Maximized Network Stream Validator 8 — Problem Statement & Solution Guide

Binary SearchMediumSearch Peak Element

TimeO(log n)

SpaceO(1)

Problem Description

You are provided with a strictly unimodal array of integers, representing a sequence of signal strengths in a distributed network. The array is guaranteed to have a single peak element, defined as an element strictly greater than its immediate neighbors. For the boundaries, the peak must be strictly greater than its only neighbor. Your objective is to identify the index of this peak element using an efficient search strategy that exploits the unimodal property, rather than scanning the entire array linearly.

The input consists of a single array nums of length N. The output should be the index (0-based) of the peak element. If multiple peaks exist (which is not possible under the strict unimodal constraint provided), return the index of the first one. The solution must run in O(log N) time complexity to handle large datasets efficiently.

DSA Pattern Breakdown

"Maximized Network Stream Validator 8"

medium

WHY DOES IT MATTER?

Binary search is a classic divide-and-conquer pattern that transforms linear search into logarithmic time, which is essential for handling large datasets efficiently. It also demonstrates a candidate’s ability to reason about array properties and exploit them for optimization.

OPTIMIZATION CHALLENGE

The key insight is that the unimodal property guarantees that the direction of the slope indicates the side containing the peak. By comparing mid and mid+1, we can discard half the array each iteration, reducing time from O(n) to O(log n).

REAL-WORLD CONNECTION

In network monitoring, finding the maximum signal strength quickly can trigger load balancing or failover decisions. A binary search approach allows routers to identify peak usage zones in real-time without scanning every packet, mirroring the algorithm’s divide-and-conquer strategy.

When explaining the algorithm, emphasize the invariant: at each step, the peak lies within the current subarray. This helps interviewers see that the search space is correctly narrowed and that the algorithm will terminate with the correct index.

COMPLEXITY AT A GLANCE

⏱ Time:O(log n)

💾 Space:O(1)

Core Theory — Why This Approach?

The problem asks for the index of the unique peak in a strictly unimodal array, where each element is strictly greater than its immediate neighbors until the peak, after which the sequence strictly decreases. A naive linear scan would examine each element and compare it to its neighbors, yielding an O(n) time complexity; this is acceptable for small inputs but becomes prohibitive when n can reach millions, as the time grows linearly and may exceed time limits in production systems.

The optimal solution leverages the unimodal property to apply a binary search. By inspecting the middle element and comparing it to its right neighbor, we can determine which side of the array contains the peak: if the middle element is less than its right neighbor, the peak lies to the right; otherwise, it lies to the left (including the middle). Repeating this halving process reduces the search space logarithmically, achieving O(log n) time.

Binary search works here because the array’s structure guarantees that any local increase must eventually lead to the global maximum, and any local decrease guarantees that the maximum lies on the other side. This monotonicity ensures that the decision at each step is correct, eliminating the need to inspect every element and thus providing a dramatic performance improvement over the linear approach.

Interview Questions on This Problem

Q1How would you modify the binary search algorithm if the array could contain multiple equal peaks?

If equal peaks are allowed, the binary search must be adapted to find any peak, not necessarily the unique one. One approach is to compare mid with mid+1; if mid < mid+1, move right; else move left. This still guarantees finding a peak because the array is still unimodal, but the returned index may not be the global maximum if duplicates exist.

Q2In a distributed system, why might you prefer a logarithmic-time peak-finding algorithm over a linear scan?

Distributed systems often process large streams of data in real-time; a logarithmic algorithm reduces latency and CPU usage, allowing more efficient use of resources and better scalability. It also enables parallelization across nodes by partitioning the array and performing local binary searches before merging results.

Q3What edge case must you handle when the peak is at index 0 or n-1, and how does the binary search algorithm account for it?

When the peak is at the boundaries, the algorithm must treat the missing neighbor as negative infinity. In practice, the binary search compares mid with mid+1; if mid is at the last index, the comparison fails and the algorithm correctly identifies the peak at the boundary without accessing out-of-bounds indices.

