What is the time complexity of Sum of Unique Elements?

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

What data structures are used to solve Sum of Unique Elements?

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

What are the key insights for solving Sum of Unique Elements?

Insight 1: We need a way to efficiently count the occurrences of each element in the array. A hash table (or a frequency array in this case, due to the constraints) is ideal for this task. Insight 2: After counting the occurrences, we need to iterate through the counts and only add the elements with a count of 1 to the sum.

Which companies ask LeetCode 1748. Sum of Unique Elements in interviews?

This problem is frequently asked at tcs.

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Sum of Unique Elements - LeetCode 1748 Solution

Sum of Unique Elements - Complete Solution Guide

Sum of Unique Elements is LeetCode problem 1748, 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

You are given an integer array nums . The unique elements of an array are the elements that appear exactly once in the array. Return the sum of all the unique elements of nums . Example 1: Input: nums = [1,2,3,2] Output: 4 Explanation: The unique elements are [1,3], and the sum is 4. Example 2: Input: nums = [1,1,1,1,1] Output: 0 Explanation: There are no unique elements, and the sum is 0. Example 3: Input: nums = [1,2,3,4,5] Output: 15 Explanation: The unique elements are [1,2,3,4,5], and the s

Detailed Explanation

The problem asks you to find the sum of all unique elements in an integer array. Unique elements are defined as those that appear only once in the array. The input is an integer array `nums`, and the output is an integer representing the sum of the unique elements. If there are no unique elements, the output should be 0. The constraints specify that the array length is between 1 and 100, and each element's value is between 1 and 100.

Solution Approach

The provided solutions all use a similar approach. They first count the occurrences of each element in the input array. Then, they iterate through the counts and sum up only the elements that appeared exactly once. The difference lies in the data structure used for counting: Python uses a dictionary (hash table), while Java, C++, and C use a fixed-size array to store the counts. Using a fixed-size array is possible and more efficient because the problem constrains the input values.

Step-by-Step Algorithm

Step 1: Initialize a data structure (dictionary or array) to store the counts of each element. This structure will map each element value to its frequency.
Step 2: Iterate through the input array `nums`. For each element, increment its count in the data structure. This counts how many times each number occurs.
Step 3: Initialize a variable `sum` to 0.
Step 4: Iterate through the data structure. For each element and its count, if the count is 1, add the element's value to `sum`.
Step 5: Return the value of `sum`.

Key Insights

Insight 1: We need a way to efficiently count the occurrences of each element in the array. A hash table (or a frequency array in this case, due to the constraints) is ideal for this task.
Insight 2: After counting the occurrences, we need to iterate through the counts and only add the elements with a count of 1 to the sum.
Insight 3: The constraints on the input array (elements between 1 and 100) allow us to use a fixed-size array (size 101) instead of a hash table, leading to slightly improved performance in this specific case. This avoids the overhead of hash function calculations.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Counting.

Companies

Asked at: tcs.

Sum of Unique Elements | LeetCode 1748

Easy

Array Hash Table Counting

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

Solution Code

Available in:

class Solution {
    public int sumOfUnique(int[] nums) {
        int[] count = new int[101];
        int sum = 0;
        for (int num : nums) {
            count[num]++;
        }
        for (int i = 1; i <= 100; i++) {
            if (count[i] == 1) {
                sum += i;
            }
        }
        return sum;
    }
}

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Problem Info

Description

Problem 1748: Sum of Unique Elements - Detailed explanation and solution approaches.

Topics

Array Hash Table Counting

Companies

tcs

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1748

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

5 steps

1
Step 1: Initialize a data structure (dictionary or array) to store the counts of each element. This structure will map each element value to its frequency.
2
Step 2: Iterate through the input array `nums`. For each element, increment its count in the data structure. This counts how many times each number occurs.
3
Step 3: Initialize a variable `sum` to 0.
4
Step 4: Iterate through the data structure. For each element and its count, if the count is 1, add the element's value to `sum`.
5
Step 5: Return the value of `sum`.

💡

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 input array once, which takes O(n) time. The second loop iterates through the count array, which has a size of 101 (constant). Hence, the dominant factor is the first loop, resulting in O(n) time complexity. In space complexity, the dictionary or array used for counting can potentially store all elements. The fixed-size array uses constant space due to the constraint on the input. Therefore, the space complexity for the fixed-size array implementation is O(1).

📖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

Using a Set

Iterate through the array and add each element to a set. Then, iterate through the array again, checking if each element is present in the set and only summing up those that are present once. This is less efficient.

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

Trade-offs: This approach is less efficient than the counting approach because it involves two iterations through the array. However, it can be more concise. This method doesn't benefit from the constraint on input values.

✨

Key Insights

1Insight 1: We need a way to efficiently count the occurrences of each element in the array. A hash table (or a frequency array in this case, due to the constraints) is ideal for this task.
2Insight 2: After counting the occurrences, we need to iterate through the counts and only add the elements with a count of 1 to the sum.
3Insight 3: The constraints on the input array (elements between 1 and 100) allow us to use a fixed-size array (size 101) instead of a hash table, leading to slightly improved performance in this specific case. This avoids the overhead of hash function calculations.

📚

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 solutions correctly handle this by returning 0 (either implicitly or explicitly).
Edge case 2: Array with all duplicate elements. The solution correctly identifies that there are no unique elements and returns 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case where there are no unique elements and returning 0.

✗Common mistake 2: Inefficiently checking for uniqueness (e.g., using nested loops instead of a hash table or array for counting).

✅

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 user data to find the number of unique users who performed a specific action.

🏦Real-world scenario 2: In data analytics, identifying the sum of unique values in a dataset for a particular feature (e.g., summing up unique product IDs to find the total number of distinct products sold).

💡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

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ 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 7, 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

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

@CodeNinjaJul 13

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

@TechLeadJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Sum of Unique Elements"?

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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