2400 - Number of Ways to Reach a Position After Exactly k Steps (Medium)

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 positionendPosstarting fromstartPos, such that you perform exactlyksteps. 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

Approach 1: nCr

Written by @wingkwong

struct comb {
 public:
  int MOD = 1e9 + 7;
  vector<long long> fac, finv, inv;
  comb(int mxN) {
    fac.resize(mxN);
    finv.resize(mxN);
    inv.resize(mxN);
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 1; i < mxN - 1; i++) {
      fac[i + 1] = fac[i] * (i + 1) % MOD;
      inv[i + 1] = MOD - inv[MOD % (i + 1)] * (MOD / (i + 1)) % MOD;
      finv[i + 1] = finv[i] * inv[i + 1] % MOD;
    }
  }

  long long ncr(long long n, long long k) {
    if (k < 0 || n < k) return 0;
    long long res = 1;
    res *= fac[n] * finv[k] % MOD * finv[n - k] % MOD;
    return res;
  }
};
class Solution {
public:
    int numberOfWays(int startPos, int endPos, int k) {
        int MOD = 1e9 + 7;
        // init comb
        // see constraints: 1 <= startPos, endPos, k <= 1000
        comb c = comb(1005);
        // if (startPos + endPos) and k have different parity, then return 0
        // otherwise we can use comb to get k choose (endPos - startPos + k) / 2
        // it can also be k choose k - (endPos - startPos + k) / 2)
        // assuming endPos >= startPos, and we know
        // left + right = k -- (1)
        // right - left = endPos - startPos -- (2)
        // (1) + (2)
        // left + right - left + right = k + endPos - startPos
        // 2 * right = endPos - startPos + k
        // right = (endPos - startPos + k) / 2
        // in other word, left would be k - (endPos - startPos + k) / 2)
        return (((startPos + endPos) ^ k) & 1) ? 0 : c.ncr(k, (endPos - startPos + k) / 2) % MOD;
    }
};

Written by @wingkwong

class Solution:
    def numberOfWays(self, startPos: int, endPos: int, k: int) -> int:
        MOD = 10 ** 9 + 7
        # if (startPos + endPos) and k have different parity, then return 0
        # otherwise we can use comb to get k choose (endPos - startPos + k) / 2
        # it can also be k choose k - (endPos - startPos + k) / 2)
        # assuming endPos >= startPos, and we know
        # left + right = k -- (1)
        # right - left = endPos - startPos -- (2)
        # (1) + (2)
        # left + right - left + right = k + endPos - startPos
        # 2 * right = endPos - startPos + k
        # right = (endPos - startPos + k) / 2
        # in other word, left would be k - (endPos - startPos + k) / 2)
        if (startPos + endPos) % 2 != k % 2:
            return 0
        else:
            return comb(k, (endPos - startPos + k) // 2) % MOD