What is the time complexity of Intervals Between Identical Elements?

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

What data structures are used to solve Intervals Between Identical Elements?

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

What are the key insights for solving Intervals Between Identical Elements?

Insight 1: Using prefix sums can efficiently compute the sums of indices of identical elements. Insight 2: Iterating from both the start and end of the array allows for calculating the intervals based on elements before and after the current index.

Which companies ask LeetCode 2121. Intervals Between Identical Elements in interviews?

This problem is frequently asked at TuSimple.

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Intervals Between Identical Elements - LeetCode 2121 Solution

Intervals Between Identical Elements - Complete Solution Guide

Intervals Between Identical Elements is LeetCode problem 2121, a Medium 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 a 0-indexed array of n integers arr . The interval between two elements in arr is defined as the absolute difference between their indices. More formally, the interval between arr[i] and arr[j] is |i - j| . Return an array intervals of length n where intervals[i] is the sum of intervals between arr[i] and each element in arr with the same value as arr[i] . Note: |x| is the absolute value of x . Example 1: Input: arr = [2,1,3,1,2,3,3] Output: [4,2,7,2,4,4,5] Explanation: - Index 0:

Detailed Explanation

The problem asks us to calculate the sum of intervals between each element in an array and all other elements that have the same value. The interval between two elements is defined as the absolute difference of their indices. For each index `i` in the input array `arr`, we need to find all other indices `j` where `arr[i] == arr[j]`, calculate the absolute difference `|i - j|` for each such `j`, and sum up these differences. The output should be an array `intervals` where `intervals[i]` is the sum of intervals for the element at index `i`.

Solution Approach

The provided solution uses a two-pass approach combined with hash maps to calculate the sum of intervals for each element. The first pass iterates through the array from left to right, and for each element, it calculates the sum of intervals with all identical elements that occur before the current element's index. This is done by maintaining a count and a prefix sum of indices for each unique value encountered. The second pass iterates through the array from right to left, and for each element, it calculates the sum of intervals with all identical elements that occur after the current element's index, using a similar count and prefix sum approach. Finally, the results from both passes are combined to get the total sum of intervals for each element.

Step-by-Step Algorithm

Step 1: Initialize an array `intervals` of the same length as the input array `arr` with all elements set to 0. This array will store the final result.
Step 2: Initialize two hash maps, `count` and `prefix_sum`. The `count` hash map will store the number of occurrences of each element encountered so far, and the `prefix_sum` hash map will store the sum of indices of each element encountered so far during the forward pass.
Step 3: Iterate through the array `arr` from left to right (index `i` from 0 to `n-1`).
Step 4: For each element `arr[i]`, retrieve its current count `c` and prefix sum `s` from the `count` and `prefix_sum` hash maps, respectively. If the element is not yet in the hash maps, initialize the count to 0 and the prefix sum to 0.
Step 5: Update `intervals[i]` by adding `c * i - s`. This represents the sum of intervals between the current element and all identical elements to its left.
Step 6: Update the `count` and `prefix_sum` hash maps for the current element `arr[i]`. Increment its count and add its index `i` to its prefix sum.
Step 7: Initialize two new hash maps, `count` and `suffix_sum` for the backward pass. The `count` hash map will store the number of occurrences of each element encountered so far, and the `suffix_sum` hash map will store the sum of indices of each element encountered so far during the backward pass.
Step 8: Iterate through the array `arr` from right to left (index `i` from `n-1` to 0).
Step 9: For each element `arr[i]`, retrieve its current count `c` and suffix sum `s` from the `count` and `suffix_sum` hash maps, respectively. If the element is not yet in the hash maps, initialize the count to 0 and the suffix sum to 0.
Step 10: Update `intervals[i]` by adding `s - c * i`. This represents the sum of intervals between the current element and all identical elements to its right.
Step 11: Update the `count` and `suffix_sum` hash maps for the current element `arr[i]`. Increment its count and add its index `i` to its suffix sum.
Step 12: Return the `intervals` array.

Key Insights

Insight 1: Using prefix sums can efficiently compute the sums of indices of identical elements.
Insight 2: Iterating from both the start and end of the array allows for calculating the intervals based on elements before and after the current index.
Insight 3: Hash maps (or equivalent data structures) are essential for storing and retrieving the counts and prefix sums of elements efficiently.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Prefix Sum.

Companies

Asked at: TuSimple.

