What is the time complexity of Degree of an Array?

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

What data structures are used to solve Degree of an Array?

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

What are the key insights for solving Degree of an Array?

Insight 1: We need to find the degree of the array, which is the maximum frequency of any element. Insight 2: We need to find the first and last occurrences of each element to determine the length of the subarray containing all occurrences.

Which companies ask LeetCode 697. Degree of an Array in interviews?

This problem is frequently asked at ByteDance, Expedia, GE Digital, IBM, J.P. Morgan and 8 more companies.

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Degree of an Array - LeetCode 697 Solution

Degree of an Array - Complete Solution Guide

Degree of an Array is LeetCode problem 697, 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 a non-empty array of non-negative integers nums , the degree of this array is defined as the maximum frequency of any one of its elements. Your task is to find the smallest possible length of a (contiguous) subarray of nums , that has the same degree as nums . Example 1: Input: nums = [1,2,2,3,1] Output: 2 Explanation: The input array has a degree of 2 because both elements 1 and 2 appear twice. Of the subarrays that have the same degree: [1, 2, 2, 3, 1], [1, 2, 2, 3], [2, 2, 3, 1], [1, 2,

Detailed Explanation

The problem asks us to find the shortest contiguous subarray within a given array `nums` that has the same degree as the entire array. The degree of an array is defined as the maximum frequency of any one of its elements. In simpler terms, we need to find the most frequent number(s) in the array and then find the shortest subarray that contains all occurrences of those most frequent number(s). The input is a non-empty array of non-negative integers. The output is the length of the shortest subarray with the same degree as the input array. The length of the array is between 1 and 50,000, and the values in the array are between 0 and 49,999.

Solution Approach

The solution involves using hash maps (or arrays, in the C implementation) to store the frequency, first occurrence, and last occurrence of each element in the array. First, iterate through the array and record these values. Next, find the degree of the array, which is the maximum frequency among all elements. Finally, iterate through the elements again and check which element has a frequency equal to the degree. For each of these elements, calculate the length of the subarray from its first occurrence to its last occurrence. Maintain the minimum length found so far and return it.

Step-by-Step Algorithm

Step 1: Initialize hash maps (or arrays) to store the frequency (count), first occurrence (first), and last occurrence (last) of each element.
Step 2: Iterate through the input array `nums`. For each element, update its frequency in the `count` map, record its first occurrence in the `first` map if it's the first time encountering it, and update its last occurrence in the `last` map.
Step 3: Calculate the degree of the array by finding the maximum frequency among all elements in the `count` map.
Step 4: Initialize a variable `minLength` to infinity (or nums.length in Java).
Step 5: Iterate through the `count` map. For each element, if its frequency is equal to the degree, calculate the length of the subarray from its first occurrence to its last occurrence (last[element] - first[element] + 1). Update `minLength` with the minimum length found so far.
Step 6: Return the `minLength`.

Key Insights

Insight 1: We need to find the degree of the array, which is the maximum frequency of any element.
Insight 2: We need to find the first and last occurrences of each element to determine the length of the subarray containing all occurrences.
Insight 3: We need to iterate through elements with the same degree as the entire array and find the minimum length among them.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table.

Companies

Asked at: ByteDance, Expedia, GE Digital, IBM, J.P. Morgan, PayPal, Paycom, Rivian, Salesforce, SoFi, Turing, Walmart Labs, ZScaler.

Degree of an Array | LeetCode 697

Easy

Array Hash Table

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

Solution Code

Available in:

class Solution {
    public int findShortestSubArray(int[] nums) {
        Map<Integer, Integer> count = new HashMap<>();
        Map<Integer, Integer> first = new HashMap<>();
        Map<Integer, Integer> last = new HashMap<>();
        int degree = 0;
        for (int i = 0; i < nums.length; i++) {
            int num = nums[i];
            count.put(num, count.getOrDefault(num, 0) + 1);
            if (!first.containsKey(num)) {
                first.put(num, i);
            }
            last.put(num, i);
            degree = Math.max(degree, count.get(num));
        }
        int minLength = nums.length;
        for (int num : count.keySet()) {
            if (count.get(num) == degree) {
                minLength = Math.min(minLength, last.get(num) - first.get(num) + 1);
            }
        }
        return minLength;
    }
}

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Problem Info

Description

Problem 697: Degree of an Array - Detailed explanation and solution approaches.

