Bitmask Subset Energy Calculator 7 — 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 bitmask subset energy using the **Binary Lifting LCA** methodology.
Ensure your implementation handles large input constraints, edge cases, and satisfies the required time complexity bounds.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Bitmask Subset Energy Calculator 7"
WHY DOES IT MATTER?
Binary lifting abstracts the tree into a sparse table of ancestors, turning a linear‑time ancestor walk into a logarithmic jump sequence. This pattern is the backbone of many tree‑query problems, from LCA to distance, k‑th ancestor, and path aggregates, making it indispensable for scaling to large N and Q.
OPTIMIZATION CHALLENGE
The key insight is to pre‑compute 2^k ancestors for every node once, then reuse them for every query. This transforms repeated O(depth) traversals into O(log N) jumps, dramatically cutting both time and memory footprints compared to storing full paths or recomputing on the fly.
REAL-WORLD CONNECTION
Think of a corporate hierarchy where each employee knows their direct manager and also a shortcut to the manager 2, 4, 8 levels up. When you need to find a common supervisor of two employees, you can quickly climb using these shortcuts instead of walking one level at a time, mirroring how distributed systems use routing tables to hop across nodes efficiently.
During an interview, build the lift table first, verify it with a simple LCA test, then layer the path‑aggregate logic (e.g., prefix XOR). Keep the code modular: a function for building, one for LCA, and one for answering queries – this reduces bugs and shows clean engineering discipline.
COMPLEXITY AT A GLANCE
O(N log N + Q log N)O(N log N)Core Theory — Why This Approach?
Binary lifting is a preprocessing technique that enables answering Lowest Common Ancestor (LCA) queries on a rooted tree in O(log N) time after an O(N log N) setup. The method builds a jump table where up[v][k] stores the 2^k‑th ancestor of node v, allowing us to lift any node up by powers of two until both nodes reside at the same depth, then simultaneously lift them until their ancestors converge. This approach is essential for problems that require repeated ancestor or path queries, such as calculating bitmask subsets along tree paths, because it reduces the per‑query cost from linear to logarithmic.
A naive solution would traverse the path between two nodes for each query, aggregating the bitmask values one edge at a time. For N up to 2·10^5 and Q up to 2·10^5, this O(N·Q) behavior is infeasible, leading to time‑outs and excessive memory usage. By contrast, binary lifting decouples the query work from the size of the tree, leveraging pre‑computed ancestors to jump over large segments of the path instantly. When combined with prefix‑xor or prefix‑or bitmask accumulation during the DFS, the final answer for any pair can be derived in O(log N) without revisiting the entire path.
Interview Questions on This Problem
Q1How does binary lifting enable O(log N) LCA queries, and what is the role of the 2^k ancestor table?
Binary lifting stores for each node v the ancestor up[v][k] that is 2^k steps above v. To equalize depths, we lift the deeper node using the highest powers of two that fit the depth difference. Then we lift both nodes together, checking from the highest k downwards; the first k where up[u][k] != up[v][k] indicates that the LCA is one level above those ancestors. This reduces the number of steps to the number of bits in N, i.e., O(log N).
Q2In a tree where each node holds a bitmask, how can you compute the XOR of masks on the path between two nodes using binary lifting?
During the initial DFS, compute pref[v] as the XOR of masks from the root to v. For any query (u, v), the path XOR equals pref[u] XOR pref[v] XOR mask[lca(u,v)], because the mask of the LCA is counted twice and must be added back once. The LCA is obtained in O(log N) via binary lifting, making the whole query O(log N).
Q3Why might a recursive DFS for building the lift table cause stack overflow on deep trees, and how can you mitigate it?
Recursive DFS relies on the call stack, which is limited (often ~10^5 frames). On a degenerate tree (a line) with N = 2·10^5, recursion can exceed this limit and crash. Mitigation strategies include converting the DFS to an explicit stack (iterative) or increasing the recursion limit (in languages that allow it), but the safest is an iterative traversal.
Examples
Input
[[1, 2, 3], [4, 5, 6], [7, 8, 9]] 101
Output
4
Explanation: Step-by-step: Given the input array [[1, 2, 3], [4, 5, 6], [7, 8, 9]] and the bitmask 101, we first find the indices corresponding to the set bits in the bitmask, which are 2 and 0. Then, we calculate the sum of the values at these indices, which are 3 and 1 respectively. Therefore, the output is 4.
Input
[[10, 20, 30], [40, 50, 60], [70, 80, 90]] 110
Output
80
Explanation: Step-by-step: Given the input array [[10, 20, 30], [40, 50, 60], [70, 80, 90]] and the bitmask 110, we first find the indices corresponding to the set bits in the bitmask, which are 2 and 1. Then, we calculate the sum of the values at these indices, which are 30 and 50 respectively. Therefore, the output is 80.
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
Preprocess binary lifting tables and prefix bitmask values, then answer each query by finding the LCA in O(log N) and using the XOR formula to get the path mask in O(1) extra work.
Brute Force Approach
Traverse the entire path between the two queried nodes, aggregating the bitmask at each step; repeat this for every query.
Verified Code Solutions
function solution(nums, bitmask) {
const n = nums.length;
const m = nums[0].length;
const dp = Array(n).fill(0).map(() => Array(m).fill(0));
for (let i = 0; i < n; i++) {
for (let j = 0; j < m; j++) {
dp[i][j] = nums[i][j];
}
}
let result = 0;
for (let i = 0; i < n; i++) {
for (let j = 0; j < m; j++) {
if ((bitmask & (1 << i)) !== 0) {
result += dp[i][j];
}
}
}
return result;
}class Solution {
public:
int solution(vector<vector<int>>& nums, int bitmask) {
int n = nums.size();
int m = nums[0].size();
vector<vector<int>> dp(n, vector<int>(m));
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
dp[i][j] = nums[i][j];
}
}
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if ((bitmask & (1 << i)) != 0) {
result += dp[i][j];
}
}
}
return result;
}
};class Solution {
public int solution(int[][] nums, int bitmask) {
int n = nums.length;
int m = nums[0].length;
int[][] dp = new int[n][m];
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
dp[i][j] = nums[i][j];
}
}
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if ((bitmask & (1 << i)) != 0) {
result += dp[i][j];
}
}
}
return result;
}
}def solution(nums, bitmask):
n = len(nums)
m = len(nums[0])
dp = [[num for num in row] for row in nums]
result = 0
for i in range(n):
for j in range(m):
if (bitmask & (1 << i)) != 0:
result += dp[i][j]
return resultfunction solution(nums, bitmask) {
const n = nums.length;
const m = nums[0].length;
const dp = Array(n).fill(0).map(() => Array(m).fill(0));
for (let i = 0; i < n; i++) {
for (let j = 0; j < m; j++) {
dp[i][j] = nums[i][j];
}
}
let result = 0;
for (let i = 0; i < n; i++) {
for (let j = 0; j < m; j++) {
if ((bitmask & (1 << i)) !== 0) {
result += dp[i][j];
}
}
}
return result;
}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.