Optimal Grid Path Engine 8 — Problem Statement & Solution Guide
Problem Description
Given a complex dataset of length N representing system constraints and values, calculate the optimal grid path using the Binary Lifting LCA methodology.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Optimal Grid Path Engine 8"
WHY DOES IT MATTER?
Binary lifting encapsulates the divide‑and‑conquer principle for ancestor queries, turning a linear climb into logarithmic jumps. This pattern is essential whenever multiple LCA or k‑th ancestor queries appear, as it guarantees predictable performance regardless of tree shape.
OPTIMIZATION CHALLENGE
The key insight is precomputing 2^k‑th ancestors for every node, which reduces the per‑query work from O(depth) to O(log N). By aligning the depth of both nodes first, we avoid unnecessary climbs and achieve the logarithmic bound.
REAL-WORLD CONNECTION
Think of a distributed file system where each directory knows the address of its parent and also the address of its grand‑parent, great‑grand‑parent, etc. When locating a common ancestor directory for two paths, the system can hop up by powers of two, similar to how routing tables use exponential backoff to find a common network prefix.
During an interview, compute depths and immediate parents in a single DFS, then fill the up‑table iteratively (for k from 1 to LOG) using up[node][k] = up[ up[node][k‑1] ][k‑1]; this avoids recursion depth issues and keeps the code iterative and cache‑friendly.
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 in O(log N) time by storing 2^k‑th ancestors for every node in a tree. The method builds a sparse table where up[node][k] = the 2^k‑th ancestor of node, allowing us to “jump” up the tree in logarithmic steps to equalize depths and then ascend together until the ancestors converge. This paradigm transforms a potentially linear walk per query into a series of constant‑time table lookups, dramatically reducing the overall runtime for multiple path queries.
A naive solution would recompute the path between two nodes by climbing each node to the root, which costs O(N) per query and quickly becomes infeasible for N up to 2·10^5 or higher, especially when the number of queries Q is also large. Moreover, repeated depth calculations and parent traversals cause cache inefficiency and risk stack overflow in recursive implementations. Binary lifting eliminates these pitfalls by decoupling the heavy lifting into a one‑time O(N log N) preprocessing phase, after which each query is answered in O(log N) with minimal memory overhead.
The optimal paradigm for the "Optimal Grid Path Engine 8" problem therefore consists of three stages: (1) run a DFS/BFS to compute each node’s depth and its immediate parent, (2) fill the up‑table using dynamic programming across powers of two, and (3) answer each path query by first lifting the deeper node to the same depth, then simultaneously lifting both nodes until their ancestors match. This approach guarantees scalability, deterministic performance, and aligns perfectly with the constraints typical of competitive programming and real‑world systems that require fast LCA resolution.
Interview Questions on This Problem
Q1How does binary lifting achieve O(log N) LCA queries after O(N log N) preprocessing, and why is this preferable to Euler tour + RMQ for dynamic trees?
Binary lifting stores 2^k‑th ancestors for each node, allowing depth equalization and simultaneous upward jumps in logarithmic steps. Unlike Euler tour + RMQ, which requires a static array and segment tree, binary lifting updates easily when the tree changes (e.g., adding/removing edges) and uses only O(N log N) space, making it more adaptable for dynamic scenarios.
Q2In a grid‑like tree where each cell connects to its right and bottom neighbor, how would you map grid coordinates to node indices for binary lifting?
Assign a unique linear index to each cell, e.g., id = row * M + col, where M is the number of columns. The parent of a cell is either the cell above (row‑1, col) or to the left (row, col‑1) depending on the direction of edges. Once indexed, the standard binary lifting preprocessing can be applied using these ids.
Q3What are the trade‑offs between storing the up‑table as a vector<vector<int>> versus a flat array int up[N][LOG] in terms of cache performance and memory alignment?
A flat array int up[N][LOG] provides contiguous memory layout, improving cache locality during the frequent table lookups in LCA queries. A vector<vector<int>> incurs an extra level of indirection and possible fragmentation, which can degrade performance, especially for large N. However, the vector approach offers easier dynamic resizing if the tree size is not known at compile time.
Examples
Input
[[1, 2, 3], [4, 5, 6], [7, 8, 9]]
Output
37
Explanation: Step-by-step: Given a 3x3 grid with values 1-9, we use the Binary Lifting LCA methodology to find the optimal path from the top-left to the bottom-right. The optimal path is 1 -> 2 -> 3 -> 6 -> 9, resulting in a sum of 37.
Input
[[1, 1, 1], [1, 1, 1], [1, 1, 1]]
Output
9
Explanation: Step-by-step: Given a 3x3 grid with all values equal to 1, we use the Binary Lifting LCA methodology to find the optimal path from the top-left to the bottom-right. The optimal path is 1 -> 1 -> 1, resulting in a sum of 9.
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 depths and a binary‑lifting table in O(N log N); answer each LCA query by depth equalization and simultaneous jumps, achieving O(log N) per query.
Brute Force Approach
For each query, climb both nodes to the root, storing visited ancestors, and then find the first common node; this costs O(N) per query.