Topics

Array Hash Table

Companies

ByteDanceExpediaGE DigitalIBMJ.P. MorganPayPalPaycomRivian

Stats

Difficulty:Easy

Languages:4

Companies:13

Problem #:697

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

6 steps

1
Step 1: Initialize hash maps (or arrays) to store the frequency (count), first occurrence (first), and last occurrence (last) of each element.
2
Step 2: Iterate through the input array `nums`. For each element, update its frequency in the `count` map, record its first occurrence in the `first` map if it's the first time encountering it, and update its last occurrence in the `last` map.
3
Step 3: Calculate the degree of the array by finding the maximum frequency among all elements in the `count` map.
4
Step 4: Initialize a variable `minLength` to infinity (or nums.length in Java).
5
Step 5: Iterate through the `count` map. For each element, if its frequency is equal to the degree, calculate the length of the subarray from its first occurrence to its last occurrence (last[element] - first[element] + 1). Update `minLength` with the minimum length found so far.
6
Step 6: Return the `minLength`.

💡

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 time complexity is dominated by the iterations through the array, which are all linear. The space complexity depends on the number of distinct elements in the array. In the worst case, the number of distinct elements can be equal to the length of the array itself.

📖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

Sliding Window

We can use a sliding window approach to iterate through all possible subarrays and check if their degree is equal to the degree of the entire array. If it is, we update the minimum length.

Time: O(n^2) in the worst case. Checking the degree of each subarray is O(n), and there are O(n^2) subarrays.Space: O(n) for storing the frequency of elements in the current window.

Trade-offs: This approach is less efficient than the hash map approach due to the nested loops. Use the hash map approach as it is significantly faster.

✨

Key Insights

1Insight 1: We need to find the degree of the array, which is the maximum frequency of any element.
2Insight 2: We need to find the first and last occurrences of each element to determine the length of the subarray containing all occurrences.
3Insight 3: We need to iterate through elements with the same degree as the entire array and find the minimum length among 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: Array with only one element. The degree is 1, and the shortest subarray length is 1. The provided code handles this case correctly.
Edge case 2: Array where all elements are the same. The degree is the length of the array, and the shortest subarray length is the length of the array. The code handles this case correctly.
Edge case 3: An array where multiple elements share the maximum frequency. The code correctly iterates through all elements with the max frequency and calculates the minimum subarray length among them.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case where multiple elements have the same maximum frequency. The code should iterate through all elements with the same degree and find the minimum length among them.

✗Common mistake 2: Not initializing the minimum length to infinity or a large value (e.g., nums.length) before finding the minimum. This can lead to incorrect results.

✗Common mistake 3: Not correctly calculating the length of the subarray (last[element] - first[element] + 1). The '+ 1' is crucial to include both endpoints.

✗Common mistake 4: Using array based solution in C, and not accounting for potential out of bounds exceptions. Since we are assuming nums[i] can be between 0 and 49999, this approach works given the constraints.

✅

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: Data compression. Identifying frequently occurring patterns in data streams and finding the shortest segments containing those patterns can be useful for compressing data.

🏦Real-world scenario 2: Bioinformatics. Analyzing DNA sequences to find the shortest sequence containing the most frequent genes or motifs can be valuable in identifying key genomic regions.

🎮Real-world scenario 3: Network traffic analysis. Identifying the shortest time window where the most common network packets are transmitted, useful in detecting anomalies or patterns.

💡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: November 5, 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 "Degree of an Array"?

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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