What is the time complexity of How Many Numbers Are Smaller Than the Current Number?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve How Many Numbers Are Smaller Than the Current Number?

This problem uses Array, Hash Table, Sorting, Counting Sort. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving How Many Numbers Are Smaller Than the Current Number?

**Direct Iteration for Each Element**: The problem explicitly asks for a count 'for each `nums[i]`'. The solution directly translates this by using an outer loop to fix `nums[i]` and an inner loop to perform the required comparison across the entire array for that specific `nums[i]` element. **Strictly Smaller and Distinct Indices**: The conditions `nums[j] < nums[i]` and `i != j` are fundamental. The solution correctly implements both, ensuring that only numbers truly smaller than the current one, and not the current number itself, contribute to the count.

Which companies ask LeetCode 1365. How Many Numbers Are Smaller Than the Current Number in interviews?

This problem is frequently asked at Accenture, tcs.

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How Many Numbers Are Smaller Than the Current Number - LeetCode 1365 Solution

Q: How to solve How Many Numbers Are Smaller Than the Current Number?

The provided solution takes a very direct, literal approach to the problem statement. It uses nested loops to achieve the required counting. The outer loop iterates through each element `nums[i]` in the input array, marking it as the 'current number' we're evaluating. For each `nums[i]`, an inner loop then iterates through *all* elements `nums[j]` in the array again. Inside this inner loop, the solution checks two conditions: first, `i != j` ensures that we are comparing `nums[i]` with a *different* number in the array, not itself. Second, `nums[j] < nums[i]` checks if the other number is strictly smaller than our current number. If both conditions are met, a counter (`count`) for `nums[i]` is incremented. Once the inner loop completes, `count` holds the total number of smaller elements for the current `nums[i]`, and this `count` is appended to our `result` list. This exhaustive pairwise comparison guarantees correctness by systematically checking every relevant pairing.

Q: Which companies ask LeetCode 1365. How Many Numbers Are Smaller Than the Current Number in interviews?

This problem is frequently asked at Accenture, tcs.

How Many Numbers Are Smaller Than the Current Number - Complete Solution Guide

How Many Numbers Are Smaller Than the Current Number is LeetCode problem 1365, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given the array nums , for each nums[i] find out how many numbers in the array are smaller than it. That is, for each nums[i] you have to count the number of valid j's such that j != i and nums[j] < nums[i] . Return the answer in an array. Example 1: Input: nums = [8,1,2,2,3] Output: [4,0,1,1,3] Explanation: For nums[0]=8 there exist four smaller numbers than it (1, 2, 2 and 3). For nums[1]=1 does not exist any smaller number than it. For nums[2]=2 there exist one smaller number than it (1). For

Detailed Explanation

This problem asks us to analyze an array of numbers, `nums`, and for each individual number within it, determine how many *other* numbers in that same array are strictly smaller. The crucial part is 'for each `nums[i]`' – meaning we need to produce an output array where each element `result[i]` corresponds to the count for `nums[i]`. The constraint `j != i` is important, ensuring we don't count a number against itself. Consider `nums = [8,1,2,2,3]` from the example. For `nums[0]=8`, we scan the rest of the array: `1` is smaller, `2` is smaller, the other `2` is smaller, and `3` is smaller. That's 4 smaller numbers. For `nums[1]=1`, there are no numbers in `[8,2,2,3]` that are smaller. So the count is 0. This process needs to be repeated for every element, maintaining their original indices in the result. The interesting aspect here isn't necessarily a hidden trick, but rather the emphasis on individual counts for each element while ensuring we only consider *other* distinct elements. The presence of duplicate numbers (like `2,2` in the example) means we can't simply count unique smaller values; we must count every instance of a smaller number.

Solution Approach

The provided solution takes a very direct, literal approach to the problem statement. It uses nested loops to achieve the required counting. The outer loop iterates through each element `nums[i]` in the input array, marking it as the 'current number' we're evaluating. For each `nums[i]`, an inner loop then iterates through *all* elements `nums[j]` in the array again. Inside this inner loop, the solution checks two conditions: first, `i != j` ensures that we are comparing `nums[i]` with a *different* number in the array, not itself. Second, `nums[j] < nums[i]` checks if the other number is strictly smaller than our current number. If both conditions are met, a counter (`count`) for `nums[i]` is incremented. Once the inner loop completes, `count` holds the total number of smaller elements for the current `nums[i]`, and this `count` is appended to our `result` list. This exhaustive pairwise comparison guarantees correctness by systematically checking every relevant pairing.

Step-by-Step Algorithm

Initialize an empty array `result` of the same size as `nums` to store the counts.
Iterate through the `nums` array using a `for` loop (outer loop).
For each number `nums[i]`, initialize a counter `count` to 0.
Iterate through the `nums` array again using another `for` loop (inner loop).
Inside the inner loop, check if `j != i` and `nums[j] < nums[i]`. If both conditions are true, increment `count`.
After the inner loop completes, append `count` to the `result` array.
After the outer loop completes, return the `result` array.

