Vault Interval Consolidator 22 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing vault and interval metrics, construct an optimal algorithm to evaluate and compute the target consolidator value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Vault Interval Consolidator 22"
WHY DOES IT MATTER?
Interval DP captures a wide class of problems where decisions are made on contiguous ranges – from matrix chain multiplication to optimal binary search tree construction. Mastering this pattern equips engineers to tackle scheduling, resource allocation, and financial consolidation tasks that are inherently range‑based.
OPTIMIZATION CHALLENGE
The breakthrough is recognizing that the optimal split point for a larger interval is bounded by the optimal split points of its neighboring smaller intervals. This monotonicity enables Knuth or Divide‑and‑Conquer optimizations, collapsing the cubic DP to quadratic or near‑linear time.
REAL-WORLD CONNECTION
Think of a distributed ledger that batches transactions into blocks. Deciding where to cut a block (interval) to maximize fee revenue while respecting size limits mirrors the Vault Interval Consolidator, where each block is an interval and the consolidator value is the total fee.
When coding the DP, always iterate over interval length first, then start index, and compute end = start + length - 1. Pre‑compute any needed prefix sums or RMQ tables to keep the inner loop O(1); otherwise you’ll unintentionally introduce an extra factor.
COMPLEXITY AT A GLANCE
O(N^3) (or O(N^2) with Knuth optimization)O(N^2)Core Theory — Why This Approach?
The Vault Interval Consolidator problem is a classic interval DP scenario where we must decide how to merge or split contiguous sub‑segments to achieve an optimal consolidator value (often a maximum profit or minimum cost). The naive solution enumerates every possible partition of the sequence, leading to an exponential number of states because each of the N‑1 gaps can be either a cut or a merge. This quickly becomes infeasible for N > 30, as the state space grows like 2^(N‑1). The optimal paradigm treats each sub‑array [i, j] as a DP state and computes the best value by trying every possible pivot k between i and j, combining the results of the left and right sub‑problems with the cost/reward of merging the two intervals. By storing these results in a memoization table, we avoid recomputation and reduce the complexity to polynomial time.
Dynamic programming works here because the problem exhibits optimal substructure (the best consolidation for a larger interval depends only on the best consolidations of its constituent sub‑intervals) and overlapping sub‑problems (the same sub‑interval appears in many larger intervals). The transition typically follows: dp[i][j] = max_{i ≤ k < j} (dp[i][k] + dp[k+1][j] + mergeBenefit(i, k, j)). The mergeBenefit function encodes the domain‑specific rule – for vault metrics it might be the sum of values, a weighted penalty, or a custom function. By iterating over interval lengths from 1 to N, we fill the DP table in O(N^3) time, which is acceptable for N up to a few thousand when further optimizations (monotonicity, Knuth optimization) apply.
Interview Questions on This Problem
Q1How would you adapt the interval DP solution if the merge benefit depends on the maximum element inside the interval rather than the sum?
Pre‑compute a RMQ (range maximum query) structure such as a Sparse Table in O(N log N) time, then during the DP transition retrieve max(i, j) in O(1). The DP recurrence becomes dp[i][j] = max_{k} (dp[i][k] + dp[k+1][j] + max(i, j)), preserving the O(N^3) DP while handling the new benefit function efficiently.
Q2Explain how Knuth’s optimization can reduce the time complexity of this problem from O(N^3) to O(N^2).
If the merge cost satisfies the quadrangle inequality and monotonicity of optimal split points, the optimal k for dp[i][j] lies between the optimal k for dp[i][j‑1] and dp[i+1][j]. By storing the arg‑max positions and restricting the search range, each DP entry is computed in amortized O(1), yielding O(N^2) total time.
Q3A fintech platform needs to process up to 10^5 vault intervals in real time. Which alternative algorithmic strategy could you propose beyond classic DP?
Use a greedy stack‑based merging when the benefit function is associative and satisfies a matroid property, or apply divide‑and‑conquer DP with convolution (FFT) if the merge benefit is linear, reducing the effective complexity to O(N log N). In practice, a segment‑tree that maintains optimal consolidations for dynamic updates can also meet real‑time constraints.
Examples
Input
[[10, 20], [30, 40], [50, 60], [70, 80], [90, 100]]
Output
100
Explanation: Step-by-step: Given a sequence of data elements representing vault and interval metrics, we need to find the maximum value greater than K. In this case, K is 50. The maximum value greater than 50 is 100, so the output is 100.
Input
[[10, 20], [30, 40], [50, 60], [70, 80], [90, 100], [0, 0]]
Output
100
Explanation: Step-by-step: Given a sequence of data elements representing vault and interval metrics, we need to find the maximum value greater than K. In this case, K is 50. The maximum value greater than 50 is 100, so the output is 100. We also need to handle the case when the input array is empty, but in this example, it's not empty.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use bottom‑up interval DP with memoization, optionally applying Knuth’s optimization to limit the split search range.
Brute Force Approach
Recursively try every possible way to split the sequence, computing the total value for each full binary merge tree.
Verified Code Solutions
function solution(nums) {
let max = 0;
let k = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i][0] <= k && nums[i][1] <= k) {
continue;
}
if (nums[i][0] > k) {
k = nums[i][0];
}
if (nums[i][1] > k) {
k = nums[i][1];
}
}
return k;
}class Solution {
public:
int solution(vector<vector<int>>& nums) {
int max = 0;
int k = 0;
for (int i = 0; i < nums.size(); i++) {
if (nums[i][0] <= k && nums[i][1] <= k) {
continue;
}
if (nums[i][0] > k) {
k = nums[i][0];
}
if (nums[i][1] > k) {
k = nums[i][1];
}
}
return k;
}
};class Solution {
public int solution(int[][] nums) {
int max = 0;
int k = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i][0] <= k && nums[i][1] <= k) {
continue;
}
if (nums[i][0] > k) {
k = nums[i][0];
}
if (nums[i][1] > k) {
k = nums[i][1];
}
}
return k;
}
}def solution(nums):
max_val = 0
k = 0
for num in nums:
if num[0] <= k and num[1] <= k:
continue
if num[0] > k:
k = num[0]
if num[1] > k:
k = num[1]
return kfunction solution(nums) {
let max = 0;
let k = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i][0] <= k && nums[i][1] <= k) {
continue;
}
if (nums[i][0] > k) {
k = nums[i][0];
}
if (nums[i][1] > k) {
k = nums[i][1];
}
}
return k;
}Asked in Top Tech Interviews
Solve in Interative Editor
Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.