LeetCode 2287: Rearrange Characters to Make Target String Solution

Master LeetCode problem 2287 (Rearrange Characters to Make Target String), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2287. Rearrange Characters to Make Target String

Easy

Hash Table String Counting

Problem Explanation

Explanation:

To solve this problem, we can count the frequency of each character in both strings s and target. We then calculate the maximum number of copies of target that can be formed by taking characters from s based on the frequency of characters in both strings.

Create frequency maps for characters in strings s and target.
Iterate over the characters in target and calculate the maximum number of copies based on the minimum frequency of each character.
Return the minimum count calculated in the previous step as the result.

Time Complexity:

The time complexity of this approach is O(n), where n is the length of string s.

Space Complexity:

The space complexity is O(1) since the frequency map will have a fixed size of 26 lowercase English letters.

Solution Code

class Solution {
    public int maxNumberOfCopies(String s, String target) {
        int[] sFreq = new int[26];
        int[] targetFreq = new int[26];
        
        for (char ch : s.toCharArray()) {
            sFreq[ch - 'a']++;
        }
        
        for (char ch : target.toCharArray()) {
            targetFreq[ch - 'a']++;
        }
        
        int maxCopies = Integer.MAX_VALUE;
        for (int i = 0; i < target.length(); i++) {
            if (targetFreq[target.charAt(i) - 'a'] > 0) {
                maxCopies = Math.min(maxCopies, sFreq[target.charAt(i) - 'a'] / targetFreq[target.charAt(i) - 'a']);
            } else {
                return 0;
            }
        }
        
        return maxCopies;
    }
}

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What is the time complexity of LeetCode 2287 (Rearrange Characters to Make Target String)?

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