Intervals Between Identical Elements | LeetCode 2121

Medium

Array Hash Table Prefix Sum

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

Solution Code

Available in:

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

class Solution {
    public long[] getDistances(int[] arr) {
        int n = arr.length;
        long[] intervals = new long[n];

        Map<Integer, Integer> count = new HashMap<>();
        Map<Integer, Long> prefixSum = new HashMap<>();

        for (int i = 0; i < n; i++) {
            int val = arr[i];
            int c = count.getOrDefault(val, 0);
            long s = prefixSum.getOrDefault(val, 0L);
            intervals[i] += (long)c * i - s;
            count.put(val, c + 1);
            prefixSum.put(val, s + i);
        }

        count = new HashMap<>();
        Map<Integer, Long> suffixSum = new HashMap<>();

        for (int i = n - 1; i >= 0; i--) {
            int val = arr[i];
            int c = count.getOrDefault(val, 0);
            long s = suffixSum.getOrDefault(val, 0L);
            intervals[i] += s - (long)c * i;
            count.put(val, c + 1);
            suffixSum.put(val, s + i);
        }

        return intervals;
    }
}

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Problem Info

Description

Problem 2121: Intervals Between Identical Elements - Detailed explanation and solution approaches.

Topics

Array Hash Table Prefix Sum

Companies

TuSimple

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2121

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

12 steps

1
Step 1: Initialize an array `intervals` of the same length as the input array `arr` with all elements set to 0. This array will store the final result.
2
Step 2: Initialize two hash maps, `count` and `prefix_sum`. The `count` hash map will store the number of occurrences of each element encountered so far, and the `prefix_sum` hash map will store the sum of indices of each element encountered so far during the forward pass.
3
Step 3: Iterate through the array `arr` from left to right (index `i` from 0 to `n-1`).
4
Step 4: For each element `arr[i]`, retrieve its current count `c` and prefix sum `s` from the `count` and `prefix_sum` hash maps, respectively. If the element is not yet in the hash maps, initialize the count to 0 and the prefix sum to 0.
5
Step 5: Update `intervals[i]` by adding `c * i - s`. This represents the sum of intervals between the current element and all identical elements to its left.
6
Step 6: Update the `count` and `prefix_sum` hash maps for the current element `arr[i]`. Increment its count and add its index `i` to its prefix sum.
7
Step 7: Initialize two new hash maps, `count` and `suffix_sum` for the backward pass. The `count` hash map will store the number of occurrences of each element encountered so far, and the `suffix_sum` hash map will store the sum of indices of each element encountered so far during the backward pass.
8
Step 8: Iterate through the array `arr` from right to left (index `i` from `n-1` to 0).
9
Step 9: For each element `arr[i]`, retrieve its current count `c` and suffix sum `s` from the `count` and `suffix_sum` hash maps, respectively. If the element is not yet in the hash maps, initialize the count to 0 and the suffix sum to 0.
10
Step 10: Update `intervals[i]` by adding `s - c * i`. This represents the sum of intervals between the current element and all identical elements to its right.
11
Step 11: Update the `count` and `suffix_sum` hash maps for the current element `arr[i]`. Increment its count and add its index `i` to its suffix sum.
12
Step 12: Return the `intervals` 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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the two linear traversals of the array. The hash map operations (get, put) take O(1) time on average. The space complexity is determined by the auxiliary space used by the `intervals` array and the hash maps. In the worst-case scenario where all elements of the array are distinct, the size of the hash maps grows linearly with the input size.

📖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

Brute Force

For each element in the array, iterate through the entire array again to find all elements with the same value and calculate the sum of intervals.

Time: O(n^2), where n is the length of the input array.Space: O(1) excluding the output array.

Trade-offs: This approach is simple to understand but highly inefficient for large arrays due to the nested loops. It is only suitable for very small input sizes. The given solution is a much better approach.

✨

Key Insights

1Insight 1: Using prefix sums can efficiently compute the sums of indices of identical elements.
2Insight 2: Iterating from both the start and end of the array allows for calculating the intervals based on elements before and after the current index.
3Insight 3: Hash maps (or equivalent data structures) are essential for storing and retrieving the counts and prefix sums of elements efficiently.

📚

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 all unique elements. In this case, the intervals[i] will be 0 for all i, since there are no other elements with the same value. The algorithm handles this correctly because if an element is unique, its count in both the prefix and suffix passes will always be 0 when calculating the interval sum for that element.
Edge case 2: Array with all identical elements. The algorithm correctly calculates the sum of intervals between each element and all other elements. The prefix and suffix calculations will work to get the sum of differences. For element at index i, it is sum(|i-j|) for all j != i.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating prefix sums or suffix sums, leading to inaccurate interval sums.

✗Common mistake 2: Not handling the case where an element appears only once in the array, leading to incorrect calculations.

✗Common mistake 3: Integer overflow when calculating prefix sums or interval sums, especially in languages like Java and C++. Using `long` data types is crucial to avoid this.

✅

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 network latency. Imagine each element in the array represents a server's response time. Calculating intervals between identical response times could help identify patterns of network congestion.

🏦Real-world scenario 2: Stock market analysis. Analyzing price fluctuations and calculating the sum of differences in time between repeated stock prices can provide insights into market volatility and trading patterns. Finding similar price patterns over time.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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 20, 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

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

@AlgoMasterJul 14

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

@InterviewProJul 7

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

@CodeNinjaJun 30

❓

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.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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 "Intervals Between Identical Elements"?

For this medium-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.

🤖 Need help with this problem?