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Count of Smaller Numbers After Self with Fenwick Tree (Binary Indexed Tree)

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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].

Note there are multiple ways to approach this problem. Some solutions use modified merge sort but this is a great example to introduce the fenwick tree.

class Solution {
public List<Integer> countSmaller(int[] a) {
    int n = a.length;
    int []arr = compressedArray(a);
    int val= arr.length;
    FenwickTree BIT = new FenwickTree(val);
    List<Integer>ls = new ArrayList<>();
    for(int i = n; i >= 1; i--) {
        ls.add(BIT.query(arr[i] - 1));
        BIT.update(arr[i], 1);
    return ls;

private int[]compressedArray (int []a){
    int n = a.length;
    Map<Integer, Integer> map = new TreeMap<>();
    for(int i : a) {
        map.put(i, 0);
    int val = 1;
    for(var entry : map.entrySet()) {
        map.put(entry.getKey(), val);
    int arr[] = new int[n+1];
    for(int i =1 ; i <=n; i++) {
        arr[i] = map.get(a[i-1]);
    return arr;
class FenwickTree {
int[] bit;
int n;
FenwickTree(int n) {
    this.n = n;
    this.bit = new int[n + 1];

public void update(int i, int val) {
    while (i < bit.length) {
        bit[i] += val;
        i += (i & (-i));

public int query(int i) {
    int sum = 0;
    while (i > 0) {
        sum += bit[i];
        i -= (i & (-i));
    return sum;

public int rangeSum(int l, int r) {
    return query(r) - query(l - 1);

Explanation of the Fenwick Tree (also called Binary Index Tree or BIT):

and here is a merge sort solution for comparison:

int[] count;
public List<Integer> countSmaller(int[] nums) {
    List<Integer> res = new ArrayList<Integer>();     

    count = new int[nums.length];
    int[] indexes = new int[nums.length];
    for(int i = 0; i < nums.length; i++){
    	indexes[i] = i;
    mergesort(nums, indexes, 0, nums.length - 1);
    for(int i = 0; i < count.length; i++){
    return res;
private void mergesort(int[] nums, int[] indexes, int start, int end){
	if(end <= start){
	int mid = (start + end) / 2;
	mergesort(nums, indexes, start, mid);
	mergesort(nums, indexes, mid + 1, end);
	merge(nums, indexes, start, end);
private void merge(int[] nums, int[] indexes, int start, int end){
	int mid = (start + end) / 2;
	int left_index = start;
	int right_index = mid+1;
	int rightcount = 0;    	
	int[] new_indexes = new int[end - start + 1];

	int sort_index = 0;
	while(left_index <= mid && right_index <= end){
		if(nums[indexes[right_index]] < nums[indexes[left_index]]){
			new_indexes[sort_index] = indexes[right_index];
			new_indexes[sort_index] = indexes[left_index];
			count[indexes[left_index]] += rightcount;
	while(left_index <= mid){
		new_indexes[sort_index] = indexes[left_index];
		count[indexes[left_index]] += rightcount;
	while(right_index <= end){
		new_indexes[sort_index++] = indexes[right_index++];
	for(int i = start; i <= end; i++){
		indexes[i] = new_indexes[i - start];

Posted by Jamie Meyer 3 months ago

Related Problems

Design and implement a data structure for a LFU cache.

Implement the LFUCache class:

LFUCache(int capacity) Initializes the object with the capacity of the data structure.

int get(int key) Gets the value of the key if the key exists in the cache. Otherwise, returns -1.

void put(int key, int value) Update the value of the key if present, or inserts the key if not already present. When the cache reaches its capacity, it should invalidate and remove the least frequently used key before inserting a new item. For this problem, when there is a tie (i.e., two or more keys with the same frequency), the least recently used key would be invalidated.

To determine the least frequently used key, a use counter is maintained for each key in the cache. The key with the smallest use counter is the least frequently used key.

When a key is first inserted into the cache, its use counter is set to 1 (due to the put operation). The use counter for a key in the cache is incremented either a get or put operation is called on it.

The functions get and put must each run in O(1) average time complexity.