LeetCode #1033 — MEDIUM

Moving Stones Until Consecutive

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

The Problem

Problem Statement

There are three stones in different positions on the X-axis. You are given three integers a, b, and c, the positions of the stones.

In one move, you pick up a stone at an endpoint (i.e., either the lowest or highest position stone), and move it to an unoccupied position between those endpoints. Formally, let's say the stones are currently at positions x, y, and z with x < y < z. You pick up the stone at either position x or position z, and move that stone to an integer position k, with x < k < z and k != y.

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: a = 1, b = 2, c = 5
Output: [1,2]
Explanation: Move the stone from 5 to 3, or move the stone from 5 to 4 to 3.

Example 2:

Input: a = 4, b = 3, c = 2
Output: [0,0]
Explanation: We cannot make any moves.

Example 3:

Input: a = 3, b = 5, c = 1
Output: [1,2]
Explanation: Move the stone from 1 to 4; or move the stone from 1 to 2 to 4.

Constraints:

1 <= a, b, c <= 100
a, b, and c have different values.

Step 01

Brute Force Baseline

Problem summary: There are three stones in different positions on the X-axis. You are given three integers a, b, and c, the positions of the stones. In one move, you pick up a stone at an endpoint (i.e., either the lowest or highest position stone), and move it to an unoccupied position between those endpoints. Formally, let's say the stones are currently at positions x, y, and z with x < y < z. You pick up the stone at either position x or position z, and move that stone to an integer position k, with x < k < z and k != y. 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: Math

Example 1

1
2
5

Example 2

4
3
2

Example 3

3
5
1

Step 02

Core Insight

What unlocks the optimal approach

For the minimum: We can always do it in at most 2 moves, by moving one stone next to another, then the third stone next to the other two. When can we do it in 1 move? 0 moves? For the maximum: Every move, the maximum position minus the minimum position must decrease by at least 1.

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 #1033: Moving Stones Until Consecutive
class Solution {
    public int[] numMovesStones(int a, int b, int c) {
        int x = Math.min(a, Math.min(b, c));
        int z = Math.max(a, Math.max(b, c));
        int y = a + b + c - x - z;
        int mi = 0, mx = 0;
        if (z - x > 2) {
            mi = y - x < 3 || z - y < 3 ? 1 : 2;
            mx = z - x - 2;
        }
        return new int[] {mi, mx};
    }
}

// Accepted solution for LeetCode #1033: Moving Stones Until Consecutive
func numMovesStones(a int, b int, c int) []int {
	x := min(a, min(b, c))
	z := max(a, max(b, c))
	y := a + b + c - x - z
	mi, mx := 0, 0
	if z-x > 2 {
		mi = 2
		if y-x < 3 || z-y < 3 {
			mi = 1
		}
		mx = z - x - 2
	}
	return []int{mi, mx}
}

# Accepted solution for LeetCode #1033: Moving Stones Until Consecutive
class Solution:
    def numMovesStones(self, a: int, b: int, c: int) -> List[int]:
        x, z = min(a, b, c), max(a, b, c)
        y = a + b + c - x - z
        mi = mx = 0
        if z - x > 2:
            mi = 1 if y - x < 3 or z - y < 3 else 2
            mx = z - x - 2
        return [mi, mx]

// Accepted solution for LeetCode #1033: Moving Stones Until Consecutive
struct Solution;

impl Solution {
    fn num_moves_stones(a: i32, b: i32, c: i32) -> Vec<i32> {
        let mut a: Vec<i32> = vec![a, b, c];
        a.sort_unstable();
        let min = if a[0] + 1 == a[1] && a[1] + 1 == a[2] {
            0
        } else {
            if a[0] + 2 >= a[1] || a[1] + 2 >= a[2] {
                1
            } else {
                2
            }
        };
        let max = a[1] - a[0] - 1 + a[2] - a[1] - 1;
        vec![min, max]
    }
}

#[test]
fn test() {
    let a = 1;
    let b = 2;
    let c = 5;
    let res = vec![1, 2];
    assert_eq!(Solution::num_moves_stones(a, b, c), res);
    let a = 4;
    let b = 3;
    let c = 2;
    let res = vec![0, 0];
    assert_eq!(Solution::num_moves_stones(a, b, c), res);
    let a = 3;
    let b = 5;
    let c = 1;
    let res = vec![1, 2];
    assert_eq!(Solution::num_moves_stones(a, b, c), res);
}

// Accepted solution for LeetCode #1033: Moving Stones Until Consecutive
function numMovesStones(a: number, b: number, c: number): number[] {
    const x = Math.min(a, Math.min(b, c));
    const z = Math.max(a, Math.max(b, c));
    const y = a + b + c - x - z;
    let mi = 0,
        mx = 0;
    if (z - x > 2) {
        mi = y - x < 3 || z - y < 3 ? 1 : 2;
        mx = z - x - 2;
    }
    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(log n)

Space

O(1)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

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.