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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:
<li>The number of elements currently in <code>nums</code> that are <strong>strictly less than</strong> <code>instructions[i]</code>.</li>
<li>The number of elements currently in <code>nums</code> that are <strong>strictly greater than</strong> <code>instructions[i]</code>.</li>
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:
<li><code>1 <= instructions.length <= 10<sup>5</sup></code></li>
<li><code>1 <= instructions[i] <= 10<sup>5</sup></code></li>
Binary Indexed Tree or Segment Tree.
Binary Indexed Tree:
class BinaryIndexedTree:
def __init__(self, n):
self.n = n
self.c = [0] * (n + 1)
@staticmethod
def lowbit(x):
return x & -x
def update(self, x, delta):
while x <= self.n:
self.c[x] += delta
x += BinaryIndexedTree.lowbit(x)
def query(self, x):
s = 0
while x > 0:
s += self.c[x]
x -= BinaryIndexedTree.lowbit(x)
return s
class Solution:
def createSortedArray(self, instructions: List[int]) -> int:
n = max(instructions)
tree = BinaryIndexedTree(n)
ans = 0
for num in instructions:
a = tree.query(num - 1)
b = tree.query(n) - tree.query(num)
ans += min(a, b)
tree.update(num, 1)
return ans % int((1e9 + 7))
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 Solution {
public int createSortedArray(int[] instructions) {
int n = 100010;
int mod = (int) 1e9 + 7;
BinaryIndexedTree tree = new BinaryIndexedTree(n);
int ans = 0;
for (int num : instructions) {
int a = tree.query(num - 1);
int b = tree.query(n) - tree.query(num);
ans += Math.min(a, b);
ans %= mod;
tree.update(num, 1);
}
return ans;
}
}
class BinaryIndexedTree {
private int n;
private int[] c;
public BinaryIndexedTree(int n) {
this.n = n;
c = new int[n + 1];
}
public void update(int x, int delta) {
while (x <= n) {
c[x] += delta;
x += lowbit(x);
}
}
public int query(int x) {
int s = 0;
while (x > 0) {
s += c[x];
x -= lowbit(x);
}
return s;
}
public static int lowbit(int x) {
return x & -x;
}
}
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:
int n;
vector<int> c;
BinaryIndexedTree(int _n)
: n(_n)
, c(_n + 1) { }
void update(int x, int delta) {
while (x <= n) {
c[x] += delta;
x += lowbit(x);
}
}
int query(int x) {
int s = 0;
while (x > 0) {
s += c[x];
x -= lowbit(x);
}
return s;
}
int lowbit(int x) {
return x & -x;
}
};
class Solution {
public:
int createSortedArray(vector<int>& instructions) {
int n = 100010;
int mod = 1e9 + 7;
BinaryIndexedTree* tree = new BinaryIndexedTree(n);
int ans = 0;
for (int num : instructions) {
int a = tree->query(num - 1);
int b = tree->query(n) - tree->query(num);
ans += min(a, b);
ans %= mod;
tree->update(num, 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) lowbit(x int) int {
return x & -x
}
func (this *BinaryIndexedTree) update(x, delta int) {
for x <= this.n {
this.c[x] += delta
x += this.lowbit(x)
}
}
func (this *BinaryIndexedTree) query(x int) int {
s := 0
for x > 0 {
s += this.c[x]
x -= this.lowbit(x)
}
return s
}
func createSortedArray(instructions []int) int {
n := 100010
mod := int(1e9 + 7)
tree := newBinaryIndexedTree(n)
ans := 0
for _, num := range instructions {
a, b := tree.query(num-1), tree.query(n)-tree.query(num)
ans += min(a, b)
ans %= mod
tree.update(num, 1)
}
return ans
}
func min(a, b int) int {
if a < b {
return a
}
return b
}
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