Sensor Packet Partition 34 — Problem Statement & Solution Guide

StringsMediumBitmasking

TimeO(n)

SpaceO(1)

Problem Description

Given a sequence of data elements representing sensor and packet metrics, and a target value K, construct an optimal algorithm to evaluate and compute the target partition value under the given operational constraints that the sum of the metrics in each partition should be as close to K as possible without exceeding it.

DSA Pattern Breakdown

"Sensor Packet Partition 34"

medium

WHY DOES IT MATTER?

Greedy partitioning turns an exponential combinatorial problem into a linear scan.

OPTIMIZATION CHALLENGE

The key is to eliminate the need for backtracking by proving the earliest cut is always safe.

REAL-WORLD CONNECTION

It mirrors buffering packets in network routers where each buffer must not exceed a size limit.

Maintain a running sum and reset it instantly when the next element would overflow K.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)

💾 Space:O(1)

Core Theory — Why This Approach?

The problem reduces to partitioning a linear sequence into the fewest contiguous groups where each group's sum does not exceed K. A greedy strategy—accumulating elements until the next addition would breach K, then starting a new group—yields the optimal solution because any earlier cut would only increase the number of groups, and any later cut would violate the constraint. Naïve exhaustive enumeration of all possible cut positions incurs exponential time (2^(n‑1) partitions) and quickly becomes infeasible for large n. The optimal paradigm leverages the monotonic nature of the cumulative sum and the fact that once a prefix exceeds K, no later element can retroactively fix that violation, enabling a single linear pass.

Interview Questions on This Problem

Q1Why does a greedy left‑to‑right scan produce the minimum number of partitions?

Because cutting earlier never reduces the current sum and only adds extra partitions, while cutting later would exceed K. Hence the earliest feasible cut is always optimal.

Q2How would you adapt the algorithm if the elements could be negative?

Negative values break the monotonic sum property, so the greedy approach fails; you would need DP or prefix‑sum with binary search to handle arbitrary signs.

Q3What is the time‑space trade‑off when using a sliding‑window versus a simple accumulator?

Both run in O(n) time, but a sliding‑window may store indices for window boundaries, using O(1) extra space, while a DP table would require O(n) space.

Examples

Example 1

Input

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

Output

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

Explanation: Step-by-step: with input [1, 2, 3, 4, 5] and target value 5, we first sort the metrics in descending order. Then, we iterate over the sorted metrics and assign each metric to the partition where the sum of the metrics is closest to the target value without exceeding it. This results in the optimal partition [[1, 2], [3], [4, 5]].

Example 2

Input

[10, 20, 30, 40, 50], 50

Output

[[10], [20], [30], [40], [50]]

Explanation: Step-by-step: with input [10, 20, 30, 40, 50] and target value 50, we first sort the metrics in descending order. Then, we iterate over the sorted metrics and assign each metric to a separate partition, since the sum of each metric exceeds the target value. This results in the optimal partition [[10], [20], [30], [40], [50]].

Constraints

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

Optimal Approach & Strategy

Iterate once, accumulate a running sum, and start a new partition whenever the next element would exceed K.

Brute Force Approach

Try every possible subset of cut positions, compute sums, and keep the best valid partitioning.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n)

function solution(nums, k) { 
      nums.sort((a, b) => b - a); 
      let partitions = []; 
      for (let num of nums) { 
         let closestPartition = null; 
         let minDiff = Infinity; 
         for (let partition of partitions) { 
            let sum = partition.reduce((a, b) => a + b, 0); 
            if (sum + num <= k) { 
               let diff = Math.abs(k - (sum + num)); 
               if (diff < minDiff) { 
                  minDiff = diff; 
                  closestPartition = partition; 
               } 
            } 
         } 
         if (closestPartition === null) { 
            closestPartition = []; 
            partitions.push(closestPartition); 
         } 
         closestPartition.push(num); 
      } 
      return partitions; 
   }

class Solution { 
   public: 
      vector<vector<int>> solution(vector<int>& nums, int k) { 
         sort(nums.rbegin(), nums.rend()); 
         vector<vector<int>> partitions; 
         for (int num : nums) { 
            int minDiff = INT_MAX; 
            vector<int>* closestPartition = nullptr; 
            for (auto& partition : partitions) { 
               int sum = 0; 
               for (int n : partition) { 
                  sum += n; 
               } 
               if (sum + num <= k) { 
                  int diff = abs(k - (sum + num)); 
                  if (diff < minDiff) { 
                     minDiff = diff; 
                     closestPartition = &partition; 
                  } 
               } 
            } 
            if (closestPartition == nullptr) { 
               closestPartition = new vector<int>(); 
               partitions.push_back(*closestPartition); 
               delete closestPartition; 
            } else { 
               closestPartition->push_back(num); 
            } 
         } 
         return partitions; 
      } 
   };

import java.util.Arrays; 
   public class Solution { 
      public int[][] solution(int[] nums, int k) { 
         Arrays.sort(nums); 
         reverse(nums); 
         ArrayList<ArrayList<Integer>> partitions = new ArrayList<>(); 
         for (int num : nums) { 
            int minDiff = Integer.MAX_VALUE; 
            ArrayList<Integer> closestPartition = null; 
            for (ArrayList<Integer> partition : partitions) { 
               int sum = 0; 
               for (int n : partition) { 
                  sum += n; 
               } 
               if (sum + num <= k) { 
                  int diff = Math.abs(k - (sum + num)); 
                  if (diff < minDiff) { 
                     minDiff = diff; 
                     closestPartition = partition; 
                  } 
               } 
            } 
            if (closestPartition == null) { 
               closestPartition = new ArrayList<>(); 
               partitions.add(closestPartition); 
            } 
            closestPartition.add(num); 
         } 
         int[][] result = new int[partitions.size()][]; 
         for (int i = 0; i < partitions.size(); i++) { 
            result[i] = new int[partitions.get(i).size()]; 
            for (int j = 0; j < partitions.get(i).size(); j++) { 
               result[i][j] = partitions.get(i).get(j); 
            } 
         } 
         return result; 
      } 
      public void reverse(int[] nums) { 
         int left = 0; 
         int right = nums.length - 1; 
         while (left < right) { 
            int temp = nums[left]; 
            nums[left] = nums[right]; 
            nums[right] = temp; 
            left++; 
            right--; 
         } 
      } 
   }

def solution(nums, k): 
      nums.sort(reverse=True) 
      partitions = [] 
      for num in nums: 
         closest_partition = None 
         min_diff = float('inf') 
         for partition in partitions: 
            total = sum(partition) 
            if total + num <= k: 
               diff = abs(k - (total + num)) 
               if diff < min_diff: 
                  min_diff = diff 
                  closest_partition = partition 
         if closest_partition is None: 
            closest_partition = [] 
            partitions.append(closest_partition) 
         closest_partition.append(num) 
      return partitions

function solution(nums, k) { 
      nums.sort((a, b) => b - a); 
      let partitions = []; 
      for (let num of nums) { 
         let closestPartition = null; 
         let minDiff = Infinity; 
         for (let partition of partitions) { 
            let sum = partition.reduce((a, b) => a + b, 0); 
            if (sum + num <= k) { 
               let diff = Math.abs(k - (sum + num)); 
               if (diff < minDiff) { 
                  minDiff = diff; 
                  closestPartition = partition; 
               } 
            } 
         } 
         if (closestPartition === null) { 
            closestPartition = []; 
            partitions.push(closestPartition); 
         } 
         closestPartition.push(num); 
      } 
      return partitions; 
   }

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Sensor Packet Partition 34 Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Strings

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Sensor Packet Partition 34 — 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