false;
            }
        }
        return false;
    }

    private void updateBestReverseFirstK(char[] arr, int k, char[] best) {
        for (int i = 0; i < k; i++) {
            best[i] = arr[k - 1 - i];
        }
        if (arr.length - k >= 0) {
            System.arraycopy(arr, k, best, k, arr.length - k);
        }
    }

    private void updateBestReverseLastK(char[] arr, int k, char[] best) {
        int n = arr.length;
        if (n - k >= 0) {
            System.arraycopy(arr, 0, best, 0, n - k);
        }
        for (int i = 0; i < k; i++) {
            best[n - k + i] = arr[n - 1 - i];
        }
    }
}

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