Hyper-Dimensional Grid Optimizer 3 — Problem Statement & Solution Guide
Problem Description
You are tasked with optimizing a high-dimensional grid system where each cell is represented by a bitmask of active features. Given an array grid of length N, where each element is an integer representing a bitmask of size K (i.e., values in range [0, 2^K - 1]), determine the maximum sum of values obtainable by selecting a subsequence of cells such that no two selected cells share any common active feature (i.e., their bitwise AND is 0).
Formally, find the maximum sum of a subset of indices i_1, i_2, ..., i_m such that for all a != b, grid[i_a] & grid[i_b] == 0. The goal is to compute this maximum sum efficiently using Bitmask Dynamic Programming.
Input: An array grid of integers, where each integer represents a bitmask of active features.
Output: A single integer representing the maximum sum of values from a valid subsequence of non-overlapping feature masks.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Hyper-Dimensional Grid Optimizer 3"
WHY DOES IT MATTER?
Bitmask DP transforms exponential combinatorial constraints into polynomial‑time state transitions.
OPTIMIZATION CHALLENGE
The key is reducing the exponential search over subsets to O(N·2^K) by reusing previously computed masks.
REAL-WORLD CONNECTION
It mirrors resource‑allocation in CPUs where each core can run only non‑conflicting instruction sets.
Iterate elements outermost and update dp in reverse (or use a copy) to avoid using the same element twice in a transition.
COMPLEXITY AT A GLANCE
O(N·2^K)O(2^K)Core Theory — Why This Approach?
The problem reduces to a maximum‑weight independent set on a hyper‑graph where each vertex is a grid cell and edges connect cells whose bitmasks share a set bit. A naive O(2^N) enumeration fails because N can be up to 10^5, making exhaustive search impossible. The optimal paradigm is DP over bitmask states: dp[mask] stores the best sum achievable using exactly the feature set represented by mask. For each cell with mask m and value v, we transition from any state s where (s & m) == 0 to s|m, updating dp[s|m] = max(dp[s|m], dp[s] + v). This leverages the fact that K ≤ 20‑22, so the state space 2^K is tractable while iterating over N elements remains linear, yielding an O(N·2^K) solution.
Interview Questions on This Problem
Q1Why is a simple longest‑increasing‑subsequence style DP insufficient for this problem?
Because the constraint is on overlapping bits, not on order or value monotonicity. The state must capture which bits are already used, which LIS DP does not.
Q2What is the significance of the condition (s & m) == 0 in the transition?
It guarantees that the new cell's active features do not conflict with those already selected. Without this check the solution could double‑count a feature, violating the problem constraint.
Q3How does the choice of K (bitmask size) affect the algorithm’s feasibility?
The DP’s memory and time are O(2^K); if K exceeds ~22 the state space explodes. Therefore the algorithm is only practical when K is small relative to N.
Examples
Input
grid = [1, 2, 3, 4, 5]
Output
7
Explanation: The bitmasks are: 1 (001), 2 (010), 3 (011), 4 (100), 5 (101). We need to select a subset where no two masks share a common bit. The optimal selection is {1, 2, 4} with sum 1+2+4=7. Alternatively, {3, 4} gives 3+4=7. Both are valid, and 7 is the maximum.
Input
grid = [7, 8, 15, 16]
Output
24
Explanation: The bitmasks are: 7 (0111), 8 (1000), 15 (1111), 16 (10000). The optimal selection is {7, 8, 16} with sum 7+8+16=31. However, 7 and 8 are disjoint, and 16 is disjoint from both. Thus, the maximum sum is 31.
Input
grid = [1, 1, 2, 2, 4]
Output
7
Explanation: The bitmasks are: 1 (001), 1 (001), 2 (010), 2 (010), 4 (100). We can select one instance of each distinct mask: {1, 2, 4} with sum 1+2+4=7. Selecting duplicates does not increase the sum since they share the same bits.
Constraints
- 1 <= grid.length <= 10^5
- 0 <= grid[i] < 2^20
- The sum of grid[i] over all i does not exceed 10^9
- Time limit: 2 seconds
- Space limit: 256 MB
Optimal Approach & Strategy
Use DP over 2^K masks, iterating through the array and updating compatible states, achieving O(N·2^K) time.
