Payload Sequence Evaluator 39 — Problem Statement & Solution Guide

Binary SearchHardDFS Traversal

TimeO(log n)

SpaceO(1)

Problem Description

Given a sequence of data elements representing payload and sequence metrics, construct an optimal algorithm to evaluate and compute the target evaluator value under given operational constraints.

DSA Pattern Breakdown

"Payload Sequence Evaluator 39"

hard

WHY DOES IT MATTER?

Binary search on a monotonic predicate is a fundamental pattern for turning exponential‑time verification problems into logarithmic‑time solutions, enabling real‑time decision making on massive data streams.

OPTIMIZATION CHALLENGE

The key insight is recognizing that the predicate f(i) is monotonic, allowing the search space to be halved each iteration instead of scanning every element, which collapses O(n) to O(log n).

REAL-WORLD CONNECTION

Think of a load balancer that must find the smallest server pool size that can handle incoming traffic without breaching latency SLAs; the pool size is analogous to the index, and the latency check is the monotonic predicate.

Always guard against overflow when computing mid = low + (high - low) / 2 and verify the loop invariant; a single off‑by‑one can flip the answer from the first true index to the last false index.

COMPLEXITY AT A GLANCE

⏱ Time:O(log n)

💾 Space:O(1)

Core Theory — Why This Approach?

Binary search is a divide‑and‑conquer technique that exploits the monotonic property of a sorted sequence to locate a target or boundary in O(log n) time. In the "Payload Sequence Evaluator 39" problem, the payload metrics form a non‑decreasing array where each element encodes a cumulative property (e.g., weight, latency, or checksum). The goal is to compute the smallest index i such that a predicate f(i) – derived from the payload and sequence metrics – becomes true. A naive linear scan would examine every element, leading to O(n) time, which is prohibitive when n reaches 10^7 or higher, especially under tight time limits typical of hard‑level coding challenges. By recognizing that f(i) is monotonic (once true it stays true), we can apply binary search to halve the search space at each step, guaranteeing logarithmic time.

The optimal paradigm combines binary search with careful handling of integer overflow and off‑by‑one errors. Instead of searching for an exact value, we search for a boundary: the first index where the predicate holds. This requires a loop invariant that maintains a low‑high window where low is always false and high is always true (or vice‑versa). The loop terminates when low + 1 == high, and high is the answer. This pattern also extends to variations such as finding the maximum feasible value under a resource constraint, which is a classic "binary search on answer" technique used in many competitive programming and interview problems.

Interview Questions on This Problem

Q1How would you modify binary search to find the first index where a cumulative payload metric exceeds a given threshold in a sorted array?

Initialize low = -1 (guaranteed false) and high = n (guaranteed true). While high - low > 1, compute mid = low + (high - low) / 2. If arr[mid] >= threshold, set high = mid; else set low = mid. After the loop, high holds the first index satisfying the condition or n if none exists.

Q2Explain why a linear scan is unacceptable for the Payload Sequence Evaluator 39 problem when n can be up to 10^8, and how binary search resolves the issue.

A linear scan would require O(n) operations, which at 10^8 elements exceeds typical time limits (often 1–2 seconds) even in optimized C++/Java. Binary search reduces the number of inspections to O(log n) ≈ 27 steps for 10^8 elements, fitting comfortably within the time budget while using constant extra space.

Q3In a distributed system that processes payload streams, how can the binary‑search‑on‑answer pattern be used to determine the maximum throughput that satisfies latency constraints?

Treat the throughput as the search variable. Define a predicate that simulates or measures latency for a given throughput; it returns true if latency ≤ SLA. Since higher throughput generally increases latency, the predicate is monotonic. Apply binary search over the feasible throughput range to locate the highest value that still satisfies the latency constraint, enabling capacity planning without exhaustive testing.

Examples

Example 1

Input

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 3

Output

Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] and target value 3, we perform a binary search. We start by finding the middle index of the array, which is 4. Since the middle element 5 is greater than the target value 3, we move the right pointer to the middle index - 1, which is 3. The element at index 3 is 4, which is still greater than the target value 3. We move the right pointer to the middle index - 2, which is 2. The element at index 2 is 3, which is equal to the target value 3. However, the problem statement asks for the index of the first element greater than the target value. Therefore, we return the index of the next element, which is 2.

Example 2

Input

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 25

Output

Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] and target value 25, we perform a binary search. We start by finding the middle index of the array, which is 4. Since the middle element 5 is less than the target value 25, we move the left pointer to the middle index + 1, which is 5. The element at index 5 is 6, which is still less than the target value 25. We move the left pointer to the middle index + 2, which is 6. The element at index 6 is 7, which is still less than the target value 25. We move the left pointer to the middle index + 3, which is 7. The element at index 7 is 8, which is still less than the target value 25. We move the left pointer to the middle index + 4, which is 8. The element at index 8 is 9, which is still less than the target value 25. We move the left pointer to the middle index + 5, which is 9. The element at index 9 is 10, which is still less than the target value 25. Therefore, we return the index of the first element greater than the target value, which is 3.

Constraints

1 <= N <= 10^5
-10^4 <= metrics[i] <= 10^4
1 <= K <= N

Optimal Approach & Strategy

Apply binary search on the index range, using the monotonic predicate to decide which half to discard each iteration, yielding the first true index in logarithmic time.

Brute Force Approach

Iterate through the payload array from start to finish, checking the predicate at each index until it becomes true, then return that index.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(log n)

function solution(nums, target) {
   let left = 0;
   let right = nums.length - 1;
   while (left <= right) {
       let mid = Math.floor((left + right) / 2);
       if (nums[mid] > target) {
           return mid;
       } else if (nums[mid] < target) {
           left = mid + 1;
       } else {
           if (mid === 0 || nums[mid - 1] < target) {
               return mid;
           }
           right = mid - 1;
       }
   }
   return -1;
}

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

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

def solution(nums, target):
   left = 0
   right = len(nums) - 1
   while left <= right:
       mid = (left + right) // 2
       if nums[mid] > target:
           return mid
       elif nums[mid] < target:
           left = mid + 1
       else:
           if mid == 0 or nums[mid - 1] < target:
               return mid
           right = mid - 1
   return -1

function solution(nums, target) {
   let left = 0;
   let right = nums.length - 1;
   while (left <= right) {
       let mid = Math.floor((left + right) / 2);
       if (nums[mid] > target) {
           return mid;
       } else if (nums[mid] < target) {
           left = mid + 1;
       } else {
           if (mid === 0 || nums[mid - 1] < target) {
               return mid;
           }
           right = mid - 1;
       }
   }
   return -1;
}

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Payload Sequence Evaluator 39 Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Binary Search

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Payload Sequence Evaluator 39 — 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