EASY NC#65 Heap / Priority Queue
1046. Last Stone Weight
📖 Problem
You are given an array of integers stones where stones[i] is the weight of the ith stone. We are playing a game with the stones. On each turn, we choose the heaviest two stones and smash them together. Suppose the heaviest two stones have weights x and y with x <= y. The result of this smash is: - If x == y, both stones are destroyed - If x != y, the stone of weight x is destroyed, and the stone of weight y has new weight y - x At the end of the game, there is at most one stone left. Return the smallest possible weight of the left stone. If there are no stones left, return 0.
🧠 Visual Learning Aid
1 Model the input into the right structure
2 Choose the core technique and invariant
3 Execute step-by-step with a sample
4 Validate complexity and edge cases
JS/TS Refreshers
- •Array methods (`push`, `pop`, `shift`, `slice`)
- •Object/Map/Set usage patterns
- •Function parameter and return typing
- •Level-order traversal
- •Queue discipline
- •Shortest-step interpretation
Logical Thinking Concepts
- •Define invariants before coding
- •Check edge cases first (`[]`, single element, duplicates)
- •Estimate time/space before implementation
- •Apply BFS reasoning pattern
- •Apply Heap reasoning pattern
- •Apply Backtracking reasoning pattern
💡 Approach
- → Use max heap to always get two heaviest stones
- → Simulate smashing until 0 or 1 stones remain
- → In TypeScript, use min heap with negation for max heap
- → Time: O(n log n), Space: O(n)
🛠️ Hints & Pitfalls
Hints
- •Use max heap to always get two heaviest stones
- •Simulate smashing until 0 or 1 stones remain
- •In TypeScript, use min heap with negation for max heap
Common Pitfalls
- •Time: O(n log n), Space: O(n)
🧪 Test Cases
Hidden tests on submit: 1
Test Case 1
Not run Input:
lastStoneWeight([2,7,4,1,8,1]); Expected:
1 Test Case 2
Not run Input:
lastStoneWeight([1]); Expected:
1 Test Case 3
Not run Input:
lastStoneWeight([10,10]); Expected:
0 📝 Code Editor
📤 Output