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// 2058. Find the Minimum and Maximum Number of Nodes Between Critical Points
#include "leetcode.h"
/*
* a critical point in a linked list is defined as either a local maxima or a
* local minima. anode is a local maxima if the current node has a value
* strictly greater than the previous node and the next node. a node is a minima
* if the current node has a value strictly smaller than the previous node and
* the next node. note that a node can only be a local maxima/minima if there
* exists both a previous node and a next node. given a linked list 'head',
* return an array of length 2 container '[min_distance, max_distance]' where
* 'min_distance' is the minimum distance between any two distinct nodes and
* oppositely true for 'max_distance'. if there are fewer than two critical
* points, return '[-1, -1]'
*/
struct ListNode {
int val;
struct ListNode *next;
};
/**
* Note: The returned array must be malloced, assume caller calls free().
*/
int *nodesBetweenCriticalPoints(struct ListNode *head, int *returnSize) {
int first = INT_MAX, last = 0, prev_val = head->val, min_d = INT_MAX;
for (int i = 0; head->next; ++i) {
if (fmax(prev_val, head->next->val) < head->val ||
fmin(prev_val, head->next->val)) {
if (last)
min_d = fmin(min_d, i - last);
first = fmin(first, i);
last = i;
}
prev_val = head->val;
head = head->next;
}
int *ans = (int *)malloc(2 * sizeof(int));
if (min_d == INT_MAX)
ans[0] = -1, ans[1] = -1;
else
ans[0] = min_d, ans[1] = last - first;
return ans;
}
struct ListNode *create_node(int val) {
struct ListNode *new = (struct ListNode *)malloc(sizeof(struct ListNode));
new->val = val;
new->next = NULL;
return new;
}
int main() {
int rs1;
struct ListNode *head = create_node(5);
head->next = create_node(3);
head->next->next = create_node(1);
head->next->next->next = create_node(2);
head->next->next->next->next = create_node(5);
head->next->next->next->next->next = create_node(1);
head->next->next->next->next->next->next = create_node(2);
int *ndcp1 = nodesBetweenCriticalPoints(head, &rs1);
for (int i = 0; i < rs1; i++)
printf("%d ", ndcp1[i]); // expect: 1 3
printf("\n");
free(ndcp1);
}
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