What is the time complexity of Count of Smaller Numbers After Self?

The optimal solution has a time complexity of O(n log n).

What data structures are used to solve Count of Smaller Numbers After Self?

This problem uses Array, Binary Search, Divide and Conquer, Binary Indexed Tree, Segment Tree, Merge Sort, Ordered Set. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Count of Smaller Numbers After Self?

Insight 1: The core idea is to efficiently count smaller elements to the right for each element in the input array. Insight 2: A naive approach (iterating through the remaining array for each element) would lead to O(n^2) time complexity, which is not efficient enough. We need a more optimized approach.

Which companies ask LeetCode 315. Count of Smaller Numbers After Self in interviews?

This problem is frequently asked at Geico.

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Count of Smaller Numbers After Self - LeetCode 315 Solution

Count of Smaller Numbers After Self - Complete Solution Guide

Count of Smaller Numbers After Self is LeetCode problem 315, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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]

Detailed Explanation

The problem asks us to find, for each element in a given array `nums`, the number of elements to its right that are smaller than it. The output should be an array `counts` where `counts[i]` represents the number of elements smaller than `nums[i]` located to the right of `nums[i]` in the original array `nums`. The constraints specify the array's length can be up to 10^5 and the values can range from -10^4 to 10^4.

Solution Approach

The provided solution uses a modified Merge Sort algorithm. The key idea is that during the merge process, when an element from the left subarray is copied back to the main array, we can determine how many elements from the right subarray were smaller than it. This number represents the number of elements to the right of the original element (from the left subarray) that were smaller than it. We maintain an array `counts` to store these counts for each element based on its original index.

Step-by-Step Algorithm

Step 1: Create a `items` array where each element is a pair consisting of a number from `nums` and its original index.
Step 2: Implement a Merge Sort algorithm. This involves recursively dividing the array into smaller sub-arrays until each sub-array contains only one element.
Step 3: Implement the `merge` function. This function merges two sorted sub-arrays. While merging, if an element from the left subarray is less than or equal to an element from the right subarray, it means that all the remaining elements in the right subarray are *not* smaller than the current left subarray element. So, we increment the count of smaller numbers to the right of the element from the left subarray by the number of elements already merged from the right subarray.
Step 4: After the Merge Sort is complete, the `counts` array will contain the number of smaller elements to the right of each element in the original array. Return this array.

Key Insights

Insight 1: The core idea is to efficiently count smaller elements to the right for each element in the input array.
Insight 2: A naive approach (iterating through the remaining array for each element) would lead to O(n^2) time complexity, which is not efficient enough. We need a more optimized approach.
Insight 3: Merge Sort can be leveraged to solve the problem in O(n log n) time complexity. The trick is to modify the merge step to count the number of smaller elements encountered from the right subarray while merging.
Insight 4: Pairing each number with its original index is crucial to correctly update the `counts` array, as the numbers are reordered during the sorting process.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Binary Search, Divide and Conquer, Binary Indexed Tree, Segment Tree, Merge Sort, Ordered Set.

Companies

Asked at: Geico.

Count of Smaller Numbers After Self | LeetCode 315

Hard

Array Binary Search Divide and Conquer Binary Indexed Tree Segment Tree Merge Sort Ordered Set

Time: O(n log n)Space: O(n)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

class Solution {
    public List<Integer> countSmaller(int[] nums) {
        int n = nums.length;
        Integer[] counts = new Integer[n];
        Arrays.fill(counts, 0);
        Pair[] items = new Pair[n];
        for (int i = 0; i < n; i++) {
            items[i] = new Pair(nums[i], i);
        }

        merge_sort(items, counts, 0, n);
        return Arrays.asList(counts);
    }

    private void merge_sort(Pair[] items, Integer[] counts, int start, int end) {
        if (end - start <= 1) {
            return;
        }

        int mid = start + (end - start) / 2;
        merge_sort(items, counts, start, mid);
        merge_sort(items, counts, mid, end);
        merge(items, counts, start, mid, end);
    }

    private void merge(Pair[] items, Integer[] counts, int start, int mid, int end) {
        Pair[] left_copy = Arrays.copyOfRange(items, start, mid);
        Pair[] right_copy = Arrays.copyOfRange(items, mid, end);

        int i = 0;  // Pointer for left_copy
        int j = 0;  // Pointer for right_copy
        int k = start;  // Pointer for main items array

        while (i < left_copy.length && j < right_copy.length) {
            if (left_copy[i].val <= right_copy[j].val) {
                counts[left_copy[i].index] += j;
                items[k] = left_copy[i];
                i++;
            } else {
                items[k] = right_copy[j];
                j++;
            }
            k++;
        }

        while (i < left_copy.length) {
            counts[left_copy[i].index] += j;
            items[k] = left_copy[i];
            i++;
            k++;
        }

        while (j < right_copy.length) {
            items[k] = right_copy[j];
            j++;
            k++;
        }
    }

    static class Pair {
        int val;
        int index;

        public Pair(int val, int index) {
            this.val = val;
            this.index = index;
        }
    }
}

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Problem Info

Description

Problem 315: Count of Smaller Numbers After Self - Detailed explanation and solution approaches.

