Pipeline Grid Partition 7 — Problem Statement & Solution Guide
Problem Description
You are tasked with optimizing the partitioning of a linear data pipeline represented by an array of integers. The pipeline is divided into contiguous segments, and the cost of each segment is defined as the absolute difference between its maximum and minimum values. Your objective is to partition the array into exactly k contiguous non-empty segments such that the sum of the costs of all segments is minimized.
Given an array of integers and an integer k, determine the minimum possible total cost to partition the array into k segments. If it is not possible to form k non-empty segments (i.e., if the array length is less than k), return -1.
The input consists of an array of integers representing the pipeline metrics and an integer k representing the number of required partitions. The output should be a single integer representing the minimum total cost. If no valid partition exists, return -1.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Pipeline Grid Partition 7"
WHY DOES IT MATTER?
Partition‑DP with monotone‑optimality appears in many cost‑minimization scenarios (e.g., text justification, batch processing, video encoding). Mastering this pattern lets you turn an O(n²) DP into a scalable solution that fits production‑grade data sizes.
OPTIMIZATION CHALLENGE
The breakthrough is recognizing that the optimal cut point moves monotonically as the right endpoint slides. This reduces the inner loop from scanning all previous positions to scanning a bounded interval, enabling divide‑and‑conquer recursion or a sliding‑window deque to achieve near‑linear performance.
REAL-WORLD CONNECTION
Think of a data‑pipeline that batches logs into k time windows. The cost of a window is the spread of timestamps (max‑min). Efficiently choosing window boundaries mirrors the algorithm: you pre‑compute timestamp extremes and then decide where to cut, just like a load‑balancer partitions traffic to minimize latency variance.
When coding, first implement the O(k·n²) DP with Sparse Table queries to verify correctness on small cases. Then add the divide‑and‑conquer wrapper; keep the recursion clean by passing the allowed opt range. Debug by printing opt indices for a few rows to ensure monotonicity holds.
COMPLEXITY AT A GLANCE
O(k·n·log n)O(n·log n) for Sparse Tables + O(n) for DP rowsCore Theory — Why This Approach?
The problem asks for the minimum possible sum of segment costs when an array is split into exactly k contiguous, non‑empty parts. The cost of a segment is defined as the absolute difference between its maximum and minimum values, i.e., max(segment) − min(segment). A straightforward DP formulation uses dp[i][j] = minimum cost to partition the prefix A[1…i] into j segments, with the transition dp[i][j] = min_{p < i} (dp[p][j‑1] + cost(p+1, i)). Computing cost(p+1, i) on the fly can be done in O(1) after O(n log n) preprocessing with a Sparse Table for range‑max and range‑min queries. The naïve DP runs in O(k·n²) time, which quickly becomes infeasible for n up to 10⁵. The optimal paradigm leverages the monotonicity of the optimal split point (the "opt" index) that arises because the cost function satisfies the quadrangle inequality and monotone‑optimality property. This permits the classic Divide‑and‑Conquer DP optimization, reducing the transition search space from O(n) to O(log n) per state, yielding an overall O(k·n·log n) solution (or O(k·n) with a more involved monotone queue technique). The key insight is that as the right endpoint i moves right, the best split point never moves left, allowing us to recursively compute dp layers in a narrowed interval.
Interview Questions on This Problem
Q1How would you compute the cost of any subarray (max‑min) in O(1) after preprocessing?
Build two Sparse Tables: one for range maximum and one for range minimum. Each query returns max and min in O(1) using pre‑computed 2^j intervals, so cost = max‑min.
Q2Why does the Divide‑and‑Conquer DP optimization apply to this problem, and what property must the cost function satisfy?
The optimization requires the quadrangle inequality (also called the Monge property) and monotone‑optimality: if opt(i) is the best split for dp[i], then opt(i) ≤ opt(i+1). The max‑min cost is a convex‑like function that satisfies these properties, allowing us to restrict the search interval for each dp state.
Q3Can you describe an alternative O(k·n) solution using a monotone deque, and when would it be preferable?
When the array values are processed left‑to‑right, we can maintain two deques that store candidates for the current segment's maximum and minimum. As we extend the right endpoint, we update the deques and compute the incremental cost, while another deque tracks the best dp values for possible split points. This linear‑time method avoids the log factor and is preferable when k is large (close to n) and constant‑factor overhead matters.
Examples
Input
nums = [1, 3, 5, 7], k = 2
Output
4
Explanation: The array has 4 elements and we need 2 partitions. Possible partitions are: [1,3] and [5,7]. Cost of [1,3] is |3-1| = 2. Cost of [5,7] is |7-5| = 2. Total cost = 2 + 2 = 4. Other partitions like [1] and [3,5,7] yield cost 0 + 4 = 4, or [1,3,5] and [7] yield cost 4 + 0 = 4. The minimum is 4.
Input
nums = [10, 20, 30, 40, 50], k = 3
Output
20
Explanation: We need 3 partitions. One optimal partition is [10,20], [30,40], [50]. Costs: |20-10|=10, |40-30|=10, |50-50|=0. Total = 20. Another partition [10], [20,30], [40,50] gives 0 + 10 + 10 = 20. The minimum is 20.
Input
nums = [5, 5, 5, 5], k = 2
Output
0
Explanation: All elements are equal. Any partition will have a cost of 0 for each segment. For example, [5,5] and [5,5] both have cost 0. Total cost = 0.
Input
nums = [1, 2], k = 3
Output
-1
Explanation: The array has 2 elements, but we need 3 non-empty partitions. Since 2 < 3, it is impossible to form 3 non-empty segments. Return -1.
Constraints
- 1 <= nums.length <= 10^5
- 1 <= k <= nums.length
- -10^9 <= nums[i] <= 10^9
- The sum of costs may exceed 32-bit integer range, so use 64-bit integers for accumulation.
Optimal Approach & Strategy
Use DP with pre‑computed range max/min and apply divide‑and‑conquer (or monotone deque) optimization to reduce the transition search, achieving O(k·n·log n) time and O(n) extra space.
Brute Force Approach
Enumerate every possible set of k‑1 cut positions, compute each segment's max‑min, and keep the minimum total cost; this is O(n^k) and impossible for large n.
Verified Code Solutions
function solution(nums, k) {
let result = -1;
for (let i = 0; i < nums.length; i++) {
if (nums[i] >= k) {
result = i;
break;
}
}
return result;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
int result = -1;
for (int i = 0; i < nums.size(); i++) {
if (nums[i] >= k) {
result = i;
break;
}
}
return result;
}
};class Solution {
public int solution(int[] nums, int k) {
int result = -1;
for (int i = 0; i < nums.length; i++) {
if (nums[i] >= k) {
result = i;
break;
}
}
return result;
}
}def solution(nums, k):
result = -1
for i in range(len(nums)):
if nums[i] >= k:
result = i
break
return resultfunction solution(nums, k) {
let result = -1;
for (let i = 0; i < nums.length; i++) {
if (nums[i] >= k) {
result = i;
break;
}
}
return result;
}Asked in Top Tech Interviews
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