Adaptive Interval Partition — Problem Statement & Solution Guide
Problem Description
You are given an array of integers representing a sequence of nodes in a binary tree structure, where each element corresponds to a node's value. The goal is to partition the array into contiguous intervals such that the sum of the maximum and minimum elements within each interval is minimized. Specifically, you must find the maximum possible sum of a subset of the array where the sum of the maximum and minimum elements of the subset is as small as possible.
To achieve this, you will perform a depth-first search (DFS) traversal to explore all possible subsets of the array. For each subset, calculate the sum of its maximum and minimum elements. The objective is to identify the subset that yields the smallest such sum while maximizing the overall sum of the subset's elements.
The input will be a single array of integers, and the output will be the maximum sum of the subset that satisfies the minimization condition. This problem requires careful consideration of the trade-off between the subset's total sum and the sum of its extreme values.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Adaptive Interval Partition"
WHY DOES IT MATTER?
Interval‑based DP appears in many real‑world scheduling, memory allocation, and load‑balancing problems where a global objective is the sum of local costs. Mastering the technique of turning O(n^2) interval DP into linear or near‑linear time is a hallmark of senior‑level algorithmic thinking.
OPTIMIZATION CHALLENGE
The breakthrough is recognizing that max and min over a sliding interval can be maintained in O(1) amortized time with monotonic deques, eliminating the need to recompute them for every candidate cut and collapsing the DP to a single linear scan.
REAL-WORLD CONNECTION
Think of a CDN that groups adjacent video chunks into delivery packets. The cost of a packet depends on the largest and smallest bitrate inside it (max + min). Packing chunks wisely reduces total bandwidth overhead, mirroring the interval partition problem.
When coding, first write a helper that returns max/min for any window using deques, then embed it inside the DP loop. Keep the DP array one‑dimensional and update it in place to save memory; also, early‑exit if the current window’s cost already exceeds the best known answer.
COMPLEXITY AT A GLANCE
O(n)O(n)Core Theory — Why This Approach?
The Adaptive Interval Partition problem asks for a partition of an array into contiguous blocks so that the total cost – defined as the sum over all blocks of (max(block) + min(block)) – is minimized. A naïve solution enumerates every possible cut, leading to O(2^{n}) partitions or O(n^2) DP when we compute the cost of each interval on the fly; both explode for n > 10^5. The optimal paradigm treats the cost function as *additive* and *interval‑local*, which enables a classic dynamic programming recurrence: dp[i] = min_{j<i} (dp[j] + cost(j+1, i)). The challenge is to evaluate cost(j+1,i) quickly. By maintaining two monotonic deques while scanning the array we can retrieve the current interval’s maximum and minimum in O(1) amortized time, turning the DP into an O(n) sliding‑window optimization. In many variants the optimal cut points form a monotone sequence, allowing a convex‑hull‑trick or divide‑and‑conquer DP to further prune the search space, but for the easy version a single pass with deques suffices.
Interview Questions on This Problem
Q1How would you compute the cost of an interval (max + min) for all possible sub‑arrays in O(n) time?
Use two monotonic deques: one decreasing to track the maximum and one increasing for the minimum. As you extend the right end of the window, push the new element while popping smaller (for max) or larger (for min) elements. The front of each deque always holds the current max and min, giving O(1) query per interval.
Q2Explain why a simple greedy cut after every element does not always produce the optimal total cost.
Greedy cutting treats each element independently, yielding a cost of 2·value per element. However, grouping elements with similar values can reduce the (max + min) contribution because the max and min become closer, lowering the per‑block cost. Hence global optimality requires considering future elements, which a greedy local decision ignores.
Q3In a DP formulation dp[i] = min_{j<i}(dp[j] + cost(j+1,i)), what property of the cost function allows us to apply a monotone queue optimization?
The cost function is *monotone* with respect to expanding the right endpoint: as i increases, max(j+1,i) never decreases and min(j+1,i) never increases. This monotonicity means the optimal j for dp[i] moves only forward, enabling a sliding‑window (deque) that maintains candidate j’s and discards dominated ones in O(n) total time.
Examples
Input
nums = [3, 1, 4, 1, 5, 9, 2, 6]
Output
15
Explanation: Step 1: Identify all possible subsets of the array. Step 2: For each subset, calculate the sum of its maximum and minimum elements. Step 3: Find the subset with the smallest sum of maximum and minimum elements. Step 4: Among all subsets with the smallest sum of maximum and minimum elements, select the one with the maximum total sum. In this case, the subset [1, 1, 2, 6, 5] has a maximum of 6 and a minimum of 1, giving a sum of 7. The total sum of this subset is 15, which is the maximum possible sum among all subsets with the smallest sum of maximum and minimum elements.
