Question

2770. Maximum Number of Jumps to Reach the Last Index

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Description

You are given a 0-indexed array nums of n integers and an integer target.

You are initially positioned at index 0. In one step, you can jump from index i to any index j such that:

• 0 <= i < j < n
• -target <= nums[j] - nums[i] <= target

Return _the maximum number of jumps you can make to reach index_ n - 1.

If there is no way to reach index n - 1, return -1.

 

Example 1:

Input: nums = [1,3,6,4,1,2], target = 2
Output: 3
Explanation: To go from index 0 to index n - 1 with the maximum number of jumps, you can perform the following jumping sequence:
- Jump from index 0 to index 1. 
- Jump from index 1 to index 3.
- Jump from index 3 to index 5.
It can be proven that there is no other jumping sequence that goes from 0 to n - 1 with more than 3 jumps. Hence, the answer is 3. 

Example 2:

Input: nums = [1,3,6,4,1,2], target = 3
Output: 5
Explanation: To go from index 0 to index n - 1 with the maximum number of jumps, you can perform the following jumping sequence:
- Jump from index 0 to index 1.
- Jump from index 1 to index 2.
- Jump from index 2 to index 3.
- Jump from index 3 to index 4.
- Jump from index 4 to index 5.
It can be proven that there is no other jumping sequence that goes from 0 to n - 1 with more than 5 jumps. Hence, the answer is 5. 

Example 3:

Input: nums = [1,3,6,4,1,2], target = 0
Output: -1
Explanation: It can be proven that there is no jumping sequence that goes from 0 to n - 1. Hence, the answer is -1. 

 

Constraints:

• 2 <= nums.length == n <= 1000
• -109 <= nums[i] <= 109
• 0 <= target <= 2 * 109

Solutions

Solution 1: Memoization

For each position $i$, we consider to jump to position $j$ which satisfies $|nums[i] - nums[j]| \leq target$. Then we can jump from $i$ to $j$, and continue to jump from $j$ to the end.

Therefore, we design a function $dfs(i)$, which represents the maximum number of jumps needed to jump to the end index starting from position $i$. Then the answer is $dfs(0)$.

The calculation process of function $dfs(i)$ is as follows:

• If $i = n - 1$, then we have reached the end index and no jumps are required, so return $0$;
• Otherwise, we need to enumerate the positions $j$ that can be jumped from position $i$, and calculate the maximum number of jumps needed to jump to the end index starting from $j$, then $dfs(i)$ is equal to the maximum value of all $dfs(j)$ plus $1$. If there is no position $j$ that can be jumped from $i$, then $dfs(i) = -\infty$.

To avoid duplicate calculations, we can use memoization.

Time complexity $O(n^2)$, space complexity $O(n)$. where $n$ is the length of array.

class Solution:
  def maximumJumps(self, nums: List[int], target: int) -> int:
    @cache
    def dfs(i: int) -> int:
      if i == n - 1:
        return 0
      ans = -inf
      for j in range(i + 1, n):
        if abs(nums[i] - nums[j]) <= target:
          ans = max(ans, 1 + dfs(j))
      return ans

    n = len(nums)
    ans = dfs(0)
    return -1 if ans < 0 else ans

class Solution {
  private Integer[] f;
  private int[] nums;
  private int n;
  private int target;

  public int maximumJumps(int[] nums, int target) {
    n = nums.length;
    this.target = target;
    this.nums = nums;
    f = new Integer[n];
    int ans = dfs(0);
    return ans < 0 ? -1 : ans;
  }

  private int dfs(int i) {
    if (i == n - 1) {
      return 0;
    }
    if (f[i] != null) {
      return f[i];
    }
    int ans = -(1 << 30);
    for (int j = i + 1; j < n; ++j) {
      if (Math.abs(nums[i] - nums[j]) <= target) {
        ans = Math.max(ans, 1 + dfs(j));
      }
    }
    return f[i] = ans;
  }
}

class Solution {
public:
  int maximumJumps(vector<int>& nums, int target) {
    int n = nums.size();
    int f[n];
    memset(f, -1, sizeof(f));
    function<int(int)> dfs = [&](int i) {
      if (i == n - 1) {
        return 0;
      }
      if (f[i] != -1) {
        return f[i];
      }
      f[i] = -(1 << 30);
      for (int j = i + 1; j < n; ++j) {
        if (abs(nums[i] - nums[j]) <= target) {
          f[i] = max(f[i], 1 + dfs(j));
        }
      }
      return f[i];
    };
    int ans = dfs(0);
    return ans < 0 ? -1 : ans;
  }
};

func maximumJumps(nums []int, target int) int {
  n := len(nums)
  f := make([]int, n)
  for i := range f {
    f[i] = -1
  }
  var dfs func(int) int
  dfs = func(i int) int {
    if i == n-1 {
      return 0
    }
    if f[i] != -1 {
      return f[i]
    }
    f[i] = -(1 << 30)
    for j := i + 1; j < n; j++ {
      if abs(nums[i]-nums[j]) <= target {
        f[i] = max(f[i], 1+dfs(j))
      }
    }
    return f[i]
  }
  ans := dfs(0)
  if ans < 0 {
    return -1
  }
  return ans
}

func abs(x int) int {
  if x < 0 {
    return -x
  }
  return x
}

function maximumJumps(nums: number[], target: number): number {
  const n = nums.length;
  const f: number[] = Array(n).fill(-1);
  const dfs = (i: number): number => {
    if (i === n - 1) {
      return 0;
    }
    if (f[i] !== -1) {
      return f[i];
    }
    f[i] = -(1 << 30);
    for (let j = i + 1; j < n; ++j) {
      if (Math.abs(nums[i] - nums[j]) <= target) {
        f[i] = Math.max(f[i], 1 + dfs(j));
      }
    }
    return f[i];
  };
  const ans = dfs(0);
  return ans < 0 ? -1 : ans;
}