Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.
You have a pointer at index 0 in an array of size arrLen. At each step, you can move 1 position to the left, 1 position to the right in the array, or stay in the same place (The pointer should not be placed outside the array at any time).
Given two integers steps and arrLen, return the number of ways such that your pointer is still at index 0 after exactly steps steps. Since the answer may be too large, return it modulo 109 + 7.
Example 1:
Input: steps = 3, arrLen = 2 Output: 4 Explanation: There are 4 differents ways to stay at index 0 after 3 steps. Right, Left, Stay Stay, Right, Left Right, Stay, Left Stay, Stay, Stay
Example 2:
Input: steps = 2, arrLen = 4 Output: 2 Explanation: There are 2 differents ways to stay at index 0 after 2 steps Right, Left Stay, Stay
Example 3:
Input: steps = 4, arrLen = 2 Output: 8
Constraints:
1 <= steps <= 5001 <= arrLen <= 106Problem summary: You have a pointer at index 0 in an array of size arrLen. At each step, you can move 1 position to the left, 1 position to the right in the array, or stay in the same place (The pointer should not be placed outside the array at any time). Given two integers steps and arrLen, return the number of ways such that your pointer is still at index 0 after exactly steps steps. Since the answer may be too large, return it modulo 109 + 7.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Dynamic Programming
3 2
2 4
4 2
number-of-ways-to-reach-a-position-after-exactly-k-steps)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1269: Number of Ways to Stay in the Same Place After Some Steps
class Solution {
private Integer[][] f;
private int n;
public int numWays(int steps, int arrLen) {
f = new Integer[steps][steps + 1];
n = arrLen;
return dfs(0, steps);
}
private int dfs(int i, int j) {
if (i > j || i >= n || i < 0 || j < 0) {
return 0;
}
if (i == 0 && j == 0) {
return 1;
}
if (f[i][j] != null) {
return f[i][j];
}
int ans = 0;
final int mod = (int) 1e9 + 7;
for (int k = -1; k <= 1; ++k) {
ans = (ans + dfs(i + k, j - 1)) % mod;
}
return f[i][j] = ans;
}
}
// Accepted solution for LeetCode #1269: Number of Ways to Stay in the Same Place After Some Steps
func numWays(steps int, arrLen int) int {
const mod int = 1e9 + 7
f := make([][]int, steps)
for i := range f {
f[i] = make([]int, steps+1)
for j := range f[i] {
f[i][j] = -1
}
}
var dfs func(i, j int) int
dfs = func(i, j int) (ans int) {
if i > j || i >= arrLen || i < 0 || j < 0 {
return 0
}
if i == 0 && j == 0 {
return 1
}
if f[i][j] != -1 {
return f[i][j]
}
for k := -1; k <= 1; k++ {
ans += dfs(i+k, j-1)
ans %= mod
}
f[i][j] = ans
return
}
return dfs(0, steps)
}
# Accepted solution for LeetCode #1269: Number of Ways to Stay in the Same Place After Some Steps
class Solution:
def numWays(self, steps: int, arrLen: int) -> int:
@cache
def dfs(i, j):
if i > j or i >= arrLen or i < 0 or j < 0:
return 0
if i == 0 and j == 0:
return 1
ans = 0
for k in range(-1, 2):
ans += dfs(i + k, j - 1)
ans %= mod
return ans
mod = 10**9 + 7
return dfs(0, steps)
// Accepted solution for LeetCode #1269: Number of Ways to Stay in the Same Place After Some Steps
/**
* [1269] Number of Ways to Stay in the Same Place After Some Steps
*
* You have a pointer at index 0 in an array of size arrLen. At each step, you can move 1 position to the left, 1 position to the right in the array, or stay in the same place (The pointer should not be placed outside the array at any time).
* Given two integers steps and arrLen, return the number of ways such that your pointer is still at index 0 after exactly steps steps. Since the answer may be too large, return it modulo 10^9 + 7.
*
* Example 1:
*
* Input: steps = 3, arrLen = 2
* Output: 4
* Explanation: There are 4 differents ways to stay at index 0 after 3 steps.
* Right, Left, Stay
* Stay, Right, Left
* Right, Stay, Left
* Stay, Stay, Stay
*
* Example 2:
*
* Input: steps = 2, arrLen = 4
* Output: 2
* Explanation: There are 2 differents ways to stay at index 0 after 2 steps
* Right, Left
* Stay, Stay
*
* Example 3:
*
* Input: steps = 4, arrLen = 2
* Output: 8
*
*
* Constraints:
*
* 1 <= steps <= 500
* 1 <= arrLen <= 10^6
*
*/
pub struct Solution {}
// problem: https://leetcode.com/problems/number-of-ways-to-stay-in-the-same-place-after-some-steps/
// discuss: https://leetcode.com/problems/number-of-ways-to-stay-in-the-same-place-after-some-steps/discuss/?currentPage=1&orderBy=most_votes&query=
// submission codes start here
impl Solution {
// Credit: https://leetcode.com/problems/number-of-ways-to-stay-in-the-same-place-after-some-steps/solutions/3527268/rust-iterator-dp-bottom-up-space-optimized/
pub fn num_ways(steps: i32, arr_len: i32) -> i32 {
let k = steps as usize;
let n = k.min(arr_len as usize);
(0..k)
.fold(
{
let mut memo: Vec<usize> = vec![0; n];
memo[0] = 1;
memo
},
|memo, _| {
(0..n)
.map(|i| {
let mut ways = memo[i];
if i > 0 {
ways = (ways + memo[i - 1]) % 1_000_000_007;
}
if i + 1 < n {
ways = (ways + memo[i + 1]) % 1_000_000_007;
}
ways
})
.collect()
},
)
.get(0)
.unwrap_or(&0)
.clone() as i32
}
}
// submission codes end
#[cfg(test)]
mod tests {
use super::*;
#[test]
fn test_1269_example_1() {
let steps = 3;
let arr_len = 2;
let result = 4;
assert_eq!(Solution::num_ways(steps, arr_len), result);
}
#[test]
fn test_1269_example_2() {
let steps = 2;
let arr_len = 4;
let result = 2;
assert_eq!(Solution::num_ways(steps, arr_len), result);
}
#[test]
fn test_1269_example_3() {
let steps = 4;
let arr_len = 2;
let result = 8;
assert_eq!(Solution::num_ways(steps, arr_len), result);
}
}
// Accepted solution for LeetCode #1269: Number of Ways to Stay in the Same Place After Some Steps
function numWays(steps: number, arrLen: number): number {
const f = Array.from({ length: steps }, () => Array(steps + 1).fill(-1));
const mod = 10 ** 9 + 7;
const dfs = (i: number, j: number) => {
if (i > j || i >= arrLen || i < 0 || j < 0) {
return 0;
}
if (i == 0 && j == 0) {
return 1;
}
if (f[i][j] != -1) {
return f[i][j];
}
let ans = 0;
for (let k = -1; k <= 1; ++k) {
ans = (ans + dfs(i + k, j - 1)) % mod;
}
return (f[i][j] = ans);
};
return dfs(0, steps);
}
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.