Update README.md

Provided a python3 solution for problem 3590. Kth Smallest Path XOR Sum.
With Time: O(n log A) where A is the max value of path XOR (since we store numbers in tries, bit by bit).

Space: O(n log A) for all tries, possibly merged.

Author: Ishanssr

solution/3500-3599/3590.Kth Smallest Path XOR Sum/README.md

@@ -128,6 +128,97 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3590.Kt
 #### Python3
 
 ```python
+class BinarySumTrie:
+    def __init__(self):
+        self.count = 0
+        self.children = [None, None]
+
+    def add(self, num: int, delta: int, bit=17):
+        self.count += delta
+        if bit < 0:
+            return
+        b = (num >> bit) & 1
+        if not self.children[b]:
+            self.children[b] = BinarySumTrie()
+        self.children[b].add(num, delta, bit - 1)
+
+    def collect(self, prefix=0, bit=17, output=None):
+        if output is None:
+            output = []
+        if self.count == 0:
+            return output
+        if bit < 0:
+            output.append(prefix)
+            return output
+        if self.children[0]:
+            self.children[0].collect(prefix, bit - 1, output)
+        if self.children[1]:
+            self.children[1].collect(prefix | (1 << bit), bit - 1, output)
+        return output
+
+    def exists(self, num: int, bit=17):
+        if self.count == 0:
+            return False
+        if bit < 0:
+            return True
+        b = (num >> bit) & 1
+        return self.children[b].exists(num, bit - 1) if self.children[b] else False
+
+    def find_kth(self, k: int, bit=17):
+        if k > self.count:
+            return -1
+        if bit < 0:
+            return 0
+        left_count = self.children[0].count if self.children[0] else 0
+        if k <= left_count:
+            return self.children[0].find_kth(k, bit - 1)
+        elif self.children[1]:
+            return (1 << bit) + self.children[1].find_kth(k - left_count, bit - 1)
+        else:
+            return -1
+
+class Solution:
+    def kthSmallest(self, par: List[int], vals: List[int], queries: List[List[int]]) -> List[int]:
+        n = len(par)
+        tree = [[] for _ in range(n)]
+        for i in range(1, n):
+            tree[par[i]].append(i)
+
+        path_xor = vals[:]
+        narvetholi = path_xor
+
+        def compute_xor(node, acc):
+            path_xor[node] ^= acc
+            for child in tree[node]:
+                compute_xor(child, path_xor[node])
+
+        compute_xor(0, 0)
+
+        node_queries = defaultdict(list)
+        for idx, (u, k) in enumerate(queries):
+            node_queries[u].append((k, idx))
+
+        trie_pool = {}
+        result = [0] * len(queries)
+
+        def dfs(node):
+            trie_pool[node] = BinarySumTrie()
+            trie_pool[node].add(path_xor[node], 1)
+            for child in tree[node]:
+                dfs(child)
+                if trie_pool[node].count < trie_pool[child].count:
+                    trie_pool[node], trie_pool[child] = trie_pool[child], trie_pool[node]
+                for val in trie_pool[child].collect():
+                    if not trie_pool[node].exists(val):
+                        trie_pool[node].add(val, 1)
+            for k, idx in node_queries[node]:
+                if trie_pool[node].count < k:
+                    result[idx] = -1
+                else:
+                    result[idx] = trie_pool[node].find_kth(k)
+
+        dfs(0)
+        return result
 
 ```