45. Jump Game II

Problem¶

You are given a 0-indexed array of integers nums of length n. You are initially positioned at nums[0].

Each element nums[i] represents the maximum length of a forward jump from index i. In other words, if you are at nums[i], you can jump to any nums[i + j] where:

• 0 <= j <= nums[i] and
• i + j < n

Return the minimum number of jumps to reach nums[n - 1]. The test cases are generated such that you can reach nums[n - 1].

Example 1:

Input: nums = [2,3,1,1,4]
Output: 2
Explanation: The minimum number of jumps to reach the last index is 2. Jump 1 step from index 0 to 1, then 3 steps to the last index.

Example 2:

Input: nums = [2,3,0,1,4]
Output: 2

Constraints:

• 1 <= nums.length <= 10**4
• 0 <= nums[i] <= 1000
• It’s guaranteed that you can reach nums[n - 1].

Solve¶

Water falling¶

O(n)

Following a minimized greedy/dynamic programming minimized path problem from graph.

• Calling the minimum array is to store the total minimal step needed to reach each element, minimum[i] correspond to element[i]. At first, minimum array isn’t set, as none element is being processed.
• We start at index=0, cost step=0 jump step. As we already start at the first element. This also mean we have minimum[0] = 0
• From next, we start a loop
  • We find a element with index pos that have minimum step value, that is not visited/process
  • Flag it as already visited
  • Update all other element our current pos can jump into new minimum (step + 1) if it either uninitialized or it currently have a less optimal total of step need to jump into.

This is when I also realized we dealing with a serial of number, a special case of graph data:

• The minimum jump step to each element isn’t change value after initialized.

So that mean we can just skip most of the unnecessary loop to update the minimum.

class Solution:
    def jump(self, nums: List[int]) -> int:
        n = len(nums)
        p = 0
        minimum = [-1] * n
        minimum[0] = 0
        q = [(0, 0)]
        while q:
            pos, step = q.pop(0)
            for npos in range(p, min(n, pos + nums[pos]+1)):
                if minimum[npos] == -1:
                    minimum[npos] = step + 1
                    q.append((npos, step + 1))
                    p = npos
        return minimum[-1]

So I calling it water fall basically because the height being collapsing all at once (this is visualization process on white board)

Time complexity: O(n)

• BFS: Each minimum is updated once, visited once;
• DP update loop: By updating temporary p pointer, the nested loop only run in total n times independent with the total time BFS queue call (thus not our time complexity isn’t multiple to O(n^2) ).