Number of Ways to Stay in the Same Place After Some Steps solution and explanation
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The problem
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 <= 106
Approach overview
Problem Overview: You start at index 0 of a 1D array. For each step you can move left, move right, or stay in place. After exactly steps moves, count how many sequences keep you at index 0. The array length limits movement, and results must be returned modulo 1e9 + 7.
Approach 1: Recursive with Memoization (Top-Down DP) (Time: O(steps * min(arrLen, steps)), Space: O(steps * min(arrLen, steps)))
Model the process as a recursive state dfs(step, pos) representing the number of ways to reach the final state from a given step and position. From each state you have three transitions: stay (pos), move left (pos - 1), or move right (pos + 1). Memoize results in a cache to avoid recomputing overlapping subproblems. A key optimization: you never need positions beyond min(arrLen - 1, steps) because you cannot travel farther than the remaining steps. This drastically reduces the state space and keeps the recursion efficient.
Approach 2: Dynamic Programming (Bottom-Up) (Time: O(steps * min(arrLen, steps)), Space: O(steps * min(arrLen, steps)) or O(min(arrLen, steps)))
Define dp[s][i] as the number of ways to reach position i after s steps. The transition mirrors the allowed moves: dp[s][i] = dp[s-1][i] + dp[s-1][i-1] + dp[s-1][i+1]. Bound the position range to maxPos = min(arrLen - 1, steps) since larger indices are unreachable. Initialize dp[0][0] = 1 and iterate step by step while applying modulo arithmetic. A rolling array optimization reduces memory to one dimension because each row depends only on the previous step. This bottom-up approach avoids recursion overhead and is typically faster in production implementations of dynamic programming.
Recommended for interviews: The bottom-up dynamic programming approach is what interviewers typically expect. It clearly demonstrates understanding of state design, transition rules, and boundary constraints. Mentioning the position bound optimization (min(arrLen, steps)) shows deeper insight into reducing the DP state space. The recursive solution with memoization still helps demonstrate how the problem naturally decomposes into overlapping subproblems before converting it into iterative DP.
Approach 1: Dynamic Programming Approach
This approach leverages dynamic programming to find the number of ways to stay at index 0 after a given number of steps. We define a 2D table dp[i][j] where i represents the number of steps remaining, and j represents the current position of the pointer.
To optimize computation, we can limit the table size to the minimum of steps and arrLen since going beyond these positions is unnecessary.
MOD = 1000000007
def numWays(steps: int, arrLen: int) -> int:
maxPos = min(steps, arrLen - 1)
dp = [[0] * (maxPos + 1) for _ in range(steps + 1)]
dp[0][0] = 1
for i in range(1, steps + 1):
for j in range(maxPos + 1):
dp[i][j] = dp[i - 1][j] % MOD
if j > 0:
dp[i][j] = (dp[i][j] + dp[i - 1][j - 1]) % MOD
if j < maxPos:
dp[i][j] = (dp[i][j] + dp[i - 1][j + 1]) % MOD
return dp[steps][0]
print(numWays(3, 2)) # Output: 4public class NumWays {
private static final int MOD = 1000000007;
