LeetCode #315 — HARD

Count of Smaller Numbers After Self

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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].

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search · Segment Tree

Example 1

[5,2,6,1]

Example 2

[-1]

Example 3

[-1,-1]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #315: Count of Smaller Numbers After Self
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;
    }
}

// Accepted solution for LeetCode #315: Count of Smaller Numbers After Self
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
}

# Accepted solution for LeetCode #315: Count of Smaller Numbers After Self
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]

// Accepted solution for LeetCode #315: Count of Smaller Numbers After Self
struct Solution;

use std::collections::BTreeMap;

impl Solution {
    fn count_smaller(nums: Vec<i32>) -> Vec<i32> {
        let mut count: BTreeMap<i32, usize> = BTreeMap::new();
        let n = nums.len();
        let mut res = vec![0; n];
        for i in (0..n).rev() {
            let mut sum = 0;
            let x = nums[i];
            for (_, v) in count.range(..x) {
                sum += v;
            }
            *count.entry(x).or_default() += 1;
            res[i] = sum as i32;
        }
        res
    }
}

#[test]
fn test() {
    let nums = vec![5, 2, 6, 1];
    let res = vec![2, 1, 1, 0];
    assert_eq!(Solution::count_smaller(nums), res);
}

// Accepted solution for LeetCode #315: Count of Smaller Numbers After Self
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #315: Count of Smaller Numbers After Self
// 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;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.