Key Insights

**Direct Iteration for Each Element**: The problem explicitly asks for a count 'for each `nums[i]`'. The solution directly translates this by using an outer loop to fix `nums[i]` and an inner loop to perform the required comparison across the entire array for that specific `nums[i]` element.
**Strictly Smaller and Distinct Indices**: The conditions `nums[j] < nums[i]` and `i != j` are fundamental. The solution correctly implements both, ensuring that only numbers truly smaller than the current one, and not the current number itself, contribute to the count.
**Brute-Force Simplicity**: This approach foregoes any complex sorting or hash map optimizations. Its 'elegance' lies in its straightforwardness: it directly simulates the problem's request without any abstraction, making it easy to understand and verify for correctness, particularly for smaller input sizes.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Sorting, Counting Sort.

Companies

Asked at: Accenture, tcs.

How Many Numbers Are Smaller Than the Current Number | LeetCode 1365

Easy

Array Hash Table Sorting Counting Sort

Time: O(n^2)Space: O(n)

Solution Code

Available in:

class Solution {
    public int[] smallerNumbersThanCurrent(int[] nums) {
        int n = nums.length;
        int[] result = new int[n];
        for (int i = 0; i < n; i++) {
            int count = 0;
            for (int j = 0; j < n; j++) {
                if (i != j && nums[j] < nums[i]) {
                    count++;
                }
            }
            result[i] = count;
        }
        return result;
    }
}

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Problem Info

Description

Problem 1365: How Many Numbers Are Smaller Than the Current Number - Detailed explanation and solution approaches.

Topics

Array Hash Table Sorting Counting Sort

Companies

Accenturetcs

Stats

Difficulty:Easy

Languages:4

Companies:2

Problem #:1365

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

7 steps

1
Initialize an empty array `result` of the same size as `nums` to store the counts.
2
Iterate through the `nums` array using a `for` loop (outer loop).
3
For each number `nums[i]`, initialize a counter `count` to 0.
4
Iterate through the `nums` array again using another `for` loop (inner loop).
5
Inside the inner loop, check if `j != i` and `nums[j] < nums[i]`. If both conditions are true, increment `count`.
6
After the inner loop completes, append `count` to the `result` array.
7
After the outer loop completes, return the `result` 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^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The nested loops dominate the time complexity, resulting in O(n^2). The space complexity is O(n) because we create a new array `result` with the same size as the input array `nums` to store the output.

📖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

Sorting and Counting

Sort the input array. Then, for each number, use binary search to find the index of the number smaller than the current number to get the count of smaller numbers.

Time: O(n log n)Space: O(n) or O(1) (In-place sort)

Trade-offs: This approach is more efficient than the brute-force approach for larger input sizes because sorting takes O(n log n) time compared to O(n^2) for the nested loops. However, it requires extra space or space for sorting in place. The brute force approach is simpler to implement.

Hash Table (Frequency Count)

Create a hash table (or array) to store the frequency of each number. Iterate through the sorted array and find the sum of frequencies of numbers smaller than the current number.

Time: O(n)Space: O(101) which is O(1) because max value of the number is 100

Trade-offs: This is the most efficient approach. It achieves linear time complexity, but it requires extra space for the hash table. Ideal for constraints where numbers are within a known limited range.

✨

Key Insights

1**Direct Iteration for Each Element**: The problem explicitly asks for a count 'for each `nums[i]`'. The solution directly translates this by using an outer loop to fix `nums[i]` and an inner loop to perform the required comparison across the entire array for that specific `nums[i]` element.
2**Strictly Smaller and Distinct Indices**: The conditions `nums[j] < nums[i]` and `i != j` are fundamental. The solution correctly implements both, ensuring that only numbers truly smaller than the current one, and not the current number itself, contribute to the count.
3**Brute-Force Simplicity**: This approach foregoes any complex sorting or hash map optimizations. Its 'elegance' lies in its straightforwardness: it directly simulates the problem's request without any abstraction, making it easy to understand and verify for correctness, particularly for smaller input sizes.

📚

Key Terms Explained

For Students

Array

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

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Empty array: The code handles empty arrays gracefully; the length check and loops prevent any issues, though this is an invalid case according to the constraints
Duplicate numbers: The code correctly handles duplicates; each number is compared against all others, including its duplicates, and only the strictly smaller ones are counted.
All numbers are the same: The code correctly outputs an array filled with zeros if all numbers in the input are identical.

⚠️

Common Mistakes to Avoid

✗Forgetting the `j != i` condition in the inner loop, which leads to incorrectly counting the number itself as smaller.

✗Inefficiently using nested loops without considering alternative approaches like sorting or hash tables that can significantly improve performance.

✅

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?

💼Ranking systems: Determining the rank of each element in a dataset based on its relative value.

🏦Data analysis: Counting the number of data points that fall below a certain threshold.

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

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

👨‍💻 From Experience

This is one of the first problems I solved when learning Array. It's a great foundation for more complex problems.

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: September 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.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 "How Many Numbers Are Smaller Than the Current Number"?

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

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