public int numWays(int steps, int arrLen) {
int maxPos = Math.min(steps, arrLen - 1);
int[][] dp = new int[steps + 1][maxPos + 1];
dp[0][0] = 1;
for (int i = 1; i <= steps; ++i) {
for (int j = 0; j <= maxPos; ++j) {
dp[i][j] = dp[i - 1][j];
if (j > 0) dp[i][j] = (dp[i][j] + dp[i - 1][j - 1]) % MOD;
if (j < maxPos) dp[i][j] = (dp[i][j] + dp[i - 1][j + 1]) % MOD;
}
}
return dp[steps][0];
}
public static void main(String[] args) {
NumWays nw = new NumWays();
System.out.println(nw.numWays(3, 2)); // Output: 4
}
}#include <iostream>
#include <vector>
#define MOD 1000000007
using namespace std;
int numWays(int steps, int arrLen) {
int maxPos = min(steps, arrLen - 1);
vector<vector<int>> dp(steps + 1, vector<int>(maxPos + 1, 0));
dp[0][0] = 1;
for (int i = 1; i <= steps; ++i) {
for (int j = 0; j <= maxPos; ++j) {
dp[i][j] = dp[i - 1][j];
if (j > 0) dp[i][j] = (dp[i][j] + dp[i - 1][j - 1]) % MOD;
if (j < maxPos) dp[i][j] = (dp[i][j] + dp[i - 1][j + 1]) % MOD;
}
}
return dp[steps][0];
}
int main() {
cout << numWays(3, 2) << endl; // Output: 4
return 0;
}const MOD = 1000000007;
function numWays(steps, arrLen) {
const maxPos = Math.min(steps, arrLen - 1);
const dp = Array.from({length: steps + 1}, () => Array(maxPos + 1).fill(0));
dp[0][0] = 1;
for (let i = 1; i <= steps; i++) {
for (let j = 0; j <= maxPos; j++) {
dp[i][j] = dp[i - 1][j];
if (j > 0) dp[i][j] = (dp[i][j] + dp[i - 1][j - 1]) % MOD;
if (j < maxPos) dp[i][j] = (dp[i][j] + dp[i - 1][j + 1]) % MOD;
}
}
return dp[steps][0];
}
console.log(numWays(3, 2)); // Output: 4#include <stdio.h>
#define MOD 1000000007
int numWays(int steps, int arrLen) {
int maxPos = (arrLen < steps) ? arrLen - 1 : steps;
int dp[steps+1][maxPos+1];
for(int i = 0; i <= steps; i++)
for(int j = 0; j <= maxPos; j++)
dp[i][j] = 0;
dp[0][0] = 1;
for (int i = 1; i <= steps; i++) {
for (int j = 0; j <= maxPos; j++) {
dp[i][j] = dp[i-1][j];
if (j > 0)
dp[i][j] = (dp[i][j] + dp[i-1][j-1]) % MOD;
if (j < maxPos)
dp[i][j] = (dp[i][j] + dp[i-1][j+1]) % MOD;
}
}
return dp[steps][0];
}
int main() {
printf("%d\n", numWays(3, 2)); // Output: 4
return 0;
}using System;
class NumWays
{
const int MOD = 1000000007;
public int NumWays(int steps, int arrLen)
{
int maxPos = Math.Min(steps, arrLen - 1);
int[,] dp = new int[steps + 1, maxPos + 1];
dp[0, 0] = 1;
for (int i = 1; i <= steps; i++)
{
for (int j = 0; j <= maxPos; j++)
{
dp[i, j] = dp[i - 1, j];
if (j > 0) dp[i, j] = (dp[i, j] + dp[i - 1, j - 1]) % MOD;
if (j < maxPos) dp[i, j] = (dp[i, j] + dp[i - 1, j + 1]) % MOD;
}
}
return dp[steps, 0];
}
static void Main(string[] args)
{
NumWays nw = new NumWays();
Console.WriteLine(nw.NumWays(3, 2)); // Output: 4
}
}The Python solution utilizes dynamic programming. It uses a list to store the number of ways to reach each position after each step, and follows the same logic as the C/C++/Java counterparts.
Time Complexity: O(steps * min(steps, arrLen))
Space Complexity: O(steps * min(steps, arrLen))
Approach 2: Recursive with Memoization Approach
This approach utilizes recursion combined with memoization to optimize the recursive calls. Here, recursion is used to explore all possible paths dynamically adjusting by staying at, moving left, or moving right from each position in every step.
The results of the recursive calls are stored in a memoization table to avoid redundant calculations.
