Jump Game II [LC#45]

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. Return the minimum number of jumps to reach the last position.

Intuition

• There is a DP approach which takes $O(n^2)$ steps[i] = 1+min(steps[j], 0<=j<i)

• Say we are at a position left and max reachable range is till right.
• By using a node in this range, we can extend this reachable range by adding another jump.
• The extended range is max(nums[i]+i for i in [left, right])

Code

def min_jump(nums: List[int]) -> int:
    n = len(nums)
    if n == 1: return 0
    left, right, jumps = 0, 0, 0
    while right < n - 1:
        max_reach = max(nums[i] + i for i in range(left, right + 1))
        left = right 
        right = max_reach
        jumps += 1
    return jumps

Time complexity

• $T(n) = O(n)$
• $S(n) = O(1)$