2770. Maximum Number of Jumps to Reach the Last Index

#php #leetcode #algorithms #programming

Difficulty: Medium

Topics: Senior, Array, Dynamic Programming, Weekly Contest 353

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

Hint:

Use a dynamic programming approach.
Define a dynamic programming array dp of size n, where dp[i] represents the maximum number of jumps from index 0 to index i.
For each j iterate over all i < j. Set dp[j] = max(dp[j], dp[i] + 1) if -target <= nums[j] - nums[i] <= target.

Solution:

The solution uses dynamic programming to find the maximum number of jumps from index 0 to index n-1 in the array nums, where each jump from i to j must satisfy -target ≤ nums[j] − nums[i] ≤ target.

If the last index is unreachable, return -1. The DP array tracks the maximum jumps possible to reach each index, and the answer is the value stored at the last index.

Approach:

Dynamic Programming Array: Create a dp array of size n where dp[i] stores the maximum number of jumps to reach index i from index 0. Initialize dp[0] = 0 and all others to -inf (or PHP_INT_MIN).
Transition: For each j from 1 to n-1, check all i < j. If the difference between nums[j] and nums[i] is within [-target, target] and dp[i] is not -inf, update dp[j] as max(dp[j], dp[i] + 1).
Result: If dp[n-1] is still -inf, return -1. Otherwise, return dp[n-1].

Let's implement this solution in PHP: 2770. Maximum Number of Jumps to Reach the Last Index

<?php
/**
 * @param Integer[] $nums
 * @param Integer $target
 * @return Integer
 */
function maximumJumps(array $nums, int $target): int
{
    ...
    ...
    ...
    /**
     * go to ./solution.php
     */
}

// Test cases
echo maximumJumps([1,3,6,4,1,2], 2) . "\n";             // Output: 3
echo maximumJumps([1,3,6,4,1,2], 3) . "\n";             // Output: 5
echo maximumJumps([1,3,6,4,1,2], 0) . "\n";             // Output: -1
?>

Explanation:

Initialization: dp[0] = 0 because we start at index 0 with 0 jumps. All other indices are initially unreachable (PHP_INT_MIN).
Outer loop (j): Iterates over each target index j from 1 to n-1 to compute the best way to reach j.
Inner loop (i): Checks all previous indices i from 0 to j-1 to see if a valid jump from i to j exists.
Valid jump condition: -target <= nums[j] - nums[i] <= target ensures the jump is allowed.
State update: If i is reachable (dp[i] != PHP_INT_MIN), then dp[j] = max(dp[j], dp[i] + 1) means we can extend the path ending at i to j with one more jump.
Why this works: The DP ensures that dp[j] always holds the maximum jumps to j because we consider all possible i < j that can jump to j and keep the best value.
Return: If dp[n-1] is still -inf, no sequence exists → return -1. Otherwise, return the computed maximum jumps.

Complexity

Time Complexity: O(n²), because we have two nested loops iterating over n elements each.
Space Complexity: O(n), for the DP array.

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