LeetCode #1046 — EASY

Last Stone Weight

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

You are given an array of integers stones where stones[i] is the weight of the i^th 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, and
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 weight of the last remaining stone. If there are no stones left, return 0.

Example 1:

Input: stones = [2,7,4,1,8,1]
Output: 1
Explanation: 
We combine 7 and 8 to get 1 so the array converts to [2,4,1,1,1] then,
we combine 2 and 4 to get 2 so the array converts to [2,1,1,1] then,
we combine 2 and 1 to get 1 so the array converts to [1,1,1] then,
we combine 1 and 1 to get 0 so the array converts to [1] then that's the value of the last stone.

Example 2:

Input: stones = [1]
Output: 1

Constraints:

1 <= stones.length <= 30
1 <= stones[i] <= 1000

Step 01

Brute Force Baseline

Problem summary: 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, and 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 weight of the last remaining stone. If there are no stones left, return 0.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[2,7,4,1,8,1]

Example 2

[1]

Step 02

Core Insight

What unlocks the optimal approach

Simulate the process. We can do it with a heap, or by sorting some list of stones every time we take a turn.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1046: Last Stone Weight
class Solution {
    public int lastStoneWeight(int[] stones) {
        PriorityQueue<Integer> q = new PriorityQueue<>((a, b) -> b - a);
        for (int x : stones) {
            q.offer(x);
        }
        while (q.size() > 1) {
            int y = q.poll();
            int x = q.poll();
            if (x != y) {
                q.offer(y - x);
            }
        }
        return q.isEmpty() ? 0 : q.poll();
    }
}

// Accepted solution for LeetCode #1046: Last Stone Weight
func lastStoneWeight(stones []int) int {
	q := &hp{stones}
	heap.Init(q)
	for q.Len() > 1 {
		y, x := q.pop(), q.pop()
		if x != y {
			q.push(y - x)
		}
	}
	if q.Len() > 0 {
		return q.IntSlice[0]
	}
	return 0
}

type hp struct{ sort.IntSlice }

func (h hp) Less(i, j int) bool { return h.IntSlice[i] > h.IntSlice[j] }
func (h *hp) Push(v any)        { h.IntSlice = append(h.IntSlice, v.(int)) }
func (h *hp) Pop() any {
	a := h.IntSlice
	v := a[len(a)-1]
	h.IntSlice = a[:len(a)-1]
	return v
}
func (h *hp) push(v int) { heap.Push(h, v) }
func (h *hp) pop() int   { return heap.Pop(h).(int) }

# Accepted solution for LeetCode #1046: Last Stone Weight
class Solution:
    def lastStoneWeight(self, stones: List[int]) -> int:
        h = [-x for x in stones]
        heapify(h)
        while len(h) > 1:
            y, x = -heappop(h), -heappop(h)
            if x != y:
                heappush(h, x - y)
        return 0 if not h else -h[0]

// Accepted solution for LeetCode #1046: Last Stone Weight
impl Solution {
    pub fn last_stone_weight(stones: Vec<i32>) -> i32 {
        let mut stones_heap = std::collections::BinaryHeap::new();
        for stone in stones {
            stones_heap.push(stone);
        }

        while stones_heap.len() > 1 {
            let first = stones_heap.pop().unwrap();
            let second = stones_heap.pop().unwrap();
            stones_heap.push(first - second);
        }

        match stones_heap.peek() {
            Some(val) => *val,
            None => 0,
        }
    }
}

// Accepted solution for LeetCode #1046: Last Stone Weight
function lastStoneWeight(stones: number[]): number {
    const pq = new MaxPriorityQueue<number>();
    for (const x of stones) {
        pq.enqueue(x);
    }
    while (pq.size() > 1) {
        const y = pq.dequeue();
        const x = pq.dequeue();
        if (x !== y) {
            pq.enqueue(y - x);
        }
    }
    return pq.isEmpty() ? 0 : pq.dequeue();
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.