Jump Game II

Signal

"Minimum number of jumps" on top of a reachability array is a level-by-level expansion — the same shape as BFS on an implicit graph, but collapsible into a single greedy pass since only the farthest reach per level matters.

Approach

Think of each jump as covering a "level": everything reachable within the current jump count. Track two things while scanning left to right — the boundary of the current jump (current_end), and the farthest index reachable using one more jump (farthest). When the scan reaches current_end, that level is exhausted: increment the jump count and advance current_end to farthest.

Skeleton

jumps = current_end = farthest = 0
for i in range(len(nums) - 1):
    farthest = max(farthest, i + nums[i])
    if i == current_end:
        jumps += 1
        current_end = farthest
return jumps

Solution

def jump(nums: list[int]) -> int:
    jumps = current_end = farthest = 0
    for i in range(len(nums) - 1):
        farthest = max(farthest, i + nums[i])
        if i == current_end:
            jumps += 1
            current_end = farthest
    return jumps

Complexity

O(n) time — one pass. O(1) space.

Pitfalls

• The loop bound is len(nums) - 1, not len(nums) — you never need to "jump" from the last index itself, and including it can over-count by triggering one extra boundary increment.
• Incrementing jumps based on array index boundaries (current_end) rather than actual BFS queue levels is the whole trick — reproducing this with an actual BFS over positions works but is needless overhead.
• Confusing this with Jump Game (reachability only) is easy since the setup looks identical — the tell is "minimum number of jumps," which needs the extra current_end/jumps bookkeeping this problem adds.