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-rw-r--r--frog_jump.c78
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+// 403. Frog Jump
+#include <stdbool.h>
+#include <stdio.h>
+#include <stdlib.h>
+
+/*
+ * a frog is crossing a river. the river is divided into some number of units
+ * and at each unit, there may or may not exist a stone. the frog can jump on a
+ * stone, but it must not jump into the water. given a list of 'stones'
+ * positions in (units) in sorted ascending order, determine if the frog can
+ * cross the river by landing on the last stone. initially, the frog is on the
+ * first stone and assumes the frist just must be 1 unit. if the frog's last
+ * jump was k units, its next jujp must be either 'k - 1' or 'k + 1' units. the
+ * frog can only jump in the forward direction.
+ */
+
+bool dfs(int *stones, int stones_size, int t, long long **hash, int x) {
+ if (stones_size == 1)
+ return true;
+ int p = -1, c = -1, n = -1;
+ long long val = (stones[0] << 10) + t;
+ long long d = val;
+ while (1) {
+ if (!hash[d % x])
+ break;
+ else if (hash[d % x][0] == val)
+ return false;
+ else
+ d++;
+ }
+ for (int i = 1; i < stones_size; i++) {
+ if (stones[i] == stones[0] + t + 1)
+ n = i;
+ if (stones[i] == stones[0] + t)
+ c = i;
+ if (stones[i] == stones[0] + t - 1)
+ p = i;
+ if (stones[i] > stones[0] + t + 1)
+ break;
+ }
+ if (n != -1 && dfs(stones + n, stones_size - n, t + 1, hash, x))
+ return true;
+ if (c != -1 && dfs(stones + c, stones_size - c, t, hash, x))
+ return true;
+ if (p != -1 && dfs(stones + p, stones_size - p, t - 1, hash, x))
+ return true;
+ d = val;
+ while (1) {
+ if (!hash[d % x]) {
+ hash[d % x] = malloc(sizeof(long long));
+ hash[d % x][0] = val;
+ break;
+ } else
+ d++;
+ }
+ return false;
+}
+
+bool canCross(int *stones, int stones_size) {
+ if (stones[0] || stones[1] != 1)
+ return false;
+ int x = 100 * stones_size;
+ long long **hash = calloc(x, sizeof(long long *));
+ bool ans = dfs(stones + 1, stones_size - 1, 1, hash, x);
+ for (int i = 0; i < x; i++) {
+ if (!hash[i])
+ continue;
+ free(hash[i]);
+ }
+ free(hash);
+ return ans;
+}
+
+int main() {
+ int s1[] = {0, 1, 3, 5, 6, 8, 12, 17}, s2[] = {0, 1, 2, 3, 4, 8, 9, 11};
+ printf("%d\n", canCross(s1, 8)); // expect: 1
+ printf("%d\n", canCross(s2, 8)); // expect: 0
+}