1649. Create Sorted Array through Instructions (H)

Create Sorted Array through Instructions (H)

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 109 + 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 <= 10^5
• 1 <= instructions[i] <= 10^5

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题意

从左到右遍历一个数组，每次将当前元素x插入到新数组newArr中是新数组保持递增，且定义一次插入的花费是min(newArr中严格小于x的元素个数，newArr中严格大于x的元素个数)。问全部插入后的总花费。

思路

直接使用二分法去每次查找小于/大于x的个数会超时，原因在于插入到新数组的复杂度是(O(N))，总复杂度为(O(N^2))。

使用树状数组，单点修改+区间查询，tree[i]表示在已插入的元素中数字i出现的次数，每次遍历时查询小于/大于x的个数以及插入新元素的复杂度都为(O(logN))，总复杂度为(O(NlogN))。

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代码实现

Java

class Solution {
    public int createSortedArray(int[] instructions) {
        int cost = 0;
        int[] tree = new int[100001];
        for (int i = 0; i < instructions.length; i++) {
            int num = instructions[i];
            int leftLen = sum(tree, num - 1);
            int rightLen = i - sum(tree, num);
            cost = (cost + Math.min(leftLen, rightLen)) % 1000000007;
            add(tree, num, 1, 100000);
        }
        return cost;
    }

    private int lowbit(int x) {
        return x & -x;
    }

    private void add(int[] tree, int index, int value, int max) {
        while (index <= max) {
            tree[index] += value;
            index += lowbit(index);
        }
    }

    private int sum(int[] tree, int index) {
        int sum = 0;
        while (index > 0) {
            sum += tree[index];
            index -= lowbit(index);
        }
        return sum;
    }
}