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 thaninstructions[i]
. - The number of elements currently in
nums
that are strictly greater thaninstructions[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 <= 105
1 <= instructions[i] <= 105
Binary Indexed Tree or Segment Tree.
Binary Indexed Tree:
class BinaryIndexedTree:
def __init__(self, n):
self.n = n
self.c = [0] * (n + 1)
def update(self, x: int, v: int):
while x <= self.n:
self.c[x] += v
x += x & -x
def query(self, x: int) -> int:
s = 0
while x:
s += self.c[x]
x -= x & -x
return s
class Solution:
def createSortedArray(self, instructions: List[int]) -> int:
m = max(instructions)
tree = BinaryIndexedTree(m)
ans = 0
mod = 10 ** 9 + 7
for i, x in enumerate(instructions):
cost = min(tree.query(x - 1), i - tree.query(x))
ans += cost
tree.update(x, 1)
return ans % mod
Segment Tree:
class Node:
def __init__(self):
self.l = 0
self.r = 0
self.v = 0
class SegmentTree:
def __init__(self, n):
self.tr = [Node() for _ in range(4 * n)]
self.build(1, 1, n)
def build(self, u, l, r):
self.tr[u].l = l
self.tr[u].r = r
if l == r:
return
mid = (l + r) >> 1
self.build(u << 1, l, mid)
self.build(u << 1 | 1, mid + 1, r)
def modify(self, u, x, v):
if self.tr[u].l == x and self.tr[u].r == x:
self.tr[u].v += v
return
mid = (self.tr[u].l + self.tr[u].r) >> 1
if x <= mid:
self.modify(u << 1, x, v)
else:
self.modify(u << 1 | 1, x, v)
self.pushup(u)
def pushup(self, u):
self.tr[u].v = self.tr[u << 1].v + self.tr[u << 1 | 1].v
def query(self, u, l, r):
if self.tr[u].l >= l and self.tr[u].r <= r:
return self.tr[u].v
mid = (self.tr[u].l + self.tr[u].r) >> 1
v = 0
if l <= mid:
v = self.query(u << 1, l, r)
if r > mid:
v += self.query(u << 1 | 1, l, r)
return v
class Solution:
def createSortedArray(self, instructions: List[int]) -> int:
n = max(instructions)
tree = SegmentTree(n)
ans = 0
for num in instructions:
a = tree.query(1, 1, num - 1)
b = tree.query(1, 1, n) - tree.query(1, 1, num)
ans += min(a, b)
tree.modify(1, num, 1)
return ans % int((1e9 + 7))
Binary Indexed Tree:
class BinaryIndexedTree {
private int n;
private int[] c;
public BinaryIndexedTree(int n) {
this.n = n;
this.c = new int[n + 1];
}
public void update(int x, int v) {
while (x <= n) {
c[x] += v;
x += x & -x;
}
}
public int query(int x) {
int s = 0;
while (x > 0) {
s += c[x];
x -= x & -x;
}
return s;
}
}
class Solution {
public int createSortedArray(int[] instructions) {
int m = 0;
for (int x : instructions) {
m = Math.max(m, x);
}
BinaryIndexedTree tree = new BinaryIndexedTree(m);
int ans = 0;
final int mod = (int) 1e9 + 7;
for (int i = 0; i < instructions.length; ++i) {
int x = instructions[i];
int cost = Math.min(tree.query(x - 1), i - tree.query(x));
ans = (ans + cost) % mod;
tree.update(x, 1);
}
return ans;
}
}
Segment Tree:
class Solution {
public int createSortedArray(int[] instructions) {
int n = 100010;
int mod = (int) 1e9 + 7;
SegmentTree tree = new SegmentTree(n);
int ans = 0;
for (int num : instructions) {
int a = tree.query(1, 1, num - 1);
int b = tree.query(1, 1, n) - tree.query(1, 1, num);
ans += Math.min(a, b);
ans %= mod;
tree.modify(1, num, 1);
}
return ans;
}
}
class Node {
int l;
int r;
int v;
}
class SegmentTree {
private Node[] tr;
public SegmentTree(int n) {
tr = new Node[4 * n];
for (int i = 0; i < tr.length; ++i) {
tr[i] = new Node();
}
build(1, 1, n);
}
public void build(int u, int l, int r) {
tr[u].l = l;
tr[u].r = r;
if (l == r) {
return;
}
int mid = (l + r) >> 1;
build(u << 1, l, mid);
build(u << 1 | 1, mid + 1, r);
}
public void modify(int u, int x, int v) {
if (tr[u].l == x && tr[u].r == x) {
tr[u].v += v;
return;
}
int mid = (tr[u].l + tr[u].r) >> 1;
if (x <= mid) {
modify(u << 1, x, v);
} else {
modify(u << 1 | 1, x, v);
}
pushup(u);
}
public void pushup(int u) {
tr[u].v = tr[u << 1].v + tr[u << 1 | 1].v;
