[LeetCode] Jump Game II

This problem has a nice BFS structure. Let's illustrate it using the example nums = [2, 3, 1, 1, 4] in the problem statement. We are initially at position 0. Then we can move at most nums[0]steps from it. So, after one move, we may reach nums[1] = 3 or nums[2] = 1. So these nodes are reachable in 1 move. From these nodes, we can further move to nums[3] = 1 and nums[4] = 4. Now you can see that the target nums[4] = 4 is reachable in 2 moves.

Putting these into codes, we keep two pointers start and end that record the current range of the starting nodes. Each time after we make a move, update start to be end + 1 and end to be the farthest index that can be reached in 1 move from the current [start, end].

To get an accepted solution, it is important to handle all the edge cases. And the following codes handle all of them in a unified way without using the unclean if statements :-)

---

C++

class Solution {
public:
    int jump(vector<int>& nums) {
        int n = nums.size(), step = 0, start = 0, end = 0;
        while (end < n - 1) {
            step++; 
            int maxend = end + 1;
            for (int i = start; i <= end; i++) {
                if (i + nums[i] >= n - 1) return step;
                maxend = max(maxend, i + nums[i]);
            }
            start = end + 1;
            end = maxend;
        }
        return step;
    }
};

---

Python

class Solution:
    # @param {integer[]} nums
    # @return {integer}
    def jump(self, nums):
        n, start, end, step = len(nums), 0, 0, 0
        while end < n - 1:
            step += 1
            maxend = end + 1
            for i in range(start, end + 1):
                if i + nums[i] >= n - 1:
                    return step
                maxend = max(maxend, i + nums[i])
            start, end = end + 1, maxend
        return step