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// 935. Knight Dialer
#include "leetcode.h"
/*
* the chess knight has a unique movement, it may move two squares vertically
* and one square horizontally, or two squares horizontally andone square
* vertically with both forming the shape of an L. the possible movements of
* chess knights ares shown in below diagram. given an integer 'n', return how
* many distinct phone number of length 'n' we can dial. you are allowed to
* place the knight on any numeric cell initially and then you should perform 'n
* - 1' jumps to dial a number of length 'n'. all jumps should be valid jumps.
* as the answer may be very large, return answer modulo 10^9 + 7
*/
int knightDialer(int n) {
long int new_jump[] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
long int old_jump[] = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1};
long long int numbers = 0;
int row0[] = {2, 4, 6};
int row1[] = {2, 6, 8};
int row2[] = {2, 7, 9};
int row3[] = {2, 8, 4};
int row4[] = {3, 3, 9, 0};
int row5[] = {0};
int row6[] = {3, 7, 0, 1};
int row7[] = {2, 2, 6};
int row8[] = {2, 1, 3};
int row9[] = {2, 4, 2};
int **dial = malloc(10 * sizeof(int *));
dial[0] = row0;
dial[1] = row1;
dial[2] = row2;
dial[3] = row3;
dial[4] = row4;
dial[5] = row5;
dial[6] = row6;
dial[7] = row7;
dial[8] = row8;
dial[9] = row9;
if (n == 1)
return 10;
for (int i = 1; i < n; i++) {
numbers = 0;
for (int j = 0; j < 10; j++)
for (int k = 1; k < dial[j][0] + 1; k++) {
if (j == 5)
continue;
new_jump[dial[j][k]] += old_jump[j];
new_jump[dial[j][k]] %= 1000000007LL;
}
memcpy(old_jump, new_jump, sizeof(long) * 10);
for (int j = 0; j < 10; j++) {
numbers += new_jump[j];
new_jump[j] = 0;
}
}
free(dial);
return numbers % 1000000007LL;
}
int main() {
printf("%d\n", knightDialer(1)); // expect: 10
printf("%d\n", knightDialer(2)); // expect: 20
printf("%d\n", knightDialer(3131)); // expect: 136006598
}
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