Description
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + xwhere:i + x < arr.lengthand0 < x <= d.i - xwhere:i - x >= 0and0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2 Output: 4 Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown. Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9. Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3 Output: 1 Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1 Output: 7 Explanation: Start at index 0. You can visit all the indicies.
Constraints:
1 <= arr.length <= 10001 <= arr[i] <= 1051 <= d <= arr.length
Solutions
This solution uses memoization and explores neighbors by stepping exactly i positions in each direction (from 1 to d), stopping early when a boundary is hit or a value greater than or equal to the current position is encountered. The dfs function initializes max to 1 (representing the current position alone) and recursively calculates the maximum jumps from each valid neighbor, taking the overall maximum. Since the count begins at 1, the final answer is returned directly without incrementing, making this approach slightly more efficient in its initialization logic.
/**
* @param {number[]} arr
* @param {number} d
* @return {number}
*/
var maxJumps = function(arr, d) {
const n = arr.length;
const dp = new Array(n);
const dfs = (pos) => {
if (dp[pos] !== undefined) {
return dp[pos];
}
let max = 1;
for (let i = 1; i <= d; i++) {
if (pos - i < 0 || arr[pos - i] >= arr[pos]) break;
max = Math.max(max, dfs(pos - i) + 1);
}
for (let i = 1; i <= d; i++) {
if (pos + i >= n || arr[pos + i] >= arr[pos]) break;
max = Math.max(max, dfs(pos + i) + 1);
}
dp[pos] = max;
return max;
};
let res = 1;
for (let i = 0; i < n; i++) {
res = Math.max(res, dfs(i));
}
return res;
};