You are given an integer array nums and you have to return a new counts array. The counts array has the property where counts[i] is the number of smaller elements to the right of nums[i].
Example 1:
Input: nums = [5,2,6,1] Output: [2,1,1,0] Explanation: To the right of 5 there are 2 smaller elements (2 and 1). To the right of 2 there is only 1 smaller element (1). To the right of 6 there is 1 smaller element (1). To the right of 1 there is 0 smaller element.
Example 2:
Input: nums = [-1] Output: [0]
Example 3:
Input: nums = [-1,-1] Output: [0,0]
Constraints:
1 <= nums.length <= 105-104 <= nums[i] <= 104
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 countSmaller(self, nums: List[int]) -> List[int]:
alls = sorted(set(nums))
m = {v: i for i, v in enumerate(alls, 1)}
tree = BinaryIndexedTree(len(m))
ans = []
for v in nums[::-1]:
x = m[v]
tree.update(x, 1)
ans.append(tree.query(x - 1))
return ans[::-1]class Solution {
public List<Integer> countSmaller(int[] nums) {
Set<Integer> s = new HashSet<>();
for (int v : nums) {
s.add(v);
}
List<Integer> alls = new ArrayList<>(s);
alls.sort(Comparator.comparingInt(a -> a));
int n = alls.size();
Map<Integer, Integer> m = new HashMap<>(n);
for (int i = 0; i < n; ++i) {
m.put(alls.get(i), i + 1);
}
BinaryIndexedTree tree = new BinaryIndexedTree(n);
LinkedList<Integer> ans = new LinkedList<>();
for (int i = nums.length - 1; i >= 0; --i) {
int x = m.get(nums[i]);
tree.update(x, 1);
ans.addFirst(tree.query(x - 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;
}
}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:
vector<int> countSmaller(vector<int>& nums) {
unordered_set<int> s(nums.begin(), nums.end());
vector<int> alls(s.begin(), s.end());
sort(alls.begin(), alls.end());
unordered_map<int, int> m;
int n = alls.size();
for (int i = 0; i < n; ++i) m[alls[i]] = i + 1;
BinaryIndexedTree* tree = new BinaryIndexedTree(n);
vector<int> ans(nums.size());
for (int i = nums.size() - 1; i >= 0; --i)
{
int x = m[nums[i]];
tree->update(x, 1);
ans[i] = tree->query(x - 1);
}
return ans;
}
};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 countSmaller(nums []int) []int {
s := make(map[int]bool)
for _, v := range nums {
s[v] = true
}
var alls []int
for v := range s {
alls = append(alls, v)
}
sort.Ints(alls)
m := make(map[int]int)
for i, v := range alls {
m[v] = i + 1
}
ans := make([]int, len(nums))
tree := newBinaryIndexedTree(len(alls))
for i := len(nums) - 1; i >= 0; i-- {
x := m[nums[i]]
tree.update(x, 1)
ans[i] = tree.query(x - 1)
}
return ans
}