LeetCode #2400 — MEDIUM

Number of Ways to Reach a Position After Exactly k Steps

Move from brute-force thinking to an efficient approach using math strategy.

The Problem

Problem Statement

You are given two positive integers startPos and endPos. Initially, you are standing at position startPos on an infinite number line. With one step, you can move either one position to the left, or one position to the right.

Given a positive integer k, return the number of different ways to reach the position endPos starting from startPos, such that you perform exactly k steps. Since the answer may be very large, return it modulo 10⁹ + 7.

Two ways are considered different if the order of the steps made is not exactly the same.

Note that the number line includes negative integers.

Example 1:

Input: startPos = 1, endPos = 2, k = 3
Output: 3
Explanation: We can reach position 2 from 1 in exactly 3 steps in three ways:
- 1 -> 2 -> 3 -> 2.
- 1 -> 2 -> 1 -> 2.
- 1 -> 0 -> 1 -> 2.
It can be proven that no other way is possible, so we return 3.

Example 2:

Input: startPos = 2, endPos = 5, k = 10
Output: 0
Explanation: It is impossible to reach position 5 from position 2 in exactly 10 steps.

Constraints:

1 <= startPos, endPos, k <= 1000

Step 01

Brute Force Baseline

Problem summary: You are given two positive integers startPos and endPos. Initially, you are standing at position startPos on an infinite number line. With one step, you can move either one position to the left, or one position to the right. Given a positive integer k, return the number of different ways to reach the position endPos starting from startPos, such that you perform exactly k steps. Since the answer may be very large, return it modulo 109 + 7. Two ways are considered different if the order of the steps made is not exactly the same. Note that the number line includes negative integers.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math · Dynamic Programming

Example 1

1
2
3

Example 2

2
5
10

Core Insight

What unlocks the optimal approach

How many steps to the left and to the right do you need to make exactly?
Does the order of the steps matter?
Use combinatorics to find the number of ways to order the steps.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2400: Number of Ways to Reach a Position After Exactly k Steps
class Solution {
    private Integer[][] f;
    private final int mod = (int) 1e9 + 7;

    public int numberOfWays(int startPos, int endPos, int k) {
        f = new Integer[k + 1][k + 1];
        return dfs(Math.abs(startPos - endPos), k);
    }

    private int dfs(int i, int j) {
        if (i > j || j < 0) {
            return 0;
        }
        if (j == 0) {
            return i == 0 ? 1 : 0;
        }
        if (f[i][j] != null) {
            return f[i][j];
        }
        int ans = dfs(i + 1, j - 1) + dfs(Math.abs(i - 1), j - 1);
        ans %= mod;
        return f[i][j] = ans;
    }
}

// Accepted solution for LeetCode #2400: Number of Ways to Reach a Position After Exactly k Steps
func numberOfWays(startPos int, endPos int, k int) int {
	const mod = 1e9 + 7
	f := make([][]int, k+1)
	for i := range f {
		f[i] = make([]int, k+1)
		for j := range f[i] {
			f[i][j] = -1
		}
	}
	var dfs func(i, j int) int
	dfs = func(i, j int) int {
		if i > j || j < 0 {
			return 0
		}
		if j == 0 {
			if i == 0 {
				return 1
			}
			return 0
		}
		if f[i][j] != -1 {
			return f[i][j]
		}
		f[i][j] = (dfs(i+1, j-1) + dfs(abs(i-1), j-1)) % mod
		return f[i][j]
	}
	return dfs(abs(startPos-endPos), k)
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

# Accepted solution for LeetCode #2400: Number of Ways to Reach a Position After Exactly k Steps
class Solution:
    def numberOfWays(self, startPos: int, endPos: int, k: int) -> int:
        @cache
        def dfs(i: int, j: int) -> int:
            if i > j or j < 0:
                return 0
            if j == 0:
                return 1 if i == 0 else 0
            return (dfs(i + 1, j - 1) + dfs(abs(i - 1), j - 1)) % mod

        mod = 10**9 + 7
        return dfs(abs(startPos - endPos), k)

// Accepted solution for LeetCode #2400: Number of Ways to Reach a Position After Exactly k Steps
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2400: Number of Ways to Reach a Position After Exactly k Steps
// class Solution {
//     private Integer[][] f;
//     private final int mod = (int) 1e9 + 7;
// 
//     public int numberOfWays(int startPos, int endPos, int k) {
//         f = new Integer[k + 1][k + 1];
//         return dfs(Math.abs(startPos - endPos), k);
//     }
// 
//     private int dfs(int i, int j) {
//         if (i > j || j < 0) {
//             return 0;
//         }
//         if (j == 0) {
//             return i == 0 ? 1 : 0;
//         }
//         if (f[i][j] != null) {
//             return f[i][j];
//         }
//         int ans = dfs(i + 1, j - 1) + dfs(Math.abs(i - 1), j - 1);
//         ans %= mod;
//         return f[i][j] = ans;
//     }
// }

// Accepted solution for LeetCode #2400: Number of Ways to Reach a Position After Exactly k Steps
function numberOfWays(startPos: number, endPos: number, k: number): number {
    const mod = 10 ** 9 + 7;
    const f = new Array(k + 1).fill(0).map(() => new Array(k + 1).fill(-1));
    const dfs = (i: number, j: number): number => {
        if (i > j || j < 0) {
            return 0;
        }
        if (j === 0) {
            return i === 0 ? 1 : 0;
        }
        if (f[i][j] !== -1) {
            return f[i][j];
        }
        f[i][j] = dfs(i + 1, j - 1) + dfs(Math.abs(i - 1), j - 1);
        f[i][j] %= mod;
        return f[i][j];
    };
    return dfs(Math.abs(startPos - endPos), k);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(k^2)

Space

O(k^2)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

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.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

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.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

State misses one required dimension

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.