}
public int query(int u, int l, int r) {
if (tr[u].l >= l && tr[u].r <= r) {
return tr[u].v;
}
int mid = (tr[u].l + tr[u].r) >> 1;
int v = 0;
if (l <= mid) {
v += query(u << 1, l, r);
}
if (r > mid) {
v += query(u << 1 | 1, l, r);
}
return v;
}
}
Binary Indexed Tree:
class BinaryIndexedTree {
public:
BinaryIndexedTree(int _n)
: n(_n)
, c(_n + 1) {}
void update(int x, int delta) {
while (x <= n) {
c[x] += delta;
x += x & -x;
}
}
int query(int x) {
int s = 0;
while (x) {
s += c[x];
x -= x & -x;
}
return s;
}
private:
int n;
vector<int> c;
};
class Solution {
public:
int createSortedArray(vector<int>& instructions) {
int m = *max_element(instructions.begin(), instructions.end());
BinaryIndexedTree tree(m);
const int mod = 1e9 + 7;
int ans = 0;
for (int i = 0; i < instructions.size(); ++i) {
int x = instructions[i];
int cost = min(tree.query(x - 1), i - tree.query(x));
ans = (ans + cost) % mod;
tree.update(x, 1);
}
return ans;
}
};
Segment Tree:
class Node {
public:
int l;
int r;
int v;
};
class SegmentTree {
public:
vector<Node*> tr;
SegmentTree(int n) {
tr.resize(4 * n);
for (int i = 0; i < tr.size(); ++i) tr[i] = new Node();
build(1, 1, n);
}
void build(int u, int l, int r) {
tr[u]->l = l;
tr[u]->r = r;
if (l == r) return;
int mid = (l + r) >> 1;
build(u << 1, l, mid);
build(u << 1 | 1, mid + 1, r);
}
void modify(int u, int x, int v) {
if (tr[u]->l == x && tr[u]->r == x)
{
tr[u]->v += v;
return;
}
int mid = (tr[u]->l + tr[u]->r) >> 1;
if (x <= mid) modify(u << 1, x, v);
else modify(u << 1 | 1, x, v);
pushup(u);
}
void pushup(int u) {
tr[u]->v = tr[u << 1]->v + tr[u << 1 | 1]->v;
}
int query(int u, int l, int r) {
if (tr[u]->l >= l && tr[u]->r <= r) return tr[u]->v;
int mid = (tr[u]->l + tr[u]->r) >> 1;
int v = 0;
if (l <= mid) v = query(u << 1, l, r);
if (r > mid) v += query(u << 1 | 1, l, r);
return v;
}
};
class Solution {
public:
int createSortedArray(vector<int>& instructions) {
int n = *max_element(instructions.begin(), instructions.end());
int mod = 1e9 + 7;
SegmentTree* tree = new SegmentTree(n);
int ans = 0;
for (int num : instructions)
{
int a = tree->query(1, 1, num - 1);
int b = tree->query(1, 1, n) - tree->query(1, 1, num);
ans += min(a, b);
ans %= mod;
tree->modify(1, num, 1);
}
return ans;
}
};
Binary Indexed Tree:
type BinaryIndexedTree struct {
n int
c []int
}
func newBinaryIndexedTree(n int) *BinaryIndexedTree {
c := make([]int, n+1)
return &BinaryIndexedTree{n, c}
}
func (this *BinaryIndexedTree) update(x, delta int) {
for x <= this.n {
this.c[x] += delta
x += x & -x
}
}
func (this *BinaryIndexedTree) query(x int) int {
s := 0
for x > 0 {
s += this.c[x]
x -= x & -x
}
return s
}
func createSortedArray(instructions []int) (ans int) {
m := 0
for _, x := range instructions {
m = max(m, x)
}
tree := newBinaryIndexedTree(m)
const mod = 1e9 + 7
for i, x := range instructions {
cost := min(tree.query(x-1), i-tree.query(x))
ans = (ans + cost) % mod
tree.update(x, 1)
}
return
}
func max(a, b int) int {
if a > b {
return a
}
return b
}
func min(a, b int) int {
if a < b {
return a
}
return b
}
class BinaryIndexedTree {
private n: number;
private c: number[];
constructor(n: number) {
this.n = n;
this.c = new Array(n + 1).fill(0);
}
public update(x: number, v: number): void {
while (x <= this.n) {
this.c[x] += v;
x += x & -x;
}
}
public query(x: number): number {
let s = 0;
while (x > 0) {
s += this.c[x];
x -= x & -x;
}
return s;
}
}
function createSortedArray(instructions: number[]): number {
const m = Math.max(...instructions);
const tree = new BinaryIndexedTree(m);
let ans = 0;
const mod = 10 ** 9 + 7;
for (let i = 0; i < instructions.length; ++i) {
const x = instructions[i];
const cost = Math.min(tree.query(x - 1), i - tree.query(x));
ans = (ans + cost) % mod;
tree.update(x, 1);
}
return ans;
}