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// 2791. Count Paths That Can Form a Palindrome in a Tree
#include "leetcode.h"
/*
* you are given a tree (ie. a connected, undirected graph that has no cycles)
* rooted at node 0 consisting of 'n' nodes numbered from 0 to 'n - 1'. the tree
* is represented by a 0-indexed array 'parent' of size 'n' where 'parent[i]' is
* the parent of node 'i'. since node 0 is the root, 'parent[0] == -1'. you are
* also given a string 's' of length 'n' where 's[i]' is the character assigned
* to the edge between 'i' and 'parent[i]'. 's[0]' can be ignored. return the
* number of pairs of nodes '(u, v)' such that 'u < v' and the characters
* assigned to edges on the path from 'u' to 'v' can rearranged to form a
* palindrome
*/
void fill_table(int *parent, int parentSize, char *s, int *dp, int *filled,
int k, int *hash_table, int *hash_vals) {
if (filled[k])
return;
if (!k) {
filled[k] = 1;
hash_table[0] = 1;
hash_vals[0] = 0;
return;
}
fill_table(parent, parentSize, s, dp, filled, parent[k], hash_table,
hash_vals);
int mask = dp[parent[k]] ^ (1 << (s[k] - 97));
dp[k] = mask;
filled[k] = 1;
int kk = parentSize * 2, idx = mask % kk;
while (hash_vals[idx] != -1 && hash_vals[idx] != mask) {
idx++;
if (idx >= kk)
idx -= kk;
}
if (hash_vals[idx] == -1)
hash_vals[idx] = mask;
hash_table[idx]++;
}
long long countPalindromePaths(int *parent, int parentSize, char *s) {
int kk = parentSize * 2, n = parentSize;
int *dp = (int *)calloc(n, sizeof(int));
int *filled = (int *)calloc(n, sizeof(int));
int *hash_table = (int *)calloc(kk, sizeof(int));
int *hash_vals = (int *)malloc(kk * sizeof(int));
for (int i = 0; i < kk; i++)
hash_vals[i] = -1;
for (int k = 0; k < n; k++)
fill_table(parent, parentSize, s, dp, filled, k, hash_table, hash_vals);
long long cnt = 0;
for (int k = 0; k < n; k++) {
int mask = dp[k];
int idx = mask % kk;
while (hash_vals[idx] != -1 && hash_vals[idx] != mask) {
idx++;
if (idx >= kk)
idx -= kk;
}
cnt += hash_table[idx] - 1;
for (int j = 0; j < 26; j++) {
int new_mask = mask ^ (1 << j);
int new_idx = new_mask % kk;
while (hash_vals[new_idx] != -1 && hash_vals[new_idx] != new_mask) {
new_idx++;
if (new_idx >= kk)
new_idx -= kk;
}
cnt += hash_table[new_idx];
}
}
free(dp), free(filled), free(hash_table), free(hash_vals);
return cnt;
}
int main() {
int p1[] = {-1, 0, 0, 1, 1, 2}, p2[] = {-1, 0, 0, 0, 0};
char *s1 = "acaabc", *s2 = "aaaaa";
printf("%lld\n", countPalindromePaths(p1, ARRAY_SIZE(p1), s1)); // expect: 8
printf("%lld\n", countPalindromePaths(p2, ARRAY_SIZE(p2), s2)); // expect: 10
}
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