Brute Force Approach
Enumerate all subsequences, check bitmask conflicts, and keep the maximum sum; this is O(2^N) and impossible for large N.
Verified Code Solutions
function solution(nums) {
const n = nums.length;
const dp = new Array(1 << n).fill(0).map(() => new Array(n).fill(0));
for (let i = 0; i < n; i++) {
dp[1 << i][i] = nums[i];
}
for (let mask = 0; mask < (1 << n); mask++) {
for (let i = 0; i < n; i++) {
if ((mask & (1 << i)) !== 0) {
for (let j = 0; j < i; j++) {
if ((mask & (1 << j)) !== 0) {
dp[mask][i] = Math.max(dp[mask][i], dp[mask ^ (1 << i)][j] + nums[i]);
}
}
}
}
}
let maxSum = 0;
for (let i = 0; i < n; i++) {
maxSum = Math.max(maxSum, dp[(1 << n) - 1][i]);
}
return maxSum;
}class Solution {
public:
int solution(vector<int>& nums) {
int n = nums.size();
vector<vector<int>> dp(1 << n, vector<int>(n, 0));
for (int i = 0; i < n; i++) {
dp[1 << i][i] = nums[i];
}
for (int mask = 0; mask < (1 << n); mask++) {
for (int i = 0; i < n; i++) {
if ((mask & (1 << i)) != 0) {
for (int j = 0; j < i; j++) {
if ((mask & (1 << j)) != 0) {
dp[mask][i] = max(dp[mask][i], dp[mask ^ (1 << i)][j] + nums[i]);
}
}
}
}
}
int maxSum = 0;
for (int i = 0; i < n; i++) {
maxSum = max(maxSum, dp[(1 << n) - 1][i]);
}
return maxSum;
}
};class Solution {
public int solution(int[] nums) {
int n = nums.length;
int[][] dp = new int[1 << n][n];
for (int i = 0; i < n; i++) {
dp[1 << i][i] = nums[i];
}
for (int mask = 0; mask < (1 << n); mask++) {
for (int i = 0; i < n; i++) {
if ((mask & (1 << i)) != 0) {
for (int j = 0; j < i; j++) {
if ((mask & (1 << j)) != 0) {
dp[mask][i] = Math.max(dp[mask][i], dp[mask ^ (1 << i)][j] + nums[i]);
}
}
}
}
}
int maxSum = 0;
for (int i = 0; i < n; i++) {
maxSum = Math.max(maxSum, dp[(1 << n) - 1][i]);
}
return maxSum;
}
}def solution(nums):
n = len(nums)
dp = [[0] * n for _ in range(1 << n)]
for i in range(n):
dp[1 << i][i] = nums[i]
for mask in range(1 << n):
for i in range(n):
if (mask & (1 << i)) != 0:
for j in range(i):
if (mask & (1 << j)) != 0:
dp[mask][i] = max(dp[mask][i], dp[mask ^ (1 << i)][j] + nums[i])
else:
dp[mask][i] = max(dp[mask][i], dp[mask][i - 1] if i > 0 else 0)
max_sum = 0
for i in range(n):
max_sum = max(max_sum, dp[(1 << n) - 1][i])
return max_sumfunction solution(nums) {
const n = nums.length;
const dp = new Array(1 << n).fill(0).map(() => new Array(n).fill(0));
for (let i = 0; i < n; i++) {
dp[1 << i][i] = nums[i];
}
for (let mask = 0; mask < (1 << n); mask++) {
for (let i = 0; i < n; i++) {
if ((mask & (1 << i)) !== 0) {
for (let j = 0; j < i; j++) {
if ((mask & (1 << j)) !== 0) {
dp[mask][i] = Math.max(dp[mask][i], dp[mask ^ (1 << i)][j] + nums[i]);
}
}
}
}
}
let maxSum = 0;
for (let i = 0; i < n; i++) {
maxSum = Math.max(maxSum, dp[(1 << n) - 1][i]);
}
return maxSum;
}Asked in Top Tech Interviews
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