from functools import lru_cache
MOD = 1000000007
class NumWaysMemo:
def __init__(self):
self.cache = {}
def numWaysHelper(self, steps: int, pos: int, maxPos: int) -> int:
if pos < 0 or pos > maxPos:
return 0
if steps == 0:
return 1 if pos == 0 else 0
key = (steps, pos)
if key in self.cache:
return self.cache[key]
stay = self.numWaysHelper(steps - 1, pos, maxPos)
left = self.numWaysHelper(steps - 1, pos - 1, maxPos)
right = self.numWaysHelper(steps - 1, pos + 1, maxPos)
result = (stay + left + right) % MOD
self.cache[key] = result
return result
def numWays(self, steps: int, arrLen: int) -> int:
maxPos = min(steps, arrLen - 1)
return self.numWaysHelper(steps, 0, maxPos)
nwm = NumWaysMemo()
print(nwm.numWays(3, 2)) # Output: 4import java.util.HashMap;
import java.util.Map;
public class NumWaysMemo {
private static final int MOD = 1000000007;
private final Map<String, Integer> cache = new HashMap<>();
public int numWaysHelper(int steps, int pos, int maxPos) {
if (pos < 0 || pos > maxPos) return 0;
if (steps == 0) return pos == 0 ? 1 : 0;
String key = steps + "," + pos;
if (cache.containsKey(key)) return cache.get(key);
int stay = numWaysHelper(steps - 1, pos, maxPos);
int left = numWaysHelper(steps - 1, pos - 1, maxPos);
int right = numWaysHelper(steps - 1, pos + 1, maxPos);
int result = ((stay + left) % MOD + right) % MOD;
cache.put(key, result);
return result;
}
public int numWays(int steps, int arrLen) {
int maxPos = Math.min(steps, arrLen - 1);
return numWaysHelper(steps, 0, maxPos);
}
public static void main(String[] args) {
NumWaysMemo nwm = new NumWaysMemo();
System.out.println(nwm.numWays(3, 2)); // Output: 4
}
}#include <iostream>
#include <vector>
#define MOD 1000000007
using namespace std;
vector<vector<int>> cache;
int numWaysHelper(int steps, int pos, int maxPos) {
if (pos < 0 || pos > maxPos) return 0;
if (steps == 0) return pos == 0 ? 1 : 0;
if (cache[steps][pos] != -1) return cache[steps][pos];
int stay = numWaysHelper(steps - 1, pos, maxPos);
int left = numWaysHelper(steps - 1, pos - 1, maxPos);
int right = numWaysHelper(steps - 1, pos + 1, maxPos);
return cache[steps][pos] = ((stay + left) % MOD + right) % MOD;
}
int numWays(int steps, int arrLen) {
int maxPos = min(steps, arrLen - 1);
cache.assign(steps + 1, vector<int>(maxPos + 1, -1));
return numWaysHelper(steps, 0, maxPos);
}
int main() {
cout << numWays(3, 2) << endl; // Output: 4
return 0;
}const MOD = 1000000007;
function numWays(steps, arrLen) {
const maxPos = Math.min(steps, arrLen - 1);
const cache = new Map();
function helper(steps, pos) {
if (pos < 0 || pos > maxPos) return 0;
if (steps === 0) return pos === 0 ? 1 : 0;
const key = `${steps},${pos}`;
if (cache.has(key)) return cache.get(key);
const stay = helper(steps - 1, pos);
const left = helper(steps - 1, pos - 1);
const right = helper(steps - 1, pos + 1);
const result = ((stay + left + right) % MOD);
cache.set(key, result);
return result;
}
return helper(steps, 0);
}
console.log(numWays(3, 2)); // Output: 4#include <stdio.h>
#define MOD 1000000007
int cache[501][501];
int numWaysHelper(int steps, int pos, int maxPos) {
if (pos < 0 || pos > maxPos) return 0;
if (steps == 0) return pos == 0 ? 1 : 0;
if (cache[steps][pos] != -1) return cache[steps][pos];
int stay = numWaysHelper(steps - 1, pos, maxPos);
int left = numWaysHelper(steps - 1, pos - 1, maxPos);
int right = numWaysHelper(steps - 1, pos + 1, maxPos);
return cache[steps][pos] = ((stay + left) % MOD + right) % MOD;
}
int numWays(int steps, int arrLen) {
int maxPos = (arrLen < steps) ? arrLen - 1 : steps;
for (int i = 0; i <= steps; ++i) {
for (int j = 0; j <= maxPos; ++j)
cache[i][j] = -1;
}
return numWaysHelper(steps, 0, maxPos);
}
int main() {
printf("%d\n", numWays(3, 2)); // Output: 4
return 0;
}using System;
using System.Collections.Generic;
class NumWaysMemo
{
const int MOD = 1000000007;
private Dictionary<string, int> cache = new Dictionary<string, int>();
public int NumWaysHelper(int steps, int pos, int maxPos)
{
if (pos < 0 || pos > maxPos) return 0;
if (steps == 0) return pos == 0 ? 1 : 0;
string key = steps + "," + pos;
if (cache.ContainsKey(key)) return cache[key];
int stay = NumWaysHelper(steps - 1, pos, maxPos);
int left = NumWaysHelper(steps - 1, pos - 1, maxPos);
int right = NumWaysHelper(steps - 1, pos + 1, maxPos);
int result = ((stay + left) % MOD + right) % MOD;
cache[key] = result;
return result;
}
public int NumWays(int steps, int arrLen)
{
int maxPos = Math.Min(steps, arrLen - 1);
return NumWaysHelper(steps, 0, maxPos);
}
static void Main(string[] args)
{
NumWaysMemo nwm = new NumWaysMemo();
Console.WriteLine(nwm.NumWays(3, 2)); // Output: 4
}
}This Python solution applies memoization by caching computed results for recursive calls using a dictionary, strengthening the efficiency by avoiding repetitive calculations.
