Cumulative Interval Partition — Problem Statement & Solution Guide
Problem Description
You are provided with a linear sequence of integers. The objective is to determine the maximum possible cumulative sum achievable by decomposing the sequence into a set of non-overlapping, contiguous subarrays. Each subarray contributes its internal sum to the total cumulative value. The partitioning must cover the entire original sequence without gaps or overlaps, and the order of subarrays must strictly follow the original sequence order.
Formally, given an array nums of length N, you must select a partition of indices such that the array is divided into segments [0, i1-1], [i1, i2-1], ..., [ik, N-1]. The value of a partition is the sum of the sums of these segments. Since the sum of the sums of all segments in a complete partition is mathematically equivalent to the sum of all elements in the original array, the problem reduces to identifying the global sum of the input array. However, in a generalized context where segment weights might vary (e.g., if a segment's contribution is defined as its sum multiplied by its length or a specific function), the dynamic programming approach becomes critical. For this specific 'Cumulative Interval Partition' variant, the task is to compute the total sum of the array, which represents the maximum cumulative sum under the constraint that all elements must be included in exactly one interval.
Input: An array nums containing integers.
Output: A single integer representing the maximum cumulative sum derived from the optimal partition of the array into contiguous intervals.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Cumulative Interval Partition"
WHY DOES IT MATTER?
Interval‑partition DP captures a broad class of optimization problems where a global objective is built from locally optimal contiguous segments. Mastering this pattern enables you to solve scheduling, resource allocation, and profit‑maximization tasks that appear in many system‑design interviews.
OPTIMIZATION CHALLENGE
The breakthrough is recognizing that the inner maximization over all previous cut positions can be expressed as a single running maximum (DP[j‑1]‑pref[j‑1]). This eliminates the quadratic loop and collapses the state to O(1) per element.
REAL-WORLD CONNECTION
Think of a data pipeline that batches incoming logs into variable‑size chunks for compression. The total compression gain depends on the sum of chunk sizes; deciding where to cut the stream mirrors the interval partition DP, balancing batch size against processing overhead.
During an interview, compute the prefix sum on the fly and keep a variable ‘bestSoFar’. Update DP[i] = prefix + bestSoFar, then refresh bestSoFar = max(bestSoFar, DP[i]‑prefix). This one‑liner DP shows you understand both the recurrence and the optimization.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The Cumulative Interval Partition problem can be modeled as a classic one‑dimensional dynamic programming task. For each index i we define DP[i] as the maximum cumulative value achievable for the prefix ending at i when the array is partitioned into contiguous, non‑overlapping subarrays that exactly cover positions 0…i. The naive recurrence DP[i] = max_{0≤j≤i}(DP[j‑1] + sum(arr[j…i])) examines every possible cut before i, leading to O(n^2) time. By expanding the sum with prefix sums (pref[i] = sum_{0..i} arr[k]), the recurrence becomes DP[i] = pref[i] + max_{0≤j≤i}(DP[j‑1] - pref[j‑1]). The inner maximum depends only on the best value seen so far, so we can maintain a running variable best = max(best, DP[i‑1] - pref[i‑1]) while scanning the array. This reduces the transition to O(1) per element, yielding an overall O(n) solution with O(1) extra space. The optimal paradigm is therefore a DP with prefix‑sum optimization, a pattern that appears in many interval‑partitioning and “maximum sub‑array with constraints” problems.
Interview Questions on This Problem
Q1How would you modify the DP if each subarray contributed its sum multiplied by its length?
Introduce a second prefix array for weighted sums: weightedPref[i] = Σ_{k=0..i} arr[k] * (k+1). The DP transition becomes DP[i] = max_{j≤i}(DP[j‑1] + (pref[i]-pref[j‑1]) * (i-j+1)). By rearranging terms you can keep a running maximum of (DP[j‑1] - pref[j‑1]*j + weightedPref[j‑1]) to achieve O(n) time.
Q2Why does the naive O(2^n) enumeration of all partitions fail for n = 10^5, and what is the key insight that brings it down to linear time?
Enumerating all 2^{n‑1} ways to place cuts explodes combinatorially and cannot finish within any realistic time limit. The key insight is that the contribution of a cut only depends on the prefix sums up to that point, allowing us to collapse the exponential state space into a single scalar (the best DP[j]‑pref[j] seen so far). This transforms the recurrence into a constant‑time update per element.
Q3In a streaming scenario where numbers arrive one‑by‑one, can you still compute the optimal cumulative value without storing the whole array?
Yes. Maintain the running prefix sum, the current best value of DP[j]‑pref[j], and the DP for the latest index. Each new element updates the prefix sum, computes DP[i] = prefix + best, then updates best = max(best, DP[i]‑prefix). Only O(1) memory is required.
Examples
Input
nums = [1, 2, 3, 4]
Output
10
Explanation: The array is partitioned into contiguous intervals. The sum of the entire array is 1 + 2 + 3 + 4 = 10. Any partition (e.g., [1,2] and [3,4]) yields (1+2) + (3+4) = 3 + 7 = 10. The maximum cumulative sum is 10.
Input
nums = [-5, 10, -2, 7]
Output
10
Explanation: The total sum of the array is -5 + 10 + (-2) + 7 = 10. Regardless of how the array is partitioned into contiguous subarrays, the sum of the subarray sums will always equal the total sum of the original array. Thus, the maximum cumulative sum is 10.
Input
nums = [0, 0, 0]
Output
0
Explanation: The array consists of three zeros. The sum of any partition is 0 + 0 + 0 = 0. The maximum cumulative sum is 0.
Input
nums = [100, -1, 100]
Output
199
Explanation: The total sum is 100 + (-1) + 100 = 199. Partitioning into [100], [-1], [100] gives 100 + (-1) + 100 = 199. Partitioning into [100, -1, 100] gives 199. The result is consistently 199.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- The sum of all elements in nums fits within a 64-bit signed integer.
Optimal Approach & Strategy
Use DP with prefix sums and a running maximum of (DP[j]‑prefix[j]) to compute each DP[i] in constant time, achieving linear overall complexity.
Brute Force Approach
Enumerate every possible set of cut positions (2^{n‑1} partitions) and compute the total sum for each, keeping the maximum.
Verified Code Solutions
function maxCumulativeSum(nums) {
return nums.reduce((acc, val) => acc + val, 0);
}#include <vector>
#include <numeric>
int maxCumulativeSum(const std::vector<int>& nums) {
return std::accumulate(nums.begin(), nums.end(), 0);
}public class Solution {
public int maxCumulativeSum(int[] nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
}def max_cumulative_sum(nums):
return sum(nums)function maxCumulativeSum(nums) {
return nums.reduce((acc, val) => acc + val, 0);
}Asked in Top Tech Interviews
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