LeetCode-in-Java

3365. Rearrange K Substrings to Form Target String

Medium

You are given two strings s and t, both of which are anagrams of each other, and an integer k.

Your task is to determine whether it is possible to split the string s into k equal-sized substrings, rearrange the substrings, and concatenate them in any order to create a new string that matches the given string t.

Return true if this is possible, otherwise, return false.

An anagram is a word or phrase formed by rearranging the letters of a different word or phrase, using all the original letters exactly once.

A substring is a contiguous non-empty sequence of characters within a string.

Example 1:

Input: s = “abcd”, t = “cdab”, k = 2

Output: true

Explanation:

• Split s into 2 substrings of length 2: ["ab", "cd"].
• Rearranging these substrings as ["cd", "ab"], and then concatenating them results in "cdab", which matches t.

Example 2:

Input: s = “aabbcc”, t = “bbaacc”, k = 3

Output: true

Explanation:

• Split s into 3 substrings of length 2: ["aa", "bb", "cc"].
• Rearranging these substrings as ["bb", "aa", "cc"], and then concatenating them results in "bbaacc", which matches t.

Example 3:

Input: s = “aabbcc”, t = “bbaacc”, k = 2

Output: false

Explanation:

• Split s into 2 substrings of length 3: ["aab", "bcc"].
• These substrings cannot be rearranged to form t = "bbaacc", so the output is false.

Constraints:

• 1 <= s.length == t.length <= 2 * 105
• 1 <= k <= s.length
• s.length is divisible by k.
• s and t consist only of lowercase English letters.
• The input is generated such that s and t are anagrams of each other.

Solution

import java.util.HashMap;
import java.util.Map;

public class Solution {
    public boolean isPossibleToRearrange(String s, String t, int k) {
        int size = s.length();
        int div = size / k;
        Map<String, Integer> map = new HashMap<>();
        for (int i = 0; i < size; i += div) {
            String sub = s.substring(i, i + div);
            map.put(sub, map.getOrDefault(sub, 0) + 1);
        }
        for (int i = 0; i < size; i += div) {
            String sub = t.substring(i, i + div);
            if (map.getOrDefault(sub, 0) > 0) {
                map.put(sub, map.get(sub) - 1);
            } else {
                return false;
            }
        }
        return true;
    }
}

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