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2564. Substring XOR Queries

ArrayHash TableStringBit Manipulation

Explanation:

To solve this problem, we can use a prefix XOR technique along with a hashmap to store the indices of each prefix XOR value encountered. We iterate through the string s to calculate the prefix XOR values at each position. Then, for each query, we check if there exists a prefix XOR value val such that val ^ firsti = secondi in the hashmap. If found, we update the answer with the shortest valid substring.

  • Algorithmic Idea:

    1. Create a hashmap to store the indices of each prefix XOR value encountered.
    2. Iterate through the string s and calculate the prefix XOR values at each position.
    3. For each query, check if there exists a prefix XOR value val such that val ^ firsti = secondi in the hashmap.
    4. Update the answer with the shortest valid substring.

  • Time Complexity: O(N) where N is the length of the string s.

  • Space Complexity: O(N) to store the prefix XOR values in the hashmap.

:

View Leetcode Problem
class Solution {
    public int[][] findSubstrings(String s, int[][] queries) {
        int n = s.length();
        int[] prefixXor = new int[n + 1];
        for (int i = 0; i < n; i++) {
            prefixXor[i + 1] = prefixXor[i] ^ (s.charAt(i) - '0');
        }
        
        Map<Integer, Integer> map = new HashMap<>();
        int[][] ans = new int[queries.length][2];
        
        for (int i = 0; i < n; i++) {
            if (!map.containsKey(prefixXor[i])) {
                map.put(prefixXor[i], i);
            }
        }
        
        for (int i = 0; i < queries.length; i++) {
            int left = -1, right = n;
            int first = queries[i][0];
            int second = queries[i][1];
            for (int j = 0; j <= n; j++) {
                if (map.containsKey(prefixXor[j] ^ first)) {
                    int start = map.get(prefixXor[j] ^ first);
                    if (j - start < right - left) {
                        left = start;
                        right = j;
                    }
                }
            }
            ans[i][0] = left;
            ans[i][1] = right - 1;
        }
        
        return ans;
    }
}

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