LeetCode – Count of Smaller Numbers After Self (Java)

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:

Input: [5,2,6,1]
Output: [2,1,1,0]

Java Solution 1

public List<Integer> countSmaller(int[] nums) {
    List<Integer> result = new ArrayList<Integer>();
    ArrayList<Integer> sorted = new ArrayList<Integer>();
 
    for(int i=nums.length-1; i>=0; i--){
        if(sorted.isEmpty()){
            sorted.add(nums[i]);
            result.add(0);
        }else if(nums[i]>sorted.get(sorted.size()-1)){
            sorted.add(sorted.size(), nums[i]);
            result.add(sorted.size()-1);
        }else{
            int l=0; 
            int r=sorted.size()-1;
 
            while(l<r){
                int m = l + (r-l)/2;
 
                if(nums[i]>sorted.get(m)){
                    l=m+1;
                }else{
                    r=m;
                }
            }
 
            sorted.add(r, nums[i]);
            result.add(r);
        }    
    }
 
    Collections.reverse(result);
 
    return result;
}

This solution is simple. However, note that time complexity of adding an element to a list is O(n), because elements after the insertion position need to be shifted. So the time complexity is O(n^2(logn)).

Java Solution 2

If we want to use binary search, and define a structure like the following:

On average, time complexity is O(n*log(n)) and space complexity is O(n).

class Solution {
    public List<Integer> countSmaller(int[] nums) {
 
        List<Integer> result = new ArrayList<Integer>();
        if(nums==null || nums.length==0){
            return result;
        }
 
        Node root = new Node(nums[nums.length-1]);
        root.count=1;
        result.add(0);
 
        for(int i=nums.length-2; i>=0; i--){
            result.add(insertNode(root, nums[i]));
        }
 
        Collections.reverse(result);
 
        return result;
    }
 
    public int insertNode(Node root, int value){
        Node p=root;
        int result=0;
 
        while(p!=null){
            if(value>p.value){
                result+=p.count+p.numLeft;
                if(p.right==null){
                    Node t = new Node(value);
                    t.count=1;
                    p.right=t;
                    return result;
                }else{
                    p=p.right;
                }
            }else if(value==p.value){
                p.count++;
                return result+p.numLeft;
            }else{
                p.numLeft++;
 
                if(p.left==null){
                    Node t = new Node(value);
                    t.count=1;
                    p.left=t;
                    return result;
                }else{
                    p=p.left;
                }
            }
        }
 
        return 0;
    }
}
 
class Node{
    Node left;
    Node right;
 
    int value;
    int count;
    int numLeft;
    public Node(int value){
        this.value=value;
    }
}
Category >> Algorithms  
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