Topics

Array Binary Search Divide and Conquer Binary Indexed Tree Segment Tree Merge Sort Ordered Set

Companies

Geico

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:315

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

4 steps

1
Step 1: Create a `items` array where each element is a pair consisting of a number from `nums` and its original index.
2
Step 2: Implement a Merge Sort algorithm. This involves recursively dividing the array into smaller sub-arrays until each sub-array contains only one element.
3
Step 3: Implement the `merge` function. This function merges two sorted sub-arrays. While merging, if an element from the left subarray is less than or equal to an element from the right subarray, it means that all the remaining elements in the right subarray are *not* smaller than the current left subarray element. So, we increment the count of smaller numbers to the right of the element from the left subarray by the number of elements already merged from the right subarray.
4
Step 4: After the Merge Sort is complete, the `counts` array will contain the number of smaller elements to the right of each element in the original array. Return this array.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is dominated by the Merge Sort algorithm, which has a time complexity of O(n log n). The `merge` operation takes O(n) time. The recursive calls of merge sort divide the array into log n levels. Thus the overall complexity is O(n log n). The space complexity is O(n) due to the auxiliary arrays (left_copy, right_copy, and items) created during the merge process, as well as the counts array.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Binary Indexed Tree (Fenwick Tree)

Use a Binary Indexed Tree (BIT) to store the counts of elements encountered so far. Iterate through the input array from right to left. For each element, query the BIT to find the number of elements smaller than the current element and then update the BIT with the current element.

Time: O(n log n)Space: O(n)

Trade-offs: BIT can be slightly more efficient in practice than merge sort due to lower constant factors, but can be harder to implement. Requires careful handling of negative numbers using offset or coordinate compression.

Segment Tree

Similar to the BIT approach, a segment tree can be used to query the number of elements smaller than the current element and update the tree. Requires more memory overhead than a BIT.

Time: O(n log n)Space: O(n)

Trade-offs: Segment trees are generally more complex to implement than BITs. Segment tree can handle more complex range queries and updates.

Ordered Set (using AVL Tree or Red-Black Tree)

Many languages (like C++ using `std::set`) provide ordered set data structures that maintain elements in sorted order. Iterate through the array from right to left, insert each element into the ordered set, and then use the `order_of_key` (or equivalent) method to find the number of elements smaller than the current element in the set.

Time: O(n log n)Space: O(n)

Trade-offs: This approach is generally cleaner to implement but relies on the availability of an efficient ordered set implementation in the chosen language. C++ STL's `std::set` is a viable implementation. Be aware of library support in different languages.

✨

Key Insights

1Insight 1: The core idea is to efficiently count smaller elements to the right for each element in the input array.
2Insight 2: A naive approach (iterating through the remaining array for each element) would lead to O(n^2) time complexity, which is not efficient enough. We need a more optimized approach.
3Insight 3: Merge Sort can be leveraged to solve the problem in O(n log n) time complexity. The trick is to modify the merge step to count the number of smaller elements encountered from the right subarray while merging.
4Insight 4: Pairing each number with its original index is crucial to correctly update the `counts` array, as the numbers are reordered during the sorting process.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Edge Cases

Edge case 1: Empty input array. The provided code handles this implicitly since the loops won't execute if the input array is empty, and the returned array is also empty.
Edge case 2: Array with duplicate elements. The solution correctly handles duplicate elements because the comparison in the `merge` function uses '<=', which ensures that elements with the same value from the left sub-array are processed before elements from the right sub-array with the same value. This maintains the correct counts.
Edge case 3: Array with all identical elements. The result should be an array filled with zeros. The algorithm handles this correctly.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to pair the numbers with their original indices. This leads to incorrect counts because the numbers are reordered during the sorting process.

✗Common mistake 2: Not correctly implementing the `merge` function. The logic for incrementing the counts during the merge step needs to be precise.

✗Common mistake 3: Using a naive O(n^2) approach, which exceeds the time limit for larger input sizes.

✗Common mistake 4: Forgetting to allocate memory for the result array in C and not setting the return size.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Rank aggregation and leaderboards. Imagine you want to rank users based on multiple criteria and need to count how many users are 'better' than a specific user in terms of certain metrics.

🏦Real-world scenario 2: Data analysis and anomaly detection. Counting the number of data points that are smaller (or larger) than a specific threshold can be useful for identifying outliers or trends in a dataset.

🎮Real-world scenario 3: Inversion Count: This problem is closely related to calculating the inversion count in an array, which is used in genomics for comparing gene sequences and measuring the similarity between rankings.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 14, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Can binary search be applied to non-sorted arrays?

Binary search requires a monotonic property, not necessarily sorted data. It can be applied to any search space where you can determine if the answer is in the left or right half. Examples include searching rotated arrays or finding minimum in optimization problems.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Count of Smaller Numbers After Self"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?