Hyper-Dimensional Grid Optimizer — Problem Statement & Solution Guide
Problem Description
You are given a matrix A with N rows and M columns (1 ≤ N ≤ 15, 1 ≤ M ≤ 10). For each row i you may select a contiguous segment of columns [l_i, r_i] (l_i ≤ r_i) or choose to select nothing from that row. The selected segments must satisfy a global exclusivity condition: no column may belong to the selected segment of more than one row. The profit of a selection is the sum of all A[i][j] that lie inside the chosen segments. Your task is to compute the maximum possible profit. If selecting no cells yields a higher profit (e.g., when all numbers are negative), the answer is 0.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Hyper-Dimensional Grid Optimizer"
WHY DOES IT MATTER?
DP over subsets is a cornerstone technique for problems where a small set of resources (here columns) must be allocated without conflict. Recognizing that the exclusivity constraint maps naturally to bitmask disjointness unlocks exponential reductions in state space.
OPTIMIZATION CHALLENGE
The key insight is to pre‑compute all contiguous segment masks per row and to iterate only over masks that are compatible (mask & segmentMask == 0). This avoids enumerating all N‑tuples of segments and collapses the exponential blow‑up to O(N·2^M·M^2).
REAL-WORLD CONNECTION
Think of allocating frequency bands (columns) to multiple radio towers (rows) where each tower can use a contiguous band or none, and no two towers may share a frequency. Optimizing total signal strength mirrors the DP over masks used in spectrum management systems.
When coding, store the profit for each segment mask in an array indexed by the mask; this lets you retrieve the segment profit in O(1) during DP transitions and keeps the inner loop tight.
COMPLEXITY AT A GLANCE
O(N * 2^M * M^2)O(2^M)Core Theory — Why This Approach?
The problem can be modeled as a combinatorial optimization over a small universe of columns. Each row contributes a set of possible contiguous column masks (including the empty mask) together with a profit equal to the sum of the chosen cells. The global exclusivity constraint translates to a requirement that the masks of different rows be pairwise disjoint. Because M ≤ 10, the entire column space fits into a 10‑bit integer, allowing us to treat a mask as a state in a dynamic programming (DP) over rows. A naïve exhaustive search would enumerate every combination of segments for all rows, leading to O((M^2)^N) possibilities, which explodes even for modest N. By collapsing the column usage into a bitmask, we reduce the state space to 2^M (at most 1024) and iterate over rows, updating DP transitions only for masks that do not intersect. This yields a pseudo‑polynomial DP that is optimal for the given constraints. The approach exemplifies the classic "DP over subsets" pattern, where a small combinatorial dimension (columns) is enumerated exhaustively while the larger dimension (rows) is processed sequentially.
Interview Questions on This Problem
Q1How would you adapt the solution if the columns were up to 20 instead of 10?
With M=20, 2^M becomes about one million, still feasible for N≤15 if we prune aggressively. We could use meet‑in‑the‑middle: split rows into two halves, compute all achievable masks and profits for each half, then combine compatible masks using a hashmap to keep the best profit per mask. This reduces the effective DP to O(N·2^{M/2}) and fits within memory.
Q2Explain why a greedy algorithm that picks the highest‑profit segment per row fails.
Greedy selection ignores the global column exclusivity. A high‑profit segment in one row may block a slightly lower‑profit but overall more beneficial segment in another row, leading to a sub‑optimal total. Counter‑examples can be constructed where the optimal solution uses two moderate segments that together exceed the profit of the greedy choice.
Q3In a distributed system, how could you parallelize the DP over masks for this problem?
Since each DP transition for a given row depends only on the previous row's mask values, we can parallelize across masks within a row: each worker processes a subset of masks, computes candidate updates for all non‑conflicting segment masks, and writes results to a shared next‑row DP array using atomic max or reduction. After each row, a barrier synchronizes workers before proceeding to the next row.
Examples
Input
2 3 5 1 4 2 3 6
Output
14
Explanation: Row 1 can take the segment [1,1] (value 5). Row 2 can then take the segment [2,3] (values 3+6=9). Columns 1,2,3 are each used at most once, and the total profit is 5+9=14, which is optimal. Any segment covering column 1 in both rows would violate the exclusivity rule, and taking the whole first row (5+1+4=10) would prevent any selection in the second row, yielding a smaller total.
Input
3 4 1 2 3 4 4 3 2 1 5 5 5 5
Output
18
Explanation: Choose column 4 in row 1 (value 4). Choose column 1 in row 2 (value 4). In row 3, columns 2‑3 form a contiguous segment with sum 5+5=10. All chosen columns are distinct, so the profit is 4+4+10=18. Any attempt to use more columns in row 3 would block selections in rows 1 or 2 and lead to a lower total.
Input
1 5 -1 -2 -3 -4 -5
Output
0
Explanation: All numbers are negative. Selecting an empty segment (i.e., taking no cells) is allowed and yields profit 0, which is better than any negative sum.
Constraints
- 1 ≤ N ≤ 15
- 1 ≤ M ≤ 10
- -10^9 ≤ A[i][j] ≤ 10^9
- The answer fits in a signed 64‑bit integer.
Optimal Approach & Strategy
Pre‑compute contiguous segment masks per row, then use DP over rows with a 2^M mask state, transitioning only with non‑overlapping segment masks.
Brute Force Approach
Enumerate every possible segment (or empty) for each of the N rows, generate all N‑tuples, and keep the best tuple whose masks are pairwise disjoint.
Verified Code Solutions
function solution(nums) {
if (nums.length === 0) return 0;
let maxSums = new Array(nums.length).fill(0);
maxSums[0] = nums[0];
let maxSum = nums[0];
for (let i = 1; i < nums.length; i++) {
maxSums[i] = Math.max(nums[i], maxSums[i - 1] + nums[i]);
maxSum = Math.max(maxSum, maxSums[i]);
}
return maxSum;
}class Solution {
public:
int solution(vector<int> nums) {
if (nums.size() == 0) return 0;
vector<int> maxSums(nums.size());
maxSums[0] = nums[0];
int maxSum = nums[0];
for (int i = 1; i < nums.size(); i++) {
maxSums[i] = max(nums[i], maxSums[i - 1] + nums[i]);
maxSum = max(maxSum, maxSums[i]);
}
return maxSum;
}
};class Solution {
public int solution(int[] nums) {
if (nums.length == 0) return 0;
int[] maxSums = new int[nums.length];
maxSums[0] = nums[0];
int maxSum = nums[0];
for (int i = 1; i < nums.length; i++) {
maxSums[i] = Math.max(nums[i], maxSums[i - 1] + nums[i]);
maxSum = Math.max(maxSum, maxSums[i]);
}
return maxSum;
}
}def solution(nums):
if not nums:
return 0
maxSums = [0] * len(nums)
maxSums[0] = nums[0]
maxSum = nums[0]
for i in range(1, len(nums)):
maxSums[i] = max(nums[i], maxSums[i - 1] + nums[i])
maxSum = max(maxSum, maxSums[i])
return maxSumfunction solution(nums) {
if (nums.length === 0) return 0;
let maxSums = new Array(nums.length).fill(0);
maxSums[0] = nums[0];
let maxSum = nums[0];
for (let i = 1; i < nums.length; i++) {
maxSums[i] = Math.max(nums[i], maxSums[i - 1] + nums[i]);
maxSum = Math.max(maxSum, maxSums[i]);
}
return maxSum;
}Asked in Top Tech Interviews
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