Sensor Checkpoint Partition 11 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing sensor and checkpoint metrics, construct an optimal algorithm to evaluate and compute the target partition value under given operational constraints. The algorithm should consider elements that are not greater than K but are still optimal components.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Sensor Checkpoint Partition 11"
WHY DOES IT MATTER?
Partition DP patterns appear in resource allocation, load balancing, and batch processing problems.
OPTIMIZATION CHALLENGE
The key is shrinking the O(N²) naïve DP to O(N) or O(N·logN) by pruning infeasible segment starts.
REAL-WORLD CONNECTION
Think of sensors sending data packets that must be grouped into checkpoints without exceeding bandwidth K.
Pre‑compute prefix sums and use a deque to keep candidate dp states ordered by their value and feasibility.
COMPLEXITY AT A GLANCE
O(N)O(N)Core Theory — Why This Approach?
The Sensor Checkpoint Partition problem can be modeled as a one‑dimensional DP where dp[i] stores the optimal value for the prefix ending at index i. Transitioning from dp[j] to dp[i] involves checking whether the segment (j+1…i) satisfies the constraint (e.g., sum ≤ K) and then combining dp[j] with the segment’s contribution, yielding O(N·K) or O(N) with monotonic queues. Naïve enumeration of all O(2^N) partitions quickly explodes because each element can either start a new checkpoint or extend the current one, leading to exponential blow‑up on large N. The optimal paradigm leverages prefix sums and a sliding window to limit candidate j’s, turning the exponential search into linear or near‑linear DP, which is the hallmark of efficient partition‑type problems.
Interview Questions on This Problem
Q1How does prefix‑sum preprocessing simplify the DP transition for this partition problem?
It lets us compute segment sums in O(1), so we can test the K‑constraint without scanning the segment each time. This reduces the inner loop cost dramatically.
Q2Why might a monotonic queue be useful when optimizing the DP for large K?
A monotonic queue maintains the best dp values for feasible start indices while discarding those that violate the sum constraint. This yields an O(N) overall transition.
Q3What edge case must be handled when all elements exceed K?
The algorithm should detect that no valid partition exists and return an appropriate sentinel (e.g., -1 or INF). Ignoring this leads to incorrect dp values.
Examples
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10], K = 5
Output
15
Explanation: Step-by-step: with input [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], we first initialize a 2D table dp where dp[i][j] represents the maximum sum of elements not greater than K in the subarray from index i to j. We then fill the dp table in a bottom-up manner, considering each element in the array and updating the dp table accordingly. Finally, we return the maximum sum of elements not greater than K in the entire array, which is stored in dp[0][n-1].
Input
[10, 20, 30, 40, 50, 60, 70, 80, 90, 100], K = 10
Output
200
Explanation: Step-by-step: with input [10, 20, 30, 40, 50, 60, 70, 80, 90, 100], we first initialize a 2D table dp where dp[i][j] represents the maximum sum of elements not greater than K in the subarray from index i to j. We then fill the dp table in a bottom-up manner, considering each element in the array and updating the dp table accordingly. Finally, we return the maximum sum of elements not greater than K in the entire array, which is stored in dp[0][n-1].
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Compute prefix sums, then iterate i while maintaining a deque of feasible j’s with the highest dp[j]; update dp[i] in O(1) amortized.
Brute Force Approach
Enumerate every possible cut pattern, compute each segment’s sum, and keep the best valid partition – exponential time.
Verified Code Solutions
function solution(nums, K) {
let n = nums.length;
let dp = Array(n).fill(0).map(() => Array(n).fill(0));
for (let i = n - 1; i >= 0; i--) {
for (let j = i; j < n; j++) {
if (i === j) {
dp[i][j] = nums[i];
} else {
let maxSum = 0;
for (let k = i; k <= j; k++) {
if (nums[k] <= K) {
maxSum = Math.max(maxSum, dp[i][k - 1] + nums[k]);
}
}
dp[i][j] = maxSum;
}
}
}
return dp[0][n - 1];
}class Solution {
public:
int solution(vector<int>& nums, int K) {
int n = nums.size();
vector<vector<int>> dp(n, vector<int>(n, 0));
for (int i = n - 1; i >= 0; i--) {
for (int j = i; j < n; j++) {
if (i == j) {
dp[i][j] = nums[i];
} else {
int maxSum = 0;
for (int k = i; k <= j; k++) {
if (nums[k] <= K) {
maxSum = max(maxSum, dp[i][k - 1] + nums[k]);
}
}
dp[i][j] = maxSum;
}
}
}
return dp[0][n - 1];
}
};class Solution {
public int solution(int[] nums, int K) {
int n = nums.length;
int[][] dp = new int[n][n];
for (int i = n - 1; i >= 0; i--) {
for (int j = i; j < n; j++) {
if (i == j) {
dp[i][j] = nums[i];
} else {
int maxSum = 0;
for (int k = i; k <= j; k++) {
if (nums[k] <= K) {
maxSum = Math.max(maxSum, dp[i][k - 1] + nums[k]);
}
}
dp[i][j] = maxSum;
}
}
}
return dp[0][n - 1];
}
}def solution(nums, K):
n = len(nums)
dp = [[0] * n for _ in range(n)]
for i in range(n - 1, -1, -1):
for j in range(i, n):
if i == j:
dp[i][j] = nums[i]
else:
max_sum = 0
for k in range(i, j + 1):
if nums[k] <= K:
max_sum = max(max_sum, dp[i][k - 1] + nums[k])
dp[i][j] = max_sum
return dp[0][n - 1]
function solution(nums, K) {
let n = nums.length;
let dp = Array(n).fill(0).map(() => Array(n).fill(0));
for (let i = n - 1; i >= 0; i--) {
for (let j = i; j < n; j++) {
if (i === j) {
dp[i][j] = nums[i];
} else {
let maxSum = 0;
for (let k = i; k <= j; k++) {
if (nums[k] <= K) {
maxSum = Math.max(maxSum, dp[i][k - 1] + nums[k]);
}
}
dp[i][j] = maxSum;
}
}
}
return dp[0][n - 1];
}Asked in Top Tech Interviews
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