What is the time complexity of Unique Number of Occurrences?

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

What data structures are used to solve Unique Number of Occurrences?

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

What are the key insights for solving Unique Number of Occurrences?

**Two-Stage Frequency Analysis:** The problem's core lies in a two-tier counting process: first, determine the frequency of each element in the input array; then, critically, verify the uniqueness of *those resulting frequencies*. This clear separation into initial counting and subsequent uniqueness checking is fundamental. **Efficient Initial Counting with Hash Map:** Employing a hash map (like Python's `dict`) for the first pass to tally element frequencies in `arr` is crucial. It allows for O(1) average-time updates for each element, leading to an overall O(N) average time complexity for processing the entire input array, significantly more efficient than sorting for counting.

Which companies ask LeetCode 1207. Unique Number of Occurrences in interviews?

This problem is frequently asked at Datadog.

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Unique Number of Occurrences - LeetCode 1207 Solution

Q: How to solve Unique Number of Occurrences?

The provided solution elegantly breaks down this problem into two distinct, highly efficient stages, leveraging the power of hash-based data structures. The first stage is dedicated to accurately tallying the occurrences of every number in the input array `arr`. It iterates through `arr` once, using a hash map (like Python's `dict` named `counts`) where each unique number from `arr` becomes a key, and its total frequency within `arr` is its corresponding value. This pass efficiently builds a complete frequency profile of `arr` in O(N) average time. Once the `counts` map is populated, the second stage begins its crucial task: verifying if these collected frequencies (the values from the `counts` map) are themselves unique. A hash set (like Python's `set` named `occurrences`) is employed here. The solution iterates through the *values* of the `counts` map (which are the frequencies). For each frequency encountered, it attempts to add it to the `occurrences` set. The key insight of using a hash set is its O(1) average-time complexity for both insertion and membership testing. If a frequency is encountered that is *already present* in the `occurrences` set, it immediately signifies that we have found two different numbers that occurred the same number of times, allowing the function to promptly return `false`. If the loop completes without finding any duplicate frequencies in `occurrences`, it confirms that all occurrence counts are unique, and `true` is returned.

Unique Number of Occurrences - Complete Solution Guide

Unique Number of Occurrences is LeetCode problem 1207, 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 an array of integers arr , return true if the number of occurrences of each value in the array is unique or false otherwise . Example 1: Input: arr = [1,2,2,1,1,3] Output: true Explanation: The value 1 has 3 occurrences, 2 has 2 and 3 has 1. No two values have the same number of occurrences. Example 2: Input: arr = [1,2] Output: false Example 3: Input: arr = [-3,0,1,-3,1,1,1,-3,10,0] Output: true Constraints: 1 <= arr.length <= 1000 -1000 <= arr[i] <= 1000

Detailed Explanation

This problem presents a unique twist on frequency counting: we're not just tasked with counting how many times each number appears, but rather with determining if those *counts* are themselves unique. Essentially, you need to answer whether it's possible for two distinct numbers in the input array `arr` to have occurred the exact same number of times. For example, if `arr = [1,2,2,1,1,3]`, we first tally the occurrences: '1' appears 3 times, '2' appears 2 times, and '3' appears 1 time. The set of these occurrence counts is `{3, 2, 1}`. Since all values within this set are distinct, the function should return `true`.

Solution Approach

The provided solution elegantly breaks down this problem into two distinct, highly efficient stages, leveraging the power of hash-based data structures. The first stage is dedicated to accurately tallying the occurrences of every number in the input array `arr`. It iterates through `arr` once, using a hash map (like Python's `dict` named `counts`) where each unique number from `arr` becomes a key, and its total frequency within `arr` is its corresponding value. This pass efficiently builds a complete frequency profile of `arr` in O(N) average time. Once the `counts` map is populated, the second stage begins its crucial task: verifying if these collected frequencies (the values from the `counts` map) are themselves unique. A hash set (like Python's `set` named `occurrences`) is employed here. The solution iterates through the *values* of the `counts` map (which are the frequencies). For each frequency encountered, it attempts to add it to the `occurrences` set. The key insight of using a hash set is its O(1) average-time complexity for both insertion and membership testing. If a frequency is encountered that is *already present* in the `occurrences` set, it immediately signifies that we have found two different numbers that occurred the same number of times, allowing the function to promptly return `false`. If the loop completes without finding any duplicate frequencies in `occurrences`, it confirms that all occurrence counts are unique, and `true` is returned.

Step-by-Step Algorithm

Step 1: Initialize an empty hash map (or dictionary) to store the counts of each number in the input array.
Step 2: Iterate through the input array. For each number, increment its count in the hash map (using `getOrDefault` or a similar method to handle cases where the number is not yet in the map).
Step 3: Initialize an empty set.
Step 4: Iterate through the values (counts) of the hash map. For each count:
Step 5: Check if the count is already present in the set. If it is, return `false` (non-unique frequencies).
Step 6: If the count is not in the set, add it to the set.
Step 7: If the loop completes without returning `false`, return `true` (unique frequencies).

