README_EN.md

2400. Number of Ways to Reach a Position After Exactly k Steps

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Description

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

Solutions

Python3

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)

Java

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;
    }
}

C++

class Solution {
public:
    int numberOfWays(int startPos, int endPos, int k) {
        const int mod = 1e9 + 7;
        int f[k + 1][k + 1];
        memset(f, -1, sizeof(f));
        function<int(int, int)> dfs = [&](int i, int j) -> int {
            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(abs(i - 1), j - 1)) % mod;
            return f[i][j];
        };
        return dfs(abs(startPos - endPos), k);
    }
};

Go

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
}

TypeScript

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);
}

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