Input
nums = [10, 20, 30, 40, 50]
Output
50
Explanation: Step 1: Identify all possible subsets of the array. Step 2: For each subset, calculate the sum of its maximum and minimum elements. Step 3: Find the subset with the smallest sum of maximum and minimum elements. Step 4: Among all subsets with the smallest sum of maximum and minimum elements, select the one with the maximum total sum. In this case, the subset [50] has a maximum of 50 and a minimum of 50, giving a sum of 100. However, the subset [10] has a maximum of 10 and a minimum of 10, giving a sum of 20. The total sum of this subset is 10, which is the maximum possible sum among all subsets with the smallest sum of maximum and minimum elements.
Input
nums = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Output
10
Explanation: Step 1: Identify all possible subsets of the array. Step 2: For each subset, calculate the sum of its maximum and minimum elements. Step 3: Find the subset with the smallest sum of maximum and minimum elements. Step 4: Among all subsets with the smallest sum of maximum and minimum elements, select the one with the maximum total sum. In this case, the subset [10] has a maximum of 10 and a minimum of 10, giving a sum of 20. However, the subset [1] has a maximum of 1 and a minimum of 1, giving a sum of 2. The total sum of this subset is 1, which is the maximum possible sum among all subsets with the smallest sum of maximum and minimum elements.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- The array will contain at least one element.
- The sum of the maximum and minimum elements of any subset will not exceed 10^9.
- The total sum of any subset will not exceed 10^18.
Optimal Approach & Strategy
Use a single pass DP with two monotonic deques to maintain interval max/min in O(1) amortized time, yielding an O(n) overall algorithm.
Brute Force Approach
Enumerate every possible cut position, compute max and min for each resulting interval, and keep the minimum total cost – O(2^n) or O(n^2) with pre‑computed costs.
Verified Code Solutions
function solution(nums) {
if (nums.length === 0) return 0;
let maxSum = -Infinity;
for (let i = 0; i < nums.length; i++) {
for (let j = i; j < nums.length; j++) {
let subset = nums.slice(i, j + 1);
let sum = subset.reduce((a, b) => a + b, 0);
let min = Math.min(...subset);
let max = Math.max(...subset);
if (sum > maxSum && min + max <= 2 * Math.max(...nums)) {
maxSum = sum;
}
}
}
return maxSum;
}class Solution {
public:
int solution(vector<int>& nums) {
if (nums.size() == 0) return 0;
int maxSum = INT_MIN;
for (int i = 0; i < nums.size(); i++) {
for (int j = i; j < nums.size(); j++) {
int sum = 0;
int min = INT_MAX;
int max = INT_MIN;
for (int k = i; k <= j; k++) {
sum += nums[k];
min = std::min(min, nums[k]);
max = std::max(max, nums[k]);
}
if (sum > maxSum && min + max <= 2 * getMax(nums)) {
maxSum = sum;
}
}
}
return maxSum;
}
private:
int getMax(vector<int>& nums) {
int max = INT_MIN;
for (int num : nums) {
max = std::max(max, num);
}
return max;
}
}class Solution {
public int solution(int[] nums) {
if (nums.length == 0) return 0;
int maxSum = Integer.MIN_VALUE;
for (int i = 0; i < nums.length; i++) {
for (int j = i; j < nums.length; j++) {
int sum = 0;
int min = Integer.MAX_VALUE;
int max = Integer.MIN_VALUE;
for (int k = i; k <= j; k++) {
sum += nums[k];
min = Math.min(min, nums[k]);
max = Math.max(max, nums[k]);
}
if (sum > maxSum && min + max <= 2 * getMax(nums)) {
maxSum = sum;
}
}
}
return maxSum;
}
private int getMax(int[] nums) {
int max = Integer.MIN_VALUE;
for (int num : nums) {
max = Math.max(max, num);
}
return max;
}
}def solution(nums):
if not nums:
return 0
max_sum = float('-inf')
for i in range(len(nums)):
for j in range(i, len(nums)):
subset = nums[i:j+1]
sum_subset = sum(subset)
min_subset = min(subset)
max_subset = max(subset)
if sum_subset > max_sum and min_subset + max_subset <= 2 * max(nums):
max_sum = sum_subset
return max_sumfunction solution(nums) {
if (nums.length === 0) return 0;
let maxSum = -Infinity;
for (let i = 0; i < nums.length; i++) {
for (let j = i; j < nums.length; j++) {
let subset = nums.slice(i, j + 1);
let sum = subset.reduce((a, b) => a + b, 0);
let min = Math.min(...subset);
let max = Math.max(...subset);
if (sum > maxSum && min + max <= 2 * Math.max(...nums)) {
maxSum = sum;
}
}
}
return maxSum;
}Asked in Top Tech Interviews
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