LeetCode #1040 — MEDIUM

Moving Stones Until Consecutive II

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

There are some stones in different positions on the X-axis. You are given an integer array stones, the positions of the stones.

Call a stone an endpoint stone if it has the smallest or largest position. In one move, you pick up an endpoint stone and move it to an unoccupied position so that it is no longer an endpoint stone.

In particular, if the stones are at say, stones = [1,2,5], you cannot move the endpoint stone at position 5, since moving it to any position (such as 0, or 3) will still keep that stone as an endpoint stone.

The game ends when you cannot make any more moves (i.e., the stones are in three consecutive positions).

Return an integer array answer of length 2 where:

answer[0] is the minimum number of moves you can play, and
answer[1] is the maximum number of moves you can play.

Example 1:

Input: stones = [7,4,9]
Output: [1,2]
Explanation: We can move 4 -> 8 for one move to finish the game.
Or, we can move 9 -> 5, 4 -> 6 for two moves to finish the game.

Example 2:

Input: stones = [6,5,4,3,10]
Output: [2,3]
Explanation: We can move 3 -> 8 then 10 -> 7 to finish the game.
Or, we can move 3 -> 7, 4 -> 8, 5 -> 9 to finish the game.
Notice we cannot move 10 -> 2 to finish the game, because that would be an illegal move.

Constraints:

3 <= stones.length <= 10⁴
1 <= stones[i] <= 10⁹
All the values of stones are unique.

Step 01

Brute Force Baseline

Problem summary: There are some stones in different positions on the X-axis. You are given an integer array stones, the positions of the stones. Call a stone an endpoint stone if it has the smallest or largest position. In one move, you pick up an endpoint stone and move it to an unoccupied position so that it is no longer an endpoint stone. In particular, if the stones are at say, stones = [1,2,5], you cannot move the endpoint stone at position 5, since moving it to any position (such as 0, or 3) will still keep that stone as an endpoint stone. The game ends when you cannot make any more moves (i.e., the stones are in three consecutive positions). Return an integer array answer of length 2 where: answer[0] is the minimum number of moves you can play, and answer[1] is the maximum number of moves you can play.

Baseline thinking

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

Pattern signal: Array · Math · Sliding Window

Example 1

[7,4,9]

Example 2

[6,5,4,3,10]

Core Insight

What unlocks the optimal approach

For the minimum, how many stones are already in place? For the maximum, we have to lose either the gap A[1] - A[0] or A[N-1] - A[N-2] (where N = A.length), but every other space can be occupied.

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 #1040: Moving Stones Until Consecutive II
class Solution {
    public int[] numMovesStonesII(int[] stones) {
        Arrays.sort(stones);
        int n = stones.length;
        int mi = n;
        int mx = Math.max(stones[n - 1] - stones[1] + 1, stones[n - 2] - stones[0] + 1) - (n - 1);
        for (int i = 0, j = 0; j < n; ++j) {
            while (stones[j] - stones[i] + 1 > n) {
                ++i;
            }
            if (j - i + 1 == n - 1 && stones[j] - stones[i] == n - 2) {
                mi = Math.min(mi, 2);
            } else {
                mi = Math.min(mi, n - (j - i + 1));
            }
        }
        return new int[] {mi, mx};
    }
}

// Accepted solution for LeetCode #1040: Moving Stones Until Consecutive II
func numMovesStonesII(stones []int) []int {
	sort.Ints(stones)
	n := len(stones)
	mi := n
	mx := max(stones[n-1]-stones[1]+1, stones[n-2]-stones[0]+1) - (n - 1)
	i := 0
	for j, x := range stones {
		for x-stones[i]+1 > n {
			i++
		}
		if j-i+1 == n-1 && stones[j]-stones[i] == n-2 {
			mi = min(mi, 2)
		} else {
			mi = min(mi, n-(j-i+1))
		}
	}
	return []int{mi, mx}
}

# Accepted solution for LeetCode #1040: Moving Stones Until Consecutive II
class Solution:
    def numMovesStonesII(self, stones: List[int]) -> List[int]:
        stones.sort()
        mi = n = len(stones)
        mx = max(stones[-1] - stones[1] + 1, stones[-2] - stones[0] + 1) - (n - 1)
        i = 0
        for j, x in enumerate(stones):
            while x - stones[i] + 1 > n:
                i += 1
            if j - i + 1 == n - 1 and x - stones[i] == n - 2:
                mi = min(mi, 2)
            else:
                mi = min(mi, n - (j - i + 1))
        return [mi, mx]

// Accepted solution for LeetCode #1040: Moving Stones Until Consecutive II
struct Solution;

impl Solution {
    fn num_moves_stones_ii(mut stones: Vec<i32>) -> Vec<i32> {
        stones.sort_unstable();
        let n = stones.len();
        let mut min = n as i32;
        let mut max = 0;
        max = max.max(stones[n - 1] - stones[1] + 1 - n as i32 + 1);
        max = max.max(stones[n - 2] - stones[0] + 1 - n as i32 + 1);
        let mut l = 0;
        for r in 0..n {
            while stones[r] - stones[l] + 1 > n as i32 {
                l += 1;
            }
            let d = r - l + 1;
            if d == n - 1 && stones[r] - stones[l] + 1 == n as i32 - 1 {
                min = min.min(2);
            } else {
                min = min.min((n - d) as i32);
            }
        }
        vec![min, max]
    }
}

#[test]
fn test() {
    let stones = vec![7, 4, 9];
    let res = vec![1, 2];
    assert_eq!(Solution::num_moves_stones_ii(stones), res);
    let stones = vec![6, 5, 4, 3, 10];
    let res = vec![2, 3];
    assert_eq!(Solution::num_moves_stones_ii(stones), res);
    let stones = vec![100, 101, 104, 102, 103];
    let res = vec![0, 0];
    assert_eq!(Solution::num_moves_stones_ii(stones), res);
}

// Accepted solution for LeetCode #1040: Moving Stones Until Consecutive II
function numMovesStonesII(stones: number[]): number[] {
    stones.sort((a, b) => a - b);
    const n = stones.length;
    let mi = n;
    const mx = Math.max(stones[n - 1] - stones[1] + 1, stones[n - 2] - stones[0] + 1) - (n - 1);
    for (let i = 0, j = 0; j < n; ++j) {
        while (stones[j] - stones[i] + 1 > n) {
            ++i;
        }
        if (j - i + 1 === n - 1 && stones[j] - stones[i] === n - 2) {
            mi = Math.min(mi, 2);
        } else {
            mi = Math.min(mi, n - (j - i + 1));
        }
    }
    return [mi, mx];
}

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(k)

Approach Breakdown

BRUTE FORCE

O(n × k) time

O(1) space

For each starting index, scan the next k elements to compute the window aggregate. There are n−k+1 starting positions, each requiring O(k) work, giving O(n × k) total. No extra space since we recompute from scratch each time.

SLIDING WINDOW

O(n) time

O(k) space

The window expands and contracts as we scan left to right. Each element enters the window at most once and leaves at most once, giving 2n total operations = O(n). Space depends on what we track inside the window (a hash map of at most k distinct elements, or O(1) for a fixed-size window).

Shortcut: Each element enters and exits the window once → O(n) amortized, regardless of window size.

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.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.