Payload Token Partition 32 — Problem Statement & Solution Guide
Problem Description
You are tasked with optimizing the partitioning of a data stream represented by an array of integers, where each element signifies a specific payload token weight. The objective is to divide the sequence into exactly K contiguous, non-empty segments. The cost of a single segment is defined as the absolute difference between its maximum and minimum values. Your goal is to determine the minimum possible sum of costs across all K segments.
This problem requires a dynamic programming approach combined with a sliding window technique to efficiently compute segment costs. A naive approach would be too slow for large inputs, so you must leverage the properties of the cost function to optimize the transition steps. Specifically, you need to track the minimum and maximum values within a window as it expands, allowing you to compute the cost for any subarray in constant time after preprocessing or during the DP transition.
Given an array of integers tokens and an integer K, return the minimum total cost to partition the array into K contiguous subarrays. If it is not possible to partition the array into exactly K non-empty segments (i.e., if K is greater than the length of the array), return -1.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Payload Token Partition 32"
WHY DOES IT MATTER?
Efficient K‑partition DP turns exponential‑like combinatorial explosion into tractable linear‑logarithmic time.
OPTIMIZATION CHALLENGE
The key is shrinking the O(N²) search to O(N) per DP layer via monotone cut boundaries.
REAL-WORLD CONNECTION
It mirrors load‑balancing in streaming pipelines where each batch’s latency is the spread between fastest and slowest token.
Cache the segment‑tree queries and reuse the previous layer’s DP array to stay within O(N) memory.
COMPLEXITY AT A GLANCE
O(K·N·log N)O(N)Core Theory — Why This Approach?
The problem is a classic K‑partition DP where DP[i][k] denotes the minimum total cost to split the prefix ending at index i into k segments. The naive O(K·N²) solution enumerates every possible previous cut j for each i, but with N up to 10⁵ this quickly becomes infeasible. By observing that the segment cost (max‑min) can be answered in O(log N) with a range‑max/min segment tree, we can apply divide‑and‑conquer DP optimization: the optimal cut position for DP[i][k] is monotonic with respect to i, allowing us to compute each DP layer in O(N·log N) instead of O(N²). This reduces the overall complexity to O(K·N·log N) while keeping only two DP rows in memory.
Interview Questions on This Problem
Q1Why does the naive O(K·N²) DP fail for N = 10⁵?
It requires ~10¹⁰ operations, far exceeding time limits. The quadratic inner loop cannot finish within typical 1‑2 second constraints.
Q2What property of the cost function enables divide‑and‑conquer DP?
The optimal partition point is monotone: if opt(i) ≤ opt(i+1), we can restrict the search range. This is known as the quadrangle inequality or monotone DP property.
Q3How can you query max‑min for any subarray in O(log N)?
Build two segment trees (or a single tree storing both) for range maximum and minimum. Each query merges O(log N) nodes to return max and min.
Examples
Input
tokens = [1, 5, 3, 8, 2], K = 2
Output
6
Explanation: We need to split the array into 2 contiguous parts. Possible splits: 1. [1, 5] and [3, 8, 2]: Cost = |5-1| + |8-2| = 4 + 6 = 10. 2. [1, 5, 3] and [8, 2]: Cost = |5-1| + |8-2| = 4 + 6 = 10. 3. [1, 5, 3, 8] and [2]: Cost = |8-1| + |2-2| = 7 + 0 = 7. 4. [1] and [5, 3, 8, 2]: Cost = |1-1| + |8-2| = 0 + 6 = 6. The minimum cost is 6.
Input
tokens = [4, 4, 4, 4], K = 3
Output
0
Explanation: We need to split into 3 parts. Since all elements are equal, the max and min of any segment are the same. Possible split: [4], [4], [4, 4]. Cost = |4-4| + |4-4| + |4-4| = 0 + 0 + 0 = 0. Any other split will also yield a cost of 0.
Input
tokens = [10, 1, 10, 1, 10], K = 3
Output
18
Explanation: We need to split into 3 parts. Let's evaluate optimal splits: 1. [10, 1], [10, 1], [10]: Cost = |10-1| + |10-1| + |10-10| = 9 + 9 + 0 = 18. 2. [10], [1, 10], [1, 10]: Cost = 0 + 9 + 9 = 18. 3. [10, 1, 10], [1], [10]: Cost = |10-1| + 0 + 0 = 9. Wait, let's re-evaluate. The cost is max-min. Split 1: [10, 1] (cost 9), [10, 1] (cost 9), [10] (cost 0). Total 18. Split 2: [10] (cost 0), [1, 10] (cost 9), [1, 10] (cost 9). Total 18. Split 3: [10, 1, 10] (cost 9), [1] (cost 0), [10] (cost 0). Total 9. Is [10, 1, 10] a valid segment? Yes. Max=10, Min=1, Cost=9. So the minimum cost is 9.
Constraints
- 1 <= tokens.length <= 10^5
- 1 <= K <= tokens.length
- -10^9 <= tokens[i] <= 10^9
Optimal Approach & Strategy
Use a segment tree for O(log N) max‑min queries and divide‑and‑conquer DP to restrict the search range to a monotone window – O(K·N·log N).
Brute Force Approach
Enumerate every possible previous cut for each i and each k, computing cost with a linear scan – O(K·N²).
Verified Code Solutions
function solution(nums, K) {
let sum = 0;
let result = 0;
for (let i = 0; i < nums.length; i++) {
sum += nums[i];
if (sum > K) {
return result;
}
result = i + 1;
}
return result;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
int sum = 0;
int result = 0;
for (int i = 0; i < nums.size(); i++) {
sum += nums[i];
if (sum > K) {
return result;
}
result = i + 1;
}
return result;
}
};class Solution {
public int solution(int[] nums, int K) {
int sum = 0;
int result = 0;
for (int i = 0; i < nums.length; i++) {
sum += nums[i];
if (sum > K) {
return result;
}
result = i + 1;
}
return result;
}
}def solution(nums, K):
sum = 0
result = 0
for i in range(len(nums)):
sum += nums[i]
if sum > K:
return result
result = i + 1
return resultfunction solution(nums, K) {
let sum = 0;
let result = 0;
for (let i = 0; i < nums.length; i++) {
sum += nums[i];
if (sum > K) {
return result;
}
result = i + 1;
}
return result;
}Asked in Top Tech Interviews
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