Examples

Example 1

Input

nums = [1, 3, 5, 4, 2]

Output

Explanation: The array is [1, 3, 5, 4, 2]. The element at index 2 is 5. Its left neighbor is 3 and its right neighbor is 4. Since 5 > 3 and 5 > 4, index 2 is the peak. Using binary search: mid=2, nums[2]=5, nums[3]=4. Since nums[2] > nums[3], the peak is at mid or to the left. Search left half [1, 3, 5]. mid=1, nums[1]=3, nums[2]=5. Since nums[1] < nums[2], peak is to the right. Search [5]. Peak found at index 2.

Example 2

Input

nums = [10, 9, 8, 7, 6]

Output

Explanation: The array is strictly decreasing. The first element 10 is greater than its only neighbor 9. Thus, index 0 is the peak. Binary search: mid=2, nums[2]=8, nums[3]=7. Since nums[2] > nums[3], peak is left. Search [10, 9, 8]. mid=1, nums[1]=9, nums[2]=8. Since nums[1] > nums[2], peak is left. Search [10, 9]. mid=0, nums[0]=10, nums[1]=9. Since nums[0] > nums[1], peak is at index 0.

Example 3

Input

nums = [1, 2, 3, 4, 5, 6, 7]

Output

Explanation: The array is strictly increasing. The last element 7 is greater than its only neighbor 6. Thus, index 6 is the peak. Binary search: mid=3, nums[3]=4, nums[4]=5. Since nums[3] < nums[4], peak is right. Search [5, 6, 7]. mid=5, nums[5]=6, nums[6]=7. Since nums[5] < nums[6], peak is right. Search [7]. Peak found at index 6.

Example 4

Input

nums = [42, 43, 44, 45, 44, 43, 42]

Output

Explanation: The array is [42, 43, 44, 45, 44, 43, 42]. The element at index 3 is 45. Left neighbor is 44, right neighbor is 44. 45 > 44 and 45 > 44, so index 3 is the peak. Binary search: mid=3, nums[3]=45, nums[4]=44. Since nums[3] > nums[4], peak is left or at mid. Search [42, 43, 44, 45]. mid=1, nums[1]=43, nums[2]=44. Since nums[1] < nums[2], peak is right. Search [44, 45]. mid=2, nums[2]=44, nums[3]=45. Since nums[2] < nums[3], peak is right. Search [45]. Peak found at index 3.

Constraints

1 <= nums.length <= 10^5
-10^9 <= nums[i] <= 10^9
nums[i] != nums[i + 1] for all valid i
The array is strictly unimodal (single peak)

Optimal Approach & Strategy

Apply binary search by comparing the middle element with its right neighbor to decide which half contains the peak, halving the search space each step. This achieves O(log n) time and O(1) space.

Brute Force Approach

Scan the array from left to right, keeping track of the maximum value and its index. This takes O(n) time and O(1) space but is inefficient for large arrays.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(log n)

/**
 * @param {number[]} nums
 * @return {number}
 */
var findPeakElement = function(nums) {
    let left = 0;
    let right = nums.length - 1;
    
    while (left < right) {
        const mid = Math.floor((left + right) / 2);
        if (nums[mid] < nums[mid + 1]) {
            left = mid + 1;
        } else {
            right = mid;
        }
    }
    
    return left;
};

class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        int left = 0, right = nums.size() - 1;
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] < nums[mid + 1]) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        return left;
    }
};

class Solution {
    public int findPeakElement(int[] nums) {
        int left = 0;
        int right = nums.length - 1;
        
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] < nums[mid + 1]) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        
        return left;
    }
}

class Solution:
    def findPeakElement(self, nums: List[int]) -> int:
        left, right = 0, len(nums) - 1
        
        while left < right:
            mid = (left + right) // 2
            if nums[mid] < nums[mid + 1]:
                left = mid + 1
            else:
                right = mid
        
        return left

/**
 * @param {number[]} nums
 * @return {number}
 */
var findPeakElement = function(nums) {
    let left = 0;
    let right = nums.length - 1;
    
    while (left < right) {
        const mid = Math.floor((left + right) / 2);
        if (nums[mid] < nums[mid + 1]) {
            left = mid + 1;
        } else {
            right = mid;
        }
    }
    
    return left;
};

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Maximized Network Stream Validator 8 Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Input

Output

Constraints

More Binary Search

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Maximized Network Stream Validator 8 — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor