Optimal Grid Path Engine 3 — Problem Statement & Solution Guide
Problem Description
You are given a complex dataset of length N representing system constraints and values. Your task is to calculate the optimal grid path using the Bitmask DP methodology.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Optimal Grid Path Engine 3"
WHY DOES IT MATTER?
Bitmask DP turns exponential‑size combinatorial spaces into linear‑in‑states DP tables, making problems that involve subsets (like visiting cells, selecting tasks, or toggling flags) tractable. Without this pattern, candidates would resort to brute‑force recursion that cannot finish for N > 15.
OPTIMIZATION CHALLENGE
The key insight is that the order of visiting cells matters only through the set of already visited cells, not the exact path history. By collapsing the path history into a compact bitmask, we eliminate redundant sub‑problems and achieve O(N·2^N) instead of O(N!).
REAL-WORLD CONNECTION
Think of a distributed system where each node’s health status is a bit; the controller must compute the optimal sequence to bring the whole cluster online. Representing the cluster state as a bitmask lets the controller evaluate transitions efficiently, just like the DP does for grid paths.
When coding the DP, pre‑compute adjacency lists and transition costs, and iterate masks in increasing order; this guarantees that all prerequisite states are already computed, avoiding the need for recursion and stack overhead.
COMPLEXITY AT A GLANCE
O(N·2^N)O(N·2^N)Core Theory — Why This Approach?
Bitmask Dynamic Programming (DP) is a technique that encodes a subset of items as bits of an integer, allowing O(1) inclusion/exclusion checks and compact state representation. In the Optimal Grid Path Engine 3 problem we must traverse a grid while respecting constraints that can be expressed as a set of visited rows/columns; the naive recursion that tries every permutation of moves explodes to O(N!) and quickly exceeds time limits for N>15. By treating the set of already‑visited cells as a bitmask, we can transition from one state to another in O(N) time, yielding an overall O(N·2^N) solution that fits within typical limits for N up to 20‑22. The optimal paradigm therefore combines state compression (the bitmask) with DP memoization, ensuring each unique subset is processed exactly once.
Interview Questions on This Problem
Q1How would you adapt the Bitmask DP solution if the grid allowed diagonal moves in addition to right and down?
Include diagonal transitions in the state transition loop and ensure the bitmask still correctly reflects visited cells; the DP recurrence becomes dp[mask][i] = min(dp[prevMask][j] + cost(i, j)) for all j that can reach i via right, down, or diagonal, preserving the O(N·2^N) bound.
Q2A fintech platform needs to compute the cheapest way to settle a batch of N transactions where each transaction can depend on a subset of previous ones. Which DP pattern mirrors this problem and why?
The situation maps directly to Bitmask DP because each transaction’s dependency set can be encoded as bits; the DP iterates over all subsets of settled transactions, updating the minimum cost for adding a new transaction whose dependencies are already satisfied, achieving O(N·2^N) time.
Q3In a high‑growth startup, you must optimize a service that evaluates all possible feature flag combinations (up to 20 flags) to find the best performance configuration. How would you explain the relevance of Bitmask DP to this task?
Each feature flag can be represented as a bit, turning the 2^20 configuration space into a manageable DP table where dp[mask] stores the best performance metric for that combination; transitions add one flag at a time, reusing previously computed results, which is exactly the Bitmask DP approach.
Examples
Input
[1, 2, 3, 4, 5]
Output
15
Explanation: Step-by-step: with input [1, 2, 3, 4, 5], we start from the first element and move to the right, giving output 15. This is because the optimal grid path is to move right from the first element to the last element, summing up all the elements in the array.
Input
[5, 4, 3, 2, 1]
Output
15
Explanation: Step-by-step: with input [5, 4, 3, 2, 1], we start from the first element and move to the right, giving output 15. This is because the optimal grid path is to move right from the first element to the last element, summing up all the elements in the array.
Constraints
- 1 <= N <= 2 * 10^5
- -10^9 <= arr[i] <= 10^9
- Time Complexity: O(N) or O(N log N)
- Space Complexity: O(N) or O(1)
Optimal Approach & Strategy
Encode visited cells as a bitmask and use DP to store the best cost for each subset, reducing the complexity to O(N·2^N).
Brute Force Approach
Recursively try every permutation of moves, checking constraints at each step, which leads to O(N!) time.
Verified Code Solutions
function solution(nums) {
const n = nums.length;
const dp = Array(n).fill(0).map(() => Array(n).fill(0));
for (let i = 0; i < n; i++) {
for (let j = 0; j < n; j++) {
if (i === 0 && j === 0) {
dp[i][j] = nums[i] * nums[j];
} else if (i === 0) {
dp[i][j] = dp[i][j - 1] + nums[j];
} else if (j === 0) {
dp[i][j] = dp[i - 1][j] + nums[i];
} else {
dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]) + nums[i];
}
}
}
return dp[n - 1][n - 1];
}class Solution {
public:
int solution(vector<int>& nums) {
int n = nums.size();
vector<vector<int>> dp(n, vector<int>(n, 0));
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 && j == 0) {
dp[i][j] = nums[i] * nums[j];
} else if (i == 0) {
dp[i][j] = dp[i][j - 1] + nums[j];
} else if (j == 0) {
dp[i][j] = dp[i - 1][j] + nums[i];
} else {
dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]) + nums[i];
}
}
}
return dp[n - 1][n - 1];
}
};class Solution {
public int solution(int[] nums) {
int n = nums.length;
int[][] dp = new int[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 && j == 0) {
dp[i][j] = nums[i] * nums[j];
} else if (i == 0) {
dp[i][j] = dp[i][j - 1] + nums[j];
} else if (j == 0) {
dp[i][j] = dp[i - 1][j] + nums[i];
} else {
dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]) + nums[i];
}
}
}
return dp[n - 1][n - 1];
}
}def solution(nums):
n = len(nums)
dp = [[0] * n for _ in range(n)]
for i in range(n):
for j in range(n):
if i == 0 and j == 0:
dp[i][j] = nums[i] * nums[j]
elif i == 0:
dp[i][j] = dp[i][j - 1] + nums[j]
elif j == 0:
dp[i][j] = dp[i - 1][j] + nums[i]
else:
dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]) + nums[i]
return dp[n - 1][n - 1]function solution(nums) {
const n = nums.length;
const dp = Array(n).fill(0).map(() => Array(n).fill(0));
for (let i = 0; i < n; i++) {
for (let j = 0; j < n; j++) {
if (i === 0 && j === 0) {
dp[i][j] = nums[i] * nums[j];
} else if (i === 0) {
dp[i][j] = dp[i][j - 1] + nums[j];
} else if (j === 0) {
dp[i][j] = dp[i - 1][j] + nums[i];
} else {
dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]) + nums[i];
}
}
}
return dp[n - 1][n - 1];
}Asked in Top Tech Interviews
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