LeetCode 2400: Number of Ways to Reach a Position After Exactly k Steps
LeetCode 2400 Solution Explanation
Explanation:
To solve this problem, we can use dynamic programming to keep track of the number of ways to reach a certain position after a certain number of steps. We can create a 2D array dp where dp[i][j] represents the number of ways to reach position i after j steps. We can fill up this array iteratively based on the following recurrence relation:
dp[i][j] = dp[i-1][j-1] + dp[i+1][j-1]
This means that the number of ways to reach position i after j steps is the sum of the number of ways to reach position i-1 and i+1 after j-1 steps.
We need to handle the boundary conditions carefully, as we can only move left or right by one step at each move.
The final answer will be the value at endPos in the dp array after k steps.
Time Complexity:
The time complexity of this solution is O(k * endPos) where k is the number of steps and endPos is the end position.
Space Complexity:
The space complexity of this solution is O(k * endPos) for the 2D array dp.
:
LeetCode 2400 Solutions in Java, C++, Python
class Solution {
public int numWays(int startPos, int endPos, int k) {
int MOD = 1000000007;
int maxPos = Math.max(Math.abs(startPos), Math.abs(endPos));
int[][] dp = new int[2 * maxPos + 1][k + 1];
dp[startPos + maxPos][0] = 1;
for (int j = 1; j <= k; j++) {
for (int i = 0; i < dp.length; i++) {
dp[i][j] = ((i > 0 ? dp[i - 1][j - 1] : 0) + (i < dp.length - 1 ? dp[i + 1][j - 1] : 0)) % MOD;
}
}
return dp[endPos + maxPos][k];
}
}Interactive Code Editor for LeetCode 2400
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