What is the time complexity of Peaks in Array?

The optimal solution has a time complexity of O(m*log(n) + n*log(n)).

What data structures are used to solve Peaks in Array?

This problem uses Array, Binary Indexed Tree, Segment Tree. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Peaks in Array?

Insight 1: Recognize that brute force counting of peaks for each query can lead to Time Limit Exceeded (TLE). Insight 2: Use a Fenwick Tree (Binary Indexed Tree) to efficiently count peaks within a range, enabling updates in logarithmic time.

Which companies ask LeetCode 3187. Peaks in Array in interviews?

This problem is frequently asked at Siemens.

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Peaks in Array - LeetCode 3187 Solution

Peaks in Array - Complete Solution Guide

Peaks in Array is LeetCode problem 3187, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

A peak in an array arr is an element that is greater than its previous and next element in arr . You are given an integer array nums and a 2D integer array queries . You have to process queries of two types: queries[i] = [1, l i , r i ] , determine the count of peak elements in the subarray nums[l i ..r i ] . queries[i] = [2, index i , val i ] , change nums[index i ] to val i . Return an array answer containing the results of the queries of the first type in order. Notes: The first and the last

Detailed Explanation

The problem requires us to process two types of queries on an integer array `nums`. Type 1 queries ask for the number of 'peak' elements within a specified range `[l, r]` of the array. A peak element is defined as an element that is strictly greater than its adjacent elements. Type 2 queries update an element at a given index with a new value. The goal is to return a list containing the results of all Type 1 queries in the order they appear.

Solution Approach

The solution uses a Fenwick Tree (Binary Indexed Tree) to maintain a count of peak elements. Initially, the Fenwick Tree is populated by iterating through the array and identifying peak elements. When a Type 1 query is encountered, the Fenwick Tree is used to efficiently calculate the number of peaks within the specified range. When a Type 2 query is encountered, the element at the given index is updated. Before updating, the peak status of the index and its immediate neighbors is checked. If any of these positions were a peak, their corresponding values are decremented in the Fenwick Tree. After the update, the peak status of the affected indices are re-evaluated, and their corresponding values are incremented in the Fenwick Tree if they are now a peak.

Step-by-Step Algorithm

Step 1: Initialize a Fenwick Tree (BIT) with the same size as the input array `nums`.
Step 2: Iterate through the `nums` array from index 1 to n-2 (exclusive) to identify initial peak elements. If an element is a peak, update the corresponding index in the Fenwick Tree by adding 1.
Step 3: Iterate through the `queries` array. For each query, check the query type.
Step 4: If the query type is 1 (range query), extract the left and right indices `l` and `r`. Use the Fenwick Tree to calculate the sum of peaks in the range `[l+1, r-1]`. Append this count to the result list.
Step 5: If the query type is 2 (update query), extract the index `idx` and the new value `val`. Before updating nums[idx], find all affected positions (idx-1, idx, idx+1) which might be peaks, and remove their contribution from the BIT if they are peaks.
Step 6: Update `nums[idx]` with `val`.
Step 7: After updating nums[idx], find all affected positions (idx-1, idx, idx+1) which might become peaks after updating nums[idx], and add their contribution to the BIT if they are peaks.
Step 8: Append the result list after processing all queries.

Key Insights

Insight 1: Recognize that brute force counting of peaks for each query can lead to Time Limit Exceeded (TLE).
Insight 2: Use a Fenwick Tree (Binary Indexed Tree) to efficiently count peaks within a range, enabling updates in logarithmic time.
Insight 3: Identify that modifying an element affects only the peak status of itself and its immediate neighbors. Only update the BIT for these affected indices.

Complexity Analysis

Time Complexity: O(m*log(n) + n*log(n))

Space Complexity: O(n)

Topics

This problem involves: Array, Binary Indexed Tree, Segment Tree.

Companies

Asked at: Siemens.

Peaks in Array | LeetCode 3187

Hard

Array Binary Indexed Tree Segment Tree

Time: O(m*log(n) + n*log(n))Space: O(n)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public List<Integer> countOfPeaks(int[] nums, int[][] queries) {
        int n = nums.length;

        class FenwickTree {
            private int[] tree;

            public FenwickTree(int size) {
                this.tree = new int[size + 1];
            }

            public void update(int i, int delta) {
                i += 1;
                while (i < tree.length) {
                    tree[i] += delta;
                    i += i & (-i);
                }
            }

            public int query(int i) {
                i += 1;
                int s = 0;
                while (i > 0) {
                    s += tree[i];
                    i -= i & (-i);
                }
                return s;
            }

            public int rangeSum(int l, int r) {
                if (l > r) {
                    return 0;
                }
                return query(r) - query(l - 1);
            }
        }

        class Helper {
            public boolean isPeak(int j) {
                if (!(1 <= j && j < n - 1)) {
                    return false;
                }
                return nums[j - 1] < nums[j] && nums[j] > nums[j + 1];
            }
        }

        Helper helper = new Helper();
        FenwickTree ft = new FenwickTree(n);
        for (int j = 1; j < n - 1; j++) {
            if (helper.isPeak(j)) {
                ft.update(j, 1);
            }
        }

