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Jump Game II LC #45 Medium Greedy · BFS Levels
Problem

You are given a 0-indexed array of integers nums of length n. You start at nums[0]. Each element nums[i] represents the maximum jump length from that position. Return the minimum number of jumps to reach nums[n-1]. The test cases guarantee you can always reach the last index.

Example 1
Input: nums = [2,3,1,1,4]
Output: 2
Explanation: Jump 1 step from index 0 to index 1 (val 3), then jump 3 steps to the end. Min jumps = 2.
Example 2
Input: nums = [2,3,0,1,4]
Output: 2
Explanation: Jump to index 1 (val 3), then jump to index 4. Min jumps = 2.
Constraints: 1 ≤ nums.length ≤ 10⁴  |  0 ≤ nums[i] ≤ 1000  |  You can always reach the last index.
Try Examples
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Approach
Greedy BFS Levels — Minimum Jumps
Each jump covers a "level" of platforms. Scan the level, find the farthest reach, then jump to the next level.
Minimum Jumps
0
scanning level 0
curEnd: 0 farthest: 0
Jump Arena
Lv 0 (start)
Lv 1
Lv 2
Lv 3+
Variables
Index i
—
jumps
0
curEnd
0
farthest
0
Step Logic
Press ▶ Play or Next Step to begin the animation.
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Ready
0 / 0
Select an example above and press Play.
Algorithm
1
Init jumps=0, curEnd=0, farthest=0
2
Loop i = 0 .. n-2: update farthest = max(farthest, i + nums[i])
3
When i == curEnd: jumps++, advance curEnd = farthest
4
Return jumps — the minimum number of jumps to reach last index
Time
O(n)
Space
O(1)
Why Greedy BFS Works

Think of each jump as a BFS level. Level 0 = just index 0. Level 1 = all platforms reachable in 1 jump. Level 2 = all reachable in 2 jumps, and so on. Instead of tracking exact paths, we scan the current level to find the farthest reachable index. When we exhaust the current level (i == curEnd), we commit exactly one jump and the new level spans curEnd+1 to farthest. This greedy frontier expansion always yields the minimum jumps in O(n).