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1649. Create Sorted Array through Instructions

题目

Given an integer array instructions, you are asked to create a sorted array from the elements in instructions. You start with an empty container nums. For each element from left to right in instructions, insert it into nums. The cost of each insertion is the minimum of the following:

• The number of elements currently in nums that are strictly less than instructions[i].
• The number of elements currently in nums that are strictly greater than instructions[i].

For example, if inserting element 3 into nums = [1,2,3,5], the cost of insertion is min(2, 1) (elements 1 and 2 are less than 3, element 5 is greater than 3) and nums will become [1,2,3,3,5].

Return the total cost to insert all elements from instructions into nums. Since the answer may be large, return it modulo 10^9 + 7

Example 1:

Input: instructions = [1,5,6,2]
Output: 1
Explanation: Begin with nums = [].
Insert 1 with cost min(0, 0) = 0, now nums = [1].
Insert 5 with cost min(1, 0) = 0, now nums = [1,5].
Insert 6 with cost min(2, 0) = 0, now nums = [1,5,6].
Insert 2 with cost min(1, 2) = 1, now nums = [1,2,5,6].
The total cost is 0 + 0 + 0 + 1 = 1.

Example 2:

Input: instructions = [1,2,3,6,5,4]
Output: 3
Explanation: Begin with nums = [].
Insert 1 with cost min(0, 0) = 0, now nums = [1].
Insert 2 with cost min(1, 0) = 0, now nums = [1,2].
Insert 3 with cost min(2, 0) = 0, now nums = [1,2,3].
Insert 6 with cost min(3, 0) = 0, now nums = [1,2,3,6].
Insert 5 with cost min(3, 1) = 1, now nums = [1,2,3,5,6].
Insert 4 with cost min(3, 2) = 2, now nums = [1,2,3,4,5,6].
The total cost is 0 + 0 + 0 + 0 + 1 + 2 = 3.

Example 3:

Input: instructions = [1,3,3,3,2,4,2,1,2]
Output: 4
Explanation: Begin with nums = [].
Insert 1 with cost min(0, 0) = 0, now nums = [1].
Insert 3 with cost min(1, 0) = 0, now nums = [1,3].
Insert 3 with cost min(1, 0) = 0, now nums = [1,3,3].
Insert 3 with cost min(1, 0) = 0, now nums = [1,3,3,3].
Insert 2 with cost min(1, 3) = 1, now nums = [1,2,3,3,3].
Insert 4 with cost min(5, 0) = 0, now nums = [1,2,3,3,3,4].
Insert 2 with cost min(1, 4) = 1, now nums = [1,2,2,3,3,3,4].
Insert 1 with cost min(0, 6) = 0, now nums = [1,1,2,2,3,3,3,4].
Insert 2 with cost min(2, 4) = 2, now nums = [1,1,2,2,2,3,3,3,4].
The total cost is 0 + 0 + 0 + 0 + 1 + 0 + 1 + 0 + 2 = 4.

Constraints:

• 1 <= instructions.length <= 105
• 1 <= instructions[i] <= 105

题目大意

给你一个整数数组 instructions ，你需要根据 instructions 中的元素创建一个有序数组。一开始你有一个空的数组 nums ，你需要 从左到右 遍历 instructions 中的元素，将它们依次插入 nums 数组中。每一次插入操作的 代价 是以下两者的 较小值 ：

• nums 中 严格小于 instructions[i] 的数字数目。
• nums 中 严格大于 instructions[i] 的数字数目。

比方说，如果要将 3 插入到 nums = [1,2,3,5] ，那么插入操作的 代价 为 min(2, 1) (元素 1 和 2 小于 3 ，元素 5 大于 3 ），插入后 nums 变成 [1,2,3,3,5] 。请你返回将 instructions 中所有元素依次插入 nums 后的 总最小代价 。由于答案会很大，请将它对 10^9 + 7 取余 后返回。

解题思路

• 给出一个数组，要求将其中的元素从头开始往另外一个空数组中插入，每次插入前，累加代价值 cost = min(strictly less than, strictly greater than)。最后输出累加值。
• 这一题虽然是 Hard 题，但是读完题以后就可以判定这是模板题了。可以用线段树和树状数组来解决。这里简单说说线段树的思路吧，先将待插入的数组排序，获得总的区间。每次循环做 4 步：2 次 query 分别得到 strictlyLessThan 和 strictlyGreaterThan ，再比较出两者中的最小值累加，最后一步就是 update。
• 由于题目给的数据比较大，所以建立线段树之前记得要先离散化。这一题核心代码不超过 10 行，其他的都是模板代码。具体实现见代码。

代码

package leetcode

import (
	"sort"

	"github.com/halfrost/LeetCode-Go/template"
)

// 解法一 树状数组 Binary Indexed Tree
func createSortedArray(instructions []int) int {
	bit, res := template.BinaryIndexedTree{}, 0
	bit.Init(100001)
	for i, v := range instructions {
		less := bit.Query(v - 1)
		greater := i - bit.Query(v)
		res = (res + min(less, greater)) % (1e9 + 7)
		bit.Add(v, 1)
	}
	return res
}

// 解法二 线段树 SegmentTree
func createSortedArray1(instructions []int) int {
	if len(instructions) == 0 {
		return 0
	}
	st, res, mod := template.SegmentCountTree{}, 0, 1000000007
	numsMap, numsArray, tmpArray := discretization1649(instructions)
	// 初始化线段树，节点内的值都赋值为 0，即计数为 0
	st.Init(tmpArray, func(i, j int) int {
		return 0
	})
	for i := 0; i < len(instructions); i++ {
		strictlyLessThan := st.Query(0, numsMap[instructions[i]]-1)
		strictlyGreaterThan := st.Query(numsMap[instructions[i]]+1, numsArray[len(numsArray)-1])
		res = (res + min(strictlyLessThan, strictlyGreaterThan)) % mod
		st.UpdateCount(numsMap[instructions[i]])
	}
	return res
}

func discretization1649(instructions []int) (map[int]int, []int, []int) {
	tmpArray, numsArray, numsMap := []int{}, []int{}, map[int]int{}
	for i := 0; i < len(instructions); i++ {
		numsMap[instructions[i]] = instructions[i]
	}
	for _, v := range numsMap {
		numsArray = append(numsArray, v)
	}
	sort.Ints(numsArray)
	for i, num := range numsArray {
		numsMap[num] = i
	}
	for i := range numsArray {
		tmpArray = append(tmpArray, i)
	}
	return numsMap, numsArray, tmpArray
}

func min(a int, b int) int {
	if a > b {
		return b
	}
	return a
}