        List<Integer> answers = new ArrayList<>();
        for (int[] q : queries) {
            int qType = q[0];
            if (qType == 1) {
                int l = q[1];
                int r = q[2];
                if (r - l < 2) {
                    answers.add(0);
                } else {
                    int count = ft.rangeSum(l + 1, r - 1);
                    answers.add(count);
                }
            } else {
                int idx = q[1];
                int val = q[2];

                if (nums[idx] == val) {
                    continue;
                }

                java.util.Set<Integer> affected = new java.util.HashSet<>();
                for (int j = idx - 1; j <= idx + 1; j++) {
                    if (1 <= j && j < n - 1) {
                        affected.add(j);
                    }
                }

                for (int j : affected) {
                    if (helper.isPeak(j)) {
                        ft.update(j, -1);
                    }
                }

                nums[idx] = val;

                for (int j : affected) {
                    if (helper.isPeak(j)) {
                        ft.update(j, 1);
                    }
                }
            }
        }

        return answers;
    }
}

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Problem Info

Description

Problem 3187: Peaks in Array - Detailed explanation and solution approaches.

Topics

Array Binary Indexed Tree Segment Tree

Companies

Siemens

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:3187

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

8 steps

1
Step 1: Initialize a Fenwick Tree (BIT) with the same size as the input array `nums`.
2
Step 2: Iterate through the `nums` array from index 1 to n-2 (exclusive) to identify initial peak elements. If an element is a peak, update the corresponding index in the Fenwick Tree by adding 1.
3
Step 3: Iterate through the `queries` array. For each query, check the query type.
4
Step 4: If the query type is 1 (range query), extract the left and right indices `l` and `r`. Use the Fenwick Tree to calculate the sum of peaks in the range `[l+1, r-1]`. Append this count to the result list.
5
Step 5: If the query type is 2 (update query), extract the index `idx` and the new value `val`. Before updating nums[idx], find all affected positions (idx-1, idx, idx+1) which might be peaks, and remove their contribution from the BIT if they are peaks.
6
Step 6: Update `nums[idx]` with `val`.
7
Step 7: After updating nums[idx], find all affected positions (idx-1, idx, idx+1) which might become peaks after updating nums[idx], and add their contribution to the BIT if they are peaks.
8
Step 8: Append the result list after processing all queries.

💡

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(m*log(n) + n*log(n))

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The Fenwick Tree operations (update and query) take O(log n) time. Initializing the Fenwick Tree with initial peaks requires iterating through the array once. Each update query affects a constant number of elements, therefore contributes log(n) for those few elements. The space complexity is determined by the space taken by the Fenwick Tree.

📖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

Segment Tree

A Segment Tree could also be used in place of a Fenwick Tree to count the number of peaks within a range. It is similar to Fenwick Tree but can have more general range queries.

Time: O(m*log(n) + n*log(n))Space: O(n)

Trade-offs: Segment trees often have larger constant factors and require slightly more complex implementation compared to Fenwick trees. Fenwick tree is simpler to implement for range sum queries. When the queries are limited to sum queries then Fenwick tree is better. Segment tree can solve a wider variety of range queries in general.

Brute Force

For each query of type 1, iterate through nums[l...r] and count peaks. For query of type 2, simply update the value at nums[index].

Time: O(m*n), where n is the length of nums and m is the number of queries. For each query of type 1 we need O(n). Query of type 2 takes O(1).Space: O(1)

Trade-offs: Brute force will result in TLE for the given constraints since it's O(m*n) when m and n are large. Only viable for smaller inputs.

✨

Key Insights

1Insight 1: Recognize that brute force counting of peaks for each query can lead to Time Limit Exceeded (TLE).
2Insight 2: Use a Fenwick Tree (Binary Indexed Tree) to efficiently count peaks within a range, enabling updates in logarithmic time.
3Insight 3: Identify that modifying an element affects only the peak status of itself and its immediate neighbors. Only update the BIT for these affected indices.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: The first and last elements cannot be peaks. The `is_peak` function explicitly checks for this condition by ensuring the index `j` is within the range [1, n-2].
Edge case 2: `r - l < 2` for range queries. This means the range has either 0 or 1 element in the middle, not enough to form a peak. The code handles this case by directly returning 0 without calculating any peak value.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering edge cases, like range being too small to contain any peaks or the first and last element being considered peaks.

✗Common mistake 2: Not updating the Fenwick Tree when updating an element. This will lead to incorrect results for range queries.

✗Common mistake 3: Forgetting to consider affected indices when an element is updated. Only considering the current index might miss peak statuses being changed in the immediate neighboring elements.

✅

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: Identifying local maxima in sensor data, such as temperature or pressure readings over time. The queries could simulate changes in the sensor setup or require analysis of a specific time interval.

🏦Real-world scenario 2: Analyzing stock market data to find peaks and valleys in price fluctuations. Queries might involve updating stock prices based on new trades and then finding the number of peaks within a certain period.

💡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

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ 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.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 "Peaks in Array"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether binary indexed tree techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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