The Jump Game II interview question is an optimization challenge. You are given an array of non-negative integers nums, where each element represents your maximum jump distance from that position. You start at the first index and your goal is to reach the last index in the minimum number of jumps. You are guaranteed that you can always reach the end.
Companies like Amazon, Microsoft, and Google use the Jump Game II coding problem to test a candidate's ability to move beyond basic search and apply Greedy or Dynamic Programming optimizations. While DP is , the greedy solution is . It evaluations your ability to maintain multiple "boundaries" (current reach vs. next reach) in a single pass.
This problem is best solved using the Greedy (Level-by-Level) pattern.
jumps = 0, current_jump_end = 0, farthest = 0.farthest = max(farthest, i + nums[i]). This is the furthest you can go with one more jump.current_jump_end, it means you must make another jump to proceed.
jumps.current_jump_end = farthest.jumps is your answer.nums = [2, 3, 1, 1, 4]
farthest = 2. We reached current_jump_end (0). jumps = 1, current_jump_end = 2.farthest = max(2, 1+3) = 4.farthest = max(4, 2+1) = 4. We reached current_jump_end (2). jumps = 2, current_jump_end = 4.
Result: 2 jumps.dp[i] is the min jumps to reach i. This is correct but too slow for .Think of the greedy approach as a Breadth-First Search on an array. Each "jump" is a level in the BFS. You are trying to find the minimum number of levels to reach the target index. This is a vital Array interview pattern for shortest-path problems.
| Title | Difficulty | Topics | LeetCode |
|---|---|---|---|
| Jump Game | Medium | Solve | |
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