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// 3203. Find Minimum Diameter After Merging Two Trees
#include "leetcode.h"
/*
* there exists two undirected trees with 'n' and 'm' nodes numbered from 0 to
* 'n - 1' and from 0 to 'm - 1', respectively. you are given two 2d integer
* arrays 'edges1', and 'edges2' of lengths 'n - 1' and 'm - 1', respectively,
* where 'edges1[i] = [a_i, b_i]' indicates that thereis an edge between nodes
* 'a_i' and 'b_i' in the first tree, and 'edges2[i] = [u_i, v_i]' indicates
* there is an edge between nodes 'u_i' and 'v_i' in the second tree. you must
* connect one node from the first tree with another node from the second tree
* with an edge. return the minimum possible diameter of the resulting tree. the
* diameter of a tree is the length of the longest path between any two nodes in
* the tree
*/
enum state { VAL, NEXT };
int max_diameter(int **edges, int edgesSize) {
int i, n = edgesSize + 1, queue[n + 1], head = 0, tail = 0;
int hash[n], nodes[(edgesSize << 1) + 1][2], node, ptr = 0;
int cnt, prev, curr, next, diam = 0;
bool *vis = (bool *)calloc(n, sizeof(bool));
memset(hash, -1, sizeof(hash));
for (int i = 0; i < edgesSize; i++) {
node = ptr++;
nodes[node][VAL] = edges[i][0];
nodes[node][NEXT] = hash[edges[i][1]];
hash[edges[i][1]] = node;
node = ptr++;
nodes[node][VAL] = edges[i][1];
nodes[node][NEXT] = hash[edges[i][0]];
hash[edges[i][0]] = node;
}
vis[0] = true;
queue[tail++] = 0;
while (head < tail) {
cnt = tail - head;
while (cnt--) {
curr = queue[head++];
node = hash[curr];
while (node >= 0) {
next = nodes[node][VAL];
node = nodes[node][NEXT];
if (vis[next])
continue;
vis[next] = true;
queue[tail++] = next;
}
}
}
memset(vis, false, n);
head = 0;
tail = 1;
queue[0] = curr;
while (head < tail) {
cnt = tail - head;
while (cnt--) {
curr = queue[head++];
node = hash[curr];
while (node >= 0) {
next = nodes[node][VAL];
node = nodes[node][NEXT];
if (vis[next])
continue;
vis[next] = true;
queue[tail++] = next;
}
}
if (head < tail)
++diam;
}
return diam;
}
int minimumDiameterAfterMerge(int **edges1, int edges1Size, int *edges1ColSize,
int **edges2, int edges2Size,
int *edges2ColSize) {
int d1 = max_diameter(edges1, edges1Size);
int d2 = max_diameter(edges2, edges2Size);
int d3 = (d1 >> 1) + (d1 & 1);
d3 += (d2 >> 1) + (d2 & 1) + 1;
if (d1 < d2)
d1 = d2;
if (d1 < d3)
d1 = d3;
return d1;
}
int main() {
int e1i1[3][2] = {{0, 1}, {0, 2}, {0, 3}}, e2i1[1][2] = {{0, 1}};
int e1i2[7][2] = {{0, 1}, {0, 2}, {0, 3}, {2, 4}, {2, 5}, {3, 6}, {2, 7}},
e2i2[7][2] = {{0, 1}, {0, 2}, {0, 3}, {2, 4}, {2, 5}, {3, 6}, {2, 7}};
struct two_d_arr e11, e21, e12, e22;
two_d_arr_init(&e11, 3, 2, e1i1);
two_d_arr_init(&e21, 1, 2, e2i1);
two_d_arr_init(&e12, 7, 2, e1i2);
two_d_arr_init(&e22, 7, 2, e2i2);
printf("%d\n",
minimumDiameterAfterMerge(e11.arr, e11.row_size, e11.col_size, e21.arr,
e21.row_size, e21.col_size)); // expect: 3
printf("%d\n",
minimumDiameterAfterMerge(e12.arr, e12.row_size, e12.col_size, e22.arr,
e22.row_size, e22.col_size)); // expect: 5
two_d_arr_free(&e11);
two_d_arr_free(&e21);
two_d_arr_free(&e12);
two_d_arr_free(&e22);
}
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