What is the time complexity of Find Indices With Index and Value Difference I?

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

What data structures are used to solve Find Indices With Index and Value Difference I?

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

What are the key insights for solving Find Indices With Index and Value Difference I?

The brute-force approach of checking all possible pairs of indices is straightforward and easily understandable. Nested loops are the simplest way to iterate through all possible pairs of indices in the array.

Which companies ask LeetCode 2903. Find Indices With Index and Value Difference I in interviews?

This problem is frequently asked at Paytm.

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Find Indices With Index and Value Difference I - LeetCode 2903 Solution

Find Indices With Index and Value Difference I - Complete Solution Guide

Find Indices With Index and Value Difference I is LeetCode problem 2903, 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 a 0-indexed integer array nums having length n , an integer indexDifference , and an integer valueDifference . Your task is to find two indices i and j , both in the range [0, n - 1] , that satisfy the following conditions: abs(i - j) >= indexDifference , and abs(nums[i] - nums[j]) >= valueDifference Return an integer array answer , where answer = [i, j] if there are two such indices , and answer = [-1, -1] otherwise . If there are multiple choices for the two indices, return any o

Detailed Explanation

The problem asks you to find two indices, `i` and `j`, in a given integer array `nums` that satisfy two conditions: The absolute difference between the indices (`abs(i - j)`) must be greater than or equal to a given `indexDifference`, and the absolute difference between the values at those indices (`abs(nums[i] - nums[j])`) must be greater than or equal to a given `valueDifference`. The function should return the indices `[i, j]` if such a pair exists; otherwise, it should return `[-1, -1].` The indices `i` and `j` can be the same.

Solution Approach

The provided code uses a brute-force approach. It iterates through all possible pairs of indices `i` and `j` using nested loops. For each pair, it checks if both conditions (index difference and value difference) are met. If both conditions are true, the function immediately returns the pair of indices `[i, j]`. If the loops complete without finding a suitable pair, it returns `[-1, -1]`.

Step-by-Step Algorithm

Initialize `n` to the length of the input array `nums`.
Iterate through the array using nested loops: the outer loop iterates from `i = 0` to `n - 1`, and the inner loop iterates from `j = 0` to `n - 1`.
Inside the inner loop, calculate `abs(i - j)` and `abs(nums[i] - nums[j])`.
Check if both `abs(i - j) >= indexDifference` and `abs(nums[i] - nums[j]) >= valueDifference` are true.
If both conditions are true, return the array `[i, j]`.
If the loops complete without finding a suitable pair, return `[-1, -1]`.

Key Insights

The brute-force approach of checking all possible pairs of indices is straightforward and easily understandable.
Nested loops are the simplest way to iterate through all possible pairs of indices in the array.
The problem's constraints (small array size) make a brute-force solution acceptable, even though it's not the most efficient for larger inputs.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(1)

Topics

This problem involves: Array, Two Pointers.

Companies

Asked at: Paytm.

Find Indices With Index and Value Difference I | LeetCode 2903

Easy

Array Two Pointers

Time: O(n^2)Space: O(1)

Solution Code

Available in:

class Solution {
    public int[] findIndices(int[] nums, int indexDifference, int valueDifference) {
        int n = nums.length;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (Math.abs(i - j) >= indexDifference && Math.abs(nums[i] - nums[j]) >= valueDifference) {
                    return new int[]{i, j};
                }
            }
        }
        return new int[]{-1, -1};
    }
}

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Problem Info

Description

Problem 2903: Find Indices With Index and Value Difference I - Detailed explanation and solution approaches.

Topics

Array Two Pointers

Companies

Paytm

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:2903

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
Initialize `n` to the length of the input array `nums`.
2
Iterate through the array using nested loops: the outer loop iterates from `i = 0` to `n - 1`, and the inner loop iterates from `j = 0` to `n - 1`.
3
Inside the inner loop, calculate `abs(i - j)` and `abs(nums[i] - nums[j])`.
4
Check if both `abs(i - j) >= indexDifference` and `abs(nums[i] - nums[j]) >= valueDifference` are true.
5
If both conditions are true, return the array `[i, j]`.
6
If the loops complete without finding a suitable pair, return `[-1, -1]`.

💡

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(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The nested loops iterate through all possible pairs of indices, resulting in O(n^2) time complexity. The space complexity is O(1) because the algorithm uses a constant amount of extra space regardless of 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

Optimized Brute Force (Slight Improvement)

Instead of checking all pairs, we can optimize slightly by only checking pairs where `abs(i-j) >= indexDifference`. This reduces the number of comparisons but still maintains O(n^2) complexity.

Time: O(n^2)Space: O(1)

Trade-offs: This offers a small performance improvement over the fully naive brute-force approach but still isn't optimal for very large datasets. The improvement is marginal given the constraints.

✨

Key Insights

1The brute-force approach of checking all possible pairs of indices is straightforward and easily understandable.
2Nested loops are the simplest way to iterate through all possible pairs of indices in the array.
3The problem's constraints (small array size) make a brute-force solution acceptable, even though it's not the most efficient for larger inputs.

📚

Key Terms Explained

For Students

Array

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

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Edge Cases

Empty array: The code handles this implicitly. The loops will not execute, and `[-1, -1]` will be returned.
Array with only one element: The loops will execute once, comparing the element to itself. If `indexDifference` and `valueDifference` are both 0, `[0, 0]` will be returned; otherwise, `[-1,-1]` will be returned.
All elements are the same: If `valueDifference` is greater than 0, `[-1, -1]` will be returned. If `valueDifference` is 0, any pair will satisfy the conditions if the `indexDifference` condition is met.

⚠️

Common Mistakes to Avoid

✗Forgetting to handle the case where no suitable pair of indices is found and returning `[-1, -1]`.

✗Incorrectly calculating the absolute differences between indices or values using `abs()` function (or its equivalent in other languages).

✗Off-by-one errors in loop boundaries.

✅

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?

💼Finding data points in a time series that are sufficiently far apart in time and significantly different in value (e.g., detecting anomalies or significant changes).

🏦Identifying pairs of items in a dataset that differ substantially in a specific characteristic while also being separated by a certain threshold (e.g., identifying pairs of products with large price differences and not belonging to the same product category).

💡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: June 24, 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

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 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 avoid infinite loops with two pointers?

Always ensure your loop condition guarantees termination (typically left < right or slow != fast). Move at least one pointer in each iteration, and handle edge cases where pointers might not move (like skipping duplicates). Draw out pointer movement on paper first.

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 "Find Indices With Index and Value Difference I"?

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 two pointers techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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