Time Complexity: O(steps * min(steps, arrLen))
Space Complexity: O(steps * min(steps, arrLen))
Approach 3: Memoization Search
We observe the data range of the problem and find that steps does not exceed 500, which means that we can only go to the right for up to 500 steps.
We can design a function dfs(i, j), which represents the number of schemes when we are currently at position i and the remaining steps are j. So the answer is dfs(0, steps).
The execution process of the function dfs(i, j) is as follows:
- If
i \gt jori geq arrLenori \lt 0orj \lt 0, then return0. - If
i = 0andj = 0, then the pointer has stopped in place and there are no remaining steps, so return1. - Otherwise, we can choose to move one step to the left, one step to the right, or stay still, so return
dfs(i - 1, j - 1) + dfs(i + 1, j - 1) + dfs(i, j - 1). Note the modulo operation of the answer.
During the process, we can use memoization search to avoid repeated calculations.
The time complexity is O(steps times steps), and the space complexity is O(steps times steps). Where steps is the number of steps given in the problem.
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);
}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)
}Complexity comparison
| Approach | Complexity |
|---|---|
| Dynamic Programming Approach | Time Complexity: O(steps * min(steps, arrLen)) Space Complexity: O(steps * min(steps, arrLen)) |
| Recursive with Memoization Approach | Time Complexity: O(steps * min(steps, arrLen)) Space Complexity: O(steps * min(steps, arrLen)) |
| Memoization Search | – |
Complexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Recursive with Memoization | O(steps * min(arrLen, steps)) | O(steps * min(arrLen, steps)) | When reasoning about the problem recursively or converting recursion to DP |
| Bottom-Up Dynamic Programming | O(steps * min(arrLen, steps)) | O(min(arrLen, steps)) with rolling array | Preferred for interviews and production due to predictable iteration and memory optimization |
Frequently asked questions
Is Number of Ways to Stay in the Same Place After Some Steps easy or hard?
The problem is classified as Hard because it requires careful dynamic programming state design and an optimization that limits the reachable positions. Without the position bound, the DP space becomes unnecessarily large.
Number of Ways to Stay in the Same Place After Some Steps Python/Java solution
Both Python and Java implementations typically use dynamic programming with either a 2D DP table or a rolling 1D array. The transition considers three moves per step and applies modulo 1e9+7. Time complexity remains O(steps * min(arrLen, steps)).
How to solve Number of Ways to Stay in the Same Place After Some Steps in O(n)?
A strict O(n) solution is not typical because the state depends on both steps and positions. The efficient solution reduces the state space using dp[step][pos] where pos is limited to min(arrLen, steps). This produces O(steps * min(arrLen, steps)) time, which is optimal for the constraints.
What is the best approach for Number of Ways to Stay in the Same Place After Some Steps?
The optimal approach uses dynamic programming with state dp[step][position]. Each state aggregates three transitions: stay, move left, and move right. By limiting positions to min(arrLen - 1, steps), the DP state space becomes manageable. This results in O(steps * min(arrLen, steps)) time and can be optimized to O(min(arrLen, steps)) space.
Is Number of Ways to Stay in the Same Place After Some Steps asked at Google/Amazon/Meta?
Dynamic programming problems with state transitions and movement constraints frequently appear in interviews at companies like Google, Amazon, and Meta. Variants of this problem test DP state modeling, boundary pruning, and modulo arithmetic.
What data structure is used in Number of Ways to Stay in the Same Place After Some Steps?
The core data structure is a dynamic programming array that stores the number of ways to reach each index after a certain number of steps. Some implementations also use memoization tables or hash maps for caching recursive states.
What is the time complexity of Number of Ways to Stay in the Same Place After Some Steps?
The optimal dynamic programming solution runs in O(steps * min(arrLen, steps)) time. The algorithm only processes positions that are reachable within the given number of steps. Space complexity is O(min(arrLen, steps)) with a rolling DP array.