LeetCode 2983: Palindrome Rearrangement Queries Solution

Master LeetCode problem 2983 (Palindrome Rearrangement Queries), a hard 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.

2983. Palindrome Rearrangement Queries

Hard

Hash Table String Prefix Sum

Problem Explanation

Explanation

To determine if a string can be rearranged to form a palindrome, we can check if the characters in the two halves of the string are symmetric. For each query, we rearrange the characters within the specified substrings and then check if the resulting string can form a palindrome. We do this by comparing the frequency of characters in the two halves of the string. If the frequencies are symmetric, the string can be rearranged to form a palindrome.

Algorithm

Create a helper function to check if a string can be rearranged to form a palindrome.
For each query, rearrange the characters within the specified substrings.
Check if the rearranged string can form a palindrome by comparing the frequencies of characters in the two halves.
Store the result for each query in the answer array.

Time Complexity

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

Space Complexity

The space complexity is O(1) since we are using constant extra space.

Solution Code

class Solution {
    public List<Boolean> canMakePaliQueries(String s, int[][] queries) {
        List<Boolean> answer = new ArrayList<>();
        for (int[] query : queries) {
            answer.add(canFormPalindrome(s, query));
        }
        return answer;
    }
    
    private boolean canFormPalindrome(String s, int[] query) {
        String substring1 = s.substring(query[0], query[1] + 1);
        String substring2 = s.substring(query[2], query[3] + 1);
        
        String rearranged = s.substring(0, query[0]) + substring2 + s.substring(query[1] + 1, query[2]) + substring1 + s.substring(query[3] + 1);
        
        return canRearrangeToPalindrome(rearranged);
    }
    
    private boolean canRearrangeToPalindrome(String s) {
        int[] count = new int[26];
        for (char c : s.toCharArray()) {
            count[c - 'a']++;
        }
        
        int oddCount = 0;
        for (int num : count) {
            if (num % 2 != 0) {
                oddCount++;
            }
        }
        
        return oddCount / 2 <= 1;
    }
}

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