1033-moving-stones-until-consecutive.js

// 1033. Moving Stones Until Consecutive
// Easy   39%


// Three stones are on a number line at positions a, b, and c.
// Each turn, you pick up a stone at an endpoint (ie., 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, 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, ie. the stones are in
// consecutive positions.
// When the game ends, what is the minimum and maximum number of moves that you
// could have made?  Return the answer as an length 2 array: answer =
// [minimum_moves, maximum_moves]

// 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.

// Note:
//     1 <= a <= 100
//     1 <= b <= 100
//     1 <= c <= 100
//     a != b, b != c, c != a




/**
 * @param {number} a
 * @param {number} b
 * @param {number} c
 * @return {number[]}
 */
const numMovesStones = function(a, b, c) {
  [a, b, c] = [a, b, c].sort((a, b) => a - b)
  let min = 2, p = b - a, q = c - b
  if (p === 1 && q === 1) min = 0
  else if (p === 1 || q === 1 || p === 2 || q === 2) min = 1
  return [min, c - a - 2]
}

;[
  [1, 2, 5],  // [1,2]
  [4, 3, 2],  // [0,0]
  [3, 5, 1],  // [1,2]
  [5, 1, 10], // [2,7]
].forEach(([a, b, c]) => {
  console.log(numMovesStones(a, b, c))
})

// Solution:
// 假设 a < b < c
// 最大移动值为 c - a - 2
// 最小移动值为
// 1) b - a = 1 && c - b = 1, 0
// 2) b - a = 1 || c - b = 1 || b - a = 2 || c - b = 2, 1
// 3) 2

// Submission Result: Accepted