Greedy Algorithms - Jump Game II (Minimum Jumps)What is the worst-case time complexity of the BFS-based solution for Jump Game II on an input array of length n?AO(n^2)BO(n)CO(n log n)DO(n^3)Check Answer
Step-by-Step SolutionStep 1: Identify outer and inner loopsThe BFS processes each index once, but for each index, it may enqueue up to nums[pos] next positions, which can be up to n.Step 2: Analyze nested iterationIn worst case (e.g., nums[i] = n for all i), inner loop runs O(n) times for each of O(n) indices -> O(n^2) total.Final Answer:Option A -> Option AQuick Check:Nested loops over n indices and jumps -> O(n^2) [OK]Quick Trick: Nested loops over n indices and jumps -> O(n^2) [OK]Common Mistakes:MISTAKESAssuming BFS is always O(n)Confusing with greedy linear timeIgnoring inner loop over jump rangeTrap Explanation:PITFALLMany think BFS is linear, but here inner loop over jump range can be large, causing quadratic time.Interviewer Note:CONTEXTTests understanding of nested loops and worst-case input impact on complexity.
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