0045 - Jump Game II
description
You are given a 0-indexed array of integers nums
of length n
. You are initially positioned at index 0.
Each element nums[i]
represents the maximum length of a forward jump from index i
. In other words, if you are at index i
, you can jump to any index (i + j)
where:
0 <= j <= nums[i]
andi + j < n
Return the minimum number of jumps to reach index n - 1
. The test cases are generated such that you can reach index 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 <= 104
0 <= nums[i] <= 1000
- It's guaranteed that you can reach
nums[n - 1]
.
submission
impl Solution {
// BFS solution, this is somewhat easier to understand but still
// remains O(n) time complexity and O(1) space
// since the range we could jump to is continuous, we could use a
// simple [left, right) range to record it.
// we update the farthest index we could jump to at each step
// and end if n - 1 is inside range [left, right), that is:
// left <= n - 1 && n - 1 < right
// or simply: right >= n
pub fn jump(nums: Vec<i32>) -> i32 {
let (mut left, mut right) = (0, 1);
let mut steps = 0;
// !(right >= n)
while right < nums.len() {
// update the farthest index we could jump to
let reach = (left..right).fold(right, |init, val| {
// don't forget +1 since right is open
init.max(val + nums[val] as usize + 1)
});
// if we are not guaranteed to reach n - 1, this would
// be wrong, assuming this function returns -1 when
// unreachable, we could add early return to fix it
// if reach <= right {
// return -1;
// }
(left, right) = (right, reach);
steps += 1;
}
steps
}
}