Move from brute-force thinking to an efficient approach using dynamic programming strategy.
The chess knight has a unique movement, it may move two squares vertically and one square horizontally, or two squares horizontally and one square vertically (with both forming the shape of an L). The possible movements of chess knight are shown in this diagram:
A chess knight can move as indicated in the chess diagram below:
We have a chess knight and a phone pad as shown below, the knight can only stand on a numeric cell (i.e. blue cell).
Given an integer n, return how many distinct phone numbers 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 knight jumps.
As the answer may be very large, return the answer modulo 109 + 7.
Example 1:
Input: n = 1 Output: 10 Explanation: We need to dial a number of length 1, so placing the knight over any numeric cell of the 10 cells is sufficient.
Example 2:
Input: n = 2 Output: 20 Explanation: All the valid number we can dial are [04, 06, 16, 18, 27, 29, 34, 38, 40, 43, 49, 60, 61, 67, 72, 76, 81, 83, 92, 94]
Example 3:
Input: n = 3131 Output: 136006598 Explanation: Please take care of the mod.
Constraints:
1 <= n <= 5000Problem summary: The chess knight has a unique movement, it may move two squares vertically and one square horizontally, or two squares horizontally and one square vertically (with both forming the shape of an L). The possible movements of chess knight are shown in this diagram: A chess knight can move as indicated in the chess diagram below: We have a chess knight and a phone pad as shown below, the knight can only stand on a numeric cell (i.e. blue cell). Given an integer n, return how many distinct phone numbers 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 knight jumps. As the answer may be very large, return the answer modulo 109 + 7.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Dynamic Programming
1
2
3131
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #935: Knight Dialer
class Solution {
public int knightDialer(int n) {
final int mod = (int) 1e9 + 7;
long[] f = new long[10];
Arrays.fill(f, 1);
while (--n > 0) {
long[] g = new long[10];
g[0] = (f[4] + f[6]) % mod;
g[1] = (f[6] + f[8]) % mod;
g[2] = (f[7] + f[9]) % mod;
g[3] = (f[4] + f[8]) % mod;
g[4] = (f[0] + f[3] + f[9]) % mod;
g[6] = (f[0] + f[1] + f[7]) % mod;
g[7] = (f[2] + f[6]) % mod;
g[8] = (f[1] + f[3]) % mod;
g[9] = (f[2] + f[4]) % mod;
f = g;
}
return (int) (Arrays.stream(f).sum() % mod);
}
}
// Accepted solution for LeetCode #935: Knight Dialer
func knightDialer(n int) (ans int) {
f := make([]int, 10)
for i := range f {
f[i] = 1
}
const mod int = 1e9 + 7
for i := 1; i < n; i++ {
g := make([]int, 10)
g[0] = (f[4] + f[6]) % mod
g[1] = (f[6] + f[8]) % mod
g[2] = (f[7] + f[9]) % mod
g[3] = (f[4] + f[8]) % mod
g[4] = (f[0] + f[3] + f[9]) % mod
g[6] = (f[0] + f[1] + f[7]) % mod
g[7] = (f[2] + f[6]) % mod
g[8] = (f[1] + f[3]) % mod
g[9] = (f[2] + f[4]) % mod
f = g
}
for _, x := range f {
ans = (ans + x) % mod
}
return
}
# Accepted solution for LeetCode #935: Knight Dialer
class Solution:
def knightDialer(self, n: int) -> int:
f = [1] * 10
for _ in range(n - 1):
g = [0] * 10
g[0] = f[4] + f[6]
g[1] = f[6] + f[8]
g[2] = f[7] + f[9]
g[3] = f[4] + f[8]
g[4] = f[0] + f[3] + f[9]
g[6] = f[0] + f[1] + f[7]
g[7] = f[2] + f[6]
g[8] = f[1] + f[3]
g[9] = f[2] + f[4]
f = g
return sum(f) % (10**9 + 7)
// Accepted solution for LeetCode #935: Knight Dialer
struct Solution;
impl Solution {
fn knight_dialer(n: i32) -> i32 {
let max = 1_000_000_007;
let n = n as usize;
let mut dp: [[usize; 10]; 2] = [
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
];
let map = vec![
vec![4, 6],
vec![6, 8],
vec![7, 9],
vec![4, 8],
vec![0, 3, 9],
vec![],
vec![0, 1, 7],
vec![2, 6],
vec![1, 3],
vec![2, 4],
];
let mut res = 10;
for i in 1..n {
res = 0;
for j in 0..10 {
let mut sum = 0;
for &k in &map[j] {
sum += dp[(i - 1) % 2][k];
sum %= max;
}
dp[i % 2][j] = sum;
res += dp[i % 2][j];
res %= max;
}
}
res as i32
}
}
#[test]
fn test() {
let n = 1;
let res = 10;
assert_eq!(Solution::knight_dialer(n), res);
let n = 2;
let res = 20;
assert_eq!(Solution::knight_dialer(n), res);
let n = 3;
let res = 46;
assert_eq!(Solution::knight_dialer(n), res);
let n = 161;
let res = 533_302_150;
assert_eq!(Solution::knight_dialer(n), res);
}
// Accepted solution for LeetCode #935: Knight Dialer
function knightDialer(n: number): number {
const mod = 1e9 + 7;
const f: number[] = Array(10).fill(1);
while (--n) {
const g: number[] = Array(10).fill(0);
g[0] = (f[4] + f[6]) % mod;
g[1] = (f[6] + f[8]) % mod;
g[2] = (f[7] + f[9]) % mod;
g[3] = (f[4] + f[8]) % mod;
g[4] = (f[0] + f[3] + f[9]) % mod;
g[6] = (f[0] + f[1] + f[7]) % mod;
g[7] = (f[2] + f[6]) % mod;
g[8] = (f[1] + f[3]) % mod;
g[9] = (f[2] + f[4]) % mod;
f.splice(0, 10, ...g);
}
return f.reduce((a, b) => (a + b) % mod);
}
Use this to step through a reusable interview workflow for this problem.
Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.
Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.
Review these before coding to avoid predictable interview regressions.
Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.
Usually fails on: Correctness breaks on cases that differ only in hidden state.
Fix: Define state so each unique subproblem maps to one DP cell.