Key Insights

**Two-Stage Frequency Analysis:** The problem's core lies in a two-tier counting process: first, determine the frequency of each element in the input array; then, critically, verify the uniqueness of *those resulting frequencies*. This clear separation into initial counting and subsequent uniqueness checking is fundamental.
**Efficient Initial Counting with Hash Map:** Employing a hash map (like Python's `dict`) for the first pass to tally element frequencies in `arr` is crucial. It allows for O(1) average-time updates for each element, leading to an overall O(N) average time complexity for processing the entire input array, significantly more efficient than sorting for counting.
**Leveraging Hash Set for Uniqueness Validation:** The second phase, which checks the uniqueness of the *frequencies themselves*, is made highly efficient by using a hash set (like Python's `set`). Attempting to add each frequency to the set provides O(1) average-time detection of duplicates. If a frequency is already present in the set, we immediately know the condition is violated, enabling an early and efficient exit without needing to process further counts or sort them.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table.

Companies

Asked at: Datadog.

Unique Number of Occurrences | LeetCode 1207

Easy

Array Hash Table

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

Solution Code

Available in:

import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Set;

class Solution {
    public boolean uniqueOccurrences(int[] arr) {
        Map<Integer, Integer> counts = new HashMap<>();
        for (int num : arr) {
            counts.put(num, counts.getOrDefault(num, 0) + 1);
        }

        Set<Integer> occurrences = new HashSet<>();
        for (int count : counts.values()) {
            if (!occurrences.add(count)) {
                return false;
            }
        }

        return true;
    }
}

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Problem Info

Description

Problem 1207: Unique Number of Occurrences - Detailed explanation and solution approaches.

Topics

Array Hash Table

Companies

Datadog

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1207

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
Step 1: Initialize an empty hash map (or dictionary) to store the counts of each number in the input array.
2
Step 2: Iterate through the input array. For each number, increment its count in the hash map (using `getOrDefault` or a similar method to handle cases where the number is not yet in the map).
3
Step 3: Initialize an empty set.
4
Step 4: Iterate through the values (counts) of the hash map. For each count:
5
Step 5: Check if the count is already present in the set. If it is, return `false` (non-unique frequencies).
6
Step 6: If the count is not in the set, add it to the set.
7
Step 7: If the loop completes without returning `false`, return `true` (unique frequencies).

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The first loop iterates through the array once (O(n)). The second loop iterates through the hash map which, in the worst case, contains n unique elements (O(n)). The space complexity is O(n) because the hash map and set can each store up to n elements in the worst case (when all elements in the input array are unique).

📖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 array. Then iterate through the sorted array, counting consecutive identical numbers. Store these counts and check for uniqueness.

Time: O(n log n) due to sorting.Space: O(n) in the worst case to store counts.

Trade-offs: This approach is less efficient than the hash map approach when dealing with a large array with many unique numbers because sorting dominates the time complexity.

✨

Key Insights

1**Two-Stage Frequency Analysis:** The problem's core lies in a two-tier counting process: first, determine the frequency of each element in the input array; then, critically, verify the uniqueness of *those resulting frequencies*. This clear separation into initial counting and subsequent uniqueness checking is fundamental.
2**Efficient Initial Counting with Hash Map:** Employing a hash map (like Python's `dict`) for the first pass to tally element frequencies in `arr` is crucial. It allows for O(1) average-time updates for each element, leading to an overall O(N) average time complexity for processing the entire input array, significantly more efficient than sorting for counting.
3**Leveraging Hash Set for Uniqueness Validation:** The second phase, which checks the uniqueness of the *frequencies themselves*, is made highly efficient by using a hash set (like Python's `set`). Attempting to add each frequency to the set provides O(1) average-time detection of duplicates. If a frequency is already present in the set, we immediately know the condition is violated, enabling an early and efficient exit without needing to process further counts or sort them.

📚

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!

Edge Cases

Edge case 1: Empty array ([]). The code correctly handles this case because the loops will not execute, and `true` will be returned (no non-unique frequencies).
Edge case 2: Array with all same numbers ([1, 1, 1]). The hashmap will have only one entry, and the set will only contain one frequency count. The function will correctly return `true`.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case where the input array is empty or contains only one element. The provided solution implicitly handles this.

✗Common mistake 2: Incorrectly using a list or array instead of a set to check for unique frequencies. This would lead to O(n^2) time complexity in the worst case for checking uniqueness.

✅

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: Analyzing data logs to determine if the number of occurrences of different events is unique. This could be useful for debugging or anomaly detection.

🏦Real-world scenario 2: In network security, analyzing network traffic to detect unusual patterns based on the frequency of specific data packets.

💡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

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ Watch Out

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

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 6, 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

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

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 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 "Unique Number of Occurrences"?

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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