Algorithm
Problem Name: 315. Count of Smaller Numbers After Self
Given an integer array nums, return an integer array counts 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
Code Examples
#1 Code Example with C Programming
Code -
C Programming
typedef struct node_s {
int key;
int lnum;
int dup;
struct node_s *left;
struct node_s *right;
} node_t;
void insert(node_t **nodep, node_t *buff, int key, int *pi, int lnum) {
node_t *node = *nodep;
if (!node) {
*pi = lnum;
node = buff;
node->key = key;
node->dup = 1;
*nodep = node;
} else if (node->key == key) {
node->dup ++;
*pi = lnum + node->lnum;
} else if (node->key > key) {
node->lnum ++;
insert(&node->left, buff, key, pi, lnum);
} else {
insert(&node->right, buff, key, pi, lnum + node->lnum + node->dup);
}
}
int* countSmaller(int* nums, int numsSize, int* returnSize) {
int *p, i;
node_t *node, *buff;
p = malloc(numsSize * sizeof(int));
buff = calloc(numsSize, sizeof(node_t));
//assert(p && buff);
node = NULL;
for (i = numsSize - 1; i >= 0; i --) {
insert(&node, &buff[i], nums[i], &p[i], 0);
}
free(buff);
*returnSize = numsSize;
return p;
}
Input
nums = [5,2,6,1]
Output
[2,1,1,0]
#2 Code Example with C++ Programming
Code -
C++ Programming
class Solution {
struct TreeNode{
int val;
int sum;
TreeNode* left;
TreeNode* right;
TreeNode(int x, int y): val(x), sum(y), left(NULL), right(NULL){}
};
public:
vector<int> countSmaller(vector<int>& nums) {
TreeNode* root = NULL;
vector<int>res(nums.size());
for(int i = nums.size() - 1; i >= 0; i--) root = buildTree(root, i, nums[i], 0, res);
return res;
}
TreeNode* buildTree(TreeNode* root, int i, int val, int count, vector<int>& res){
if(!root){
root = new TreeNode(val, 1);
res[i] = count;
return root;
}
if(val > root->val) root->right = buildTree(root->right, i, val, count + root->sum, res);
else{
root->sum++;
root->left = buildTree(root->left, i, val, count, res);
}
return root;
}
};
Input
nums = [5,2,6,1]
Output
[2,1,1,0]
#3 Code Example with Java Programming
Code -
Java Programming
class Solution {
public List countSmaller(int[] nums) {
int n = nums.length;
int[] result = new int[n];
int[] indices = new int[n];
for (int i = 0; i < n; i++) {
indices[i] = i;
}
mergesort(indices, 0, n, result, nums);
return Arrays.stream(result).boxed().collect(Collectors.toList());
}
private void mergesort(int[] indices, int left, int right, int[] result, int[] nums) {
if (right - left < = 1) {
return;
}
int mid = (left + right) / 2;
mergesort(indices, left, mid, result, nums);
mergesort(indices, mid, right, result, nums);
merge(indices, left, right, mid, result, nums);
}
private void merge(int[] indices, int left, int right, int mid, int[] result, int[] nums) {
int i = left;
int j = mid;
List < Integer> temp = new ArrayList<>();
while (i < mid && j < right) {
if (nums[indices[i]] <= nums[indices[j]]) {
result[indices[i]] += j - mid;
temp.add(indices[i]);
i++;
} else {
temp.add(indices[j]);
j++;
}
}
while (i < mid) {
result[indices[i]] += j - mid;
temp.add(indices[i]);
i++;
}
while (j < right) {
temp.add(indices[j]);
j++;
}
for (int k = left; k < right; k++) {
indices[k] = temp.get(k - left);
}
}
}
Input
nums = [-1]
Output
[0]
#4 Code Example with Javascript Programming
Code -
Javascript Programming
const countSmaller = function(nums) {
const numsAndIndexes = nums.map((x, i) => [x, i])
const output = [...new Array(nums.length)].map(_ => 0)
mergeSort(numsAndIndexes, output)
return output
}
function mergeSort(arr, output) {
if (arr.length <= 1) return arr
const middle = Math.floor(arr.length / 2)
const left = mergeSort(arr.slice(0, middle), output),
right = mergeSort(arr.slice(middle), output)
const sorted = []
let i = 0,
j = 0
while (i < left.length || j < right.length) {
if (i >= left.length) {
sorted.push(right[j])
j++
} else if (j >= right.length) {
sorted.push(left[i])
i++
} else {
if (left[i][0] > right[j][0]) {
sorted.push(left[i])
output[left[i][1]] += right.length - j
i++
} else {
sorted.push(right[j])
j++
}
}
}
return sorted
}
Input
nums = [-1]
Output
[0]
#5 Code Example with Python Programming
Code -
Python Programming
class Solution:
def countSmaller(self, nums):
r = []
for i in range(len(nums) - 1, -1, -1):
bisect.insort(r, nums[i])
nums[i] = bisect.bisect_left(r, nums[i])
return nums
Input
nums = [-1,-1]
Output
[0,0]
#6 Code Example with C# Programming
Code -
C# Programming
using System.Collections.Generic;
namespace LeetCode
{
public class _0315_CountOfSmallerNumbersAfterSelf
{
public IList < int> CountSmaller(int[] nums)
{
if (nums.Length == 0) return nums;
int[] counts = new int[nums.Length];
Node root = new Node(nums[nums.Length - 1]);
for (int i = nums.Length - 2; i >= 0; i--)
counts[i] = Insert(root, nums[i]);
return counts;
}
private int Insert(Node root, int value)
{
if (value < root.Value)
{
root.LessCount++;
if (root.Left == null)
{
root.Left = new Node(value);
return 0;
}
return Insert(root.Left, value);
}
else if (root.Value < value)
{
if (root.Right == null)
{
root.Right = new Node(value);
return root.EqualCount + root.LessCount;
}
return root.EqualCount + root.LessCount + Insert(root.Right, value);
}
else
{
root.EqualCount++;
return root.LessCount;
}
}
public class Node
{
public int Value;
public int EqualCount = 1;
public int LessCount = 0;
public Node Left;
public Node Right;
public Node(int value)
{
this.Value = value;
}
}
}
}
Input
nums = [-1,-1]
Output
[0,0]