JumpGameV.java

/* (C) 2024 YourCompanyName */
package dynamic_programming;

/**
 * Created by gouthamvidyapradhan on 18/02/2020 Given an array of integers arr and an integer d. In
 * one step you can jump from index i to index:
 *
 * <p>i + x where: i + x < arr.length and 0 < x <= d. i - x where: i - x >= 0 and 0 < 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)).
 *
 * <p>You can choose any index of the array and start jumping. Return the maximum number of indices
 * you can visit.
 *
 * <p>Notice that you can not jump outside of the array at any time.
 *
 * <p>Example 1:
 *
 * <p>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:
 *
 * <p>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:
 *
 * <p>Input: arr = [7,6,5,4,3,2,1], d = 1 Output: 7 Explanation: Start at index 0. You can visit all
 * the indicies. Example 4:
 *
 * <p>Input: arr = [7,1,7,1,7,1], d = 2 Output: 2 Example 5:
 *
 * <p>Input: arr = [66], d = 1 Output: 1
 *
 * <p>Constraints:
 *
 * <p>1 <= arr.length <= 1000 1 <= arr[i] <= 10^5 1 <= d <= arr.length
 */
public class JumpGameV {
  public static void main(String[] args) {
    int[] A = {7, 1, 7, 1, 7, 1};
    System.out.println(new JumpGameV().maxJumps(A, 2));
  }

  int[] DP;

  public int maxJumps(int[] arr, int d) {
    DP = new int[arr.length];
    // Arrays.fill(DP, -1);
    int max = 0;
    for (int i = 0; i < arr.length; i++) {
      max = Math.max(max, dp(arr, d, i));
    }
    return max;
  }

  private int dp(int[] A, int d, int i) {
    if (DP[i] != 0) return DP[i];
    int max = 1;
    for (int j = i - 1; j >= (i - d); j--) {
      if (j < 0 || A[j] >= A[i]) break;
      max = Math.max(max, dp(A, d, j) + 1);
    }
    for (int j = i + 1; j <= (i + d); j++) {
      if (j >= A.length || A[j] >= A[i]) break;
      max = Math.max(max, dp(A, d, j) + 1);
    }
    DP[i] = max;
    return max;
  }
}