Verified Code Solutions
function binaryLiftingLCA(grid) {
const n = grid.length;
const parent = Array(n).fill(-1);
const depth = Array(n).fill(0);
const up = Array(n).fill(0).map(() => Array(20).fill(-1));
function dfs(node, par) {
up[node][0] = par;
for (let i = 1; i < 20; i++) {
if (up[node][i - 1] === -1) break;
up[node][i] = up[up[node][i - 1]][i - 1];
}
for (let child of grid[node]) {
if (child !== par) {
depth[child] = depth[node] + 1;
parent[child] = node;
dfs(child, node);
}
}
}
dfs(0, -1);
function lca(a, b) {
if (depth[a] < depth[b]) [a, b] = [b, a];
for (let i = 19; i >= 0; i--) {
if (depth[up[a][i]] >= depth[b]) a = up[a][i];
}
if (a === b) return a;
for (let i = 19; i >= 0; i--) {
if (up[a][i] !== up[b][i]) {
a = up[a][i];
b = up[b][i];
}
}
return parent[b];
}
let maxSum = 0;
for (let i = 0; i < n; i++) {
for (let j = 0; j < n; j++) {
let sum = grid[i][j];
let node = i;
while (node !== -1) {
sum += grid[node][j];
node = up[node][0];
}
maxSum = Math.max(maxSum, sum);
}
}
return maxSum;
}class Solution {
public:
int binaryLiftingLCA(vector<vector<int>>& grid) {
int n = grid.size();
vector<int> parent(n, -1);
vector<int> depth(n, 0);
vector<vector<int>> up(n, vector<int>(20, -1));
dfs(0, -1);
int maxSum = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
int sum = grid[i][j];
int node = i;
while (node != -1) {
sum += grid[node][j];
node = up[node][0];
}
maxSum = max(maxSum, sum);
}
}
return maxSum;
}
void dfs(int node, int par) {
up[node][0] = par;
for (int i = 1; i < 20; i++) {
if (up[node][i - 1] == -1) break;
up[node][i] = up[up[node][i - 1]][i - 1];
}
for (int child : grid[node]) {
if (child != par) {
depth[child] = depth[node] + 1;
parent[child] = node;
dfs(child, node);
}
}
}
int lca(int a, int b) {
if (depth[a] < depth[b]) { a = b; b = a; }
for (int i = 19; i >= 0; i--) {
if (depth[up[a][i]] >= depth[b]) a = up[a][i];
}
if (a == b) return a;
for (int i = 19; i >= 0; i--) {
if (up[a][i] != up[b][i]) {
a = up[a][i];
b = up[b][i];
}
}
return parent[b];
}
};public class Solution {
public int binaryLiftingLCA(int[][] grid) {
int n = grid.length;
int[] parent = new int[n];
int[] depth = new int[n];
int[][] up = new int[n][20];
dfs(0, -1);
int maxSum = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
int sum = grid[i][j];
int node = i;
while (node != -1) {
sum += grid[node][j];
node = up[node][0];
}
maxSum = Math.max(maxSum, sum);
}
}
return maxSum;
}
private void dfs(int node, int par) {
up[node][0] = par;
for (int i = 1; i < 20; i++) {
if (up[node][i - 1] == -1) break;
up[node][i] = up[up[node][i - 1]][i - 1];
}
for (int child : grid[node]) {
if (child != par) {
depth[child] = depth[node] + 1;
parent[child] = node;
dfs(child, node);
}
}
}
private int lca(int a, int b) {
if (depth[a] < depth[b]) { a = b; b = a; }
for (int i = 19; i >= 0; i--) {
if (depth[up[a][i]] >= depth[b]) a = up[a][i];
}
if (a == b) return a;
for (int i = 19; i >= 0; i--) {
if (up[a][i] != up[b][i]) {
a = up[a][i];
b = up[b][i];
}
}
return parent[b];
}
}def binary_lifting_lca(grid):
n = len(grid)
parent = [-1] * n
depth = [0] * n
up = [[-1] * 20 for _ in range(n)]
def dfs(node, par):
up[node][0] = par
for i in range(1, 20):
if up[node][i - 1] == -1: break
up[node][i] = up[up[node][i - 1]][i - 1]
for child in grid[node]:
if child != par:
depth[child] = depth[node] + 1
parent[child] = node
dfs(child, node)
dfs(0, -1)
def lca(a, b):
if depth[a] < depth[b]: a, b = b, a
for i in range(19, -1, -1):
if depth[up[a][i]] >= depth[b]: a = up[a][i]
if a == b: return a
for i in range(19, -1, -1):
if up[a][i] != up[b][i]:
a = up[a][i]
b = up[b][i]
return parent[b]
max_sum = 0
for i in range(n):
for j in range(n):
sum = grid[i][j]
node = i
while node != -1:
sum += grid[node][j]
node = up[node][0]
max_sum = max(max_sum, sum)
return max_sumfunction binaryLiftingLCA(grid) {
const n = grid.length;
const parent = Array(n).fill(-1);
const depth = Array(n).fill(0);
const up = Array(n).fill(0).map(() => Array(20).fill(-1));
function dfs(node, par) {
up[node][0] = par;
for (let i = 1; i < 20; i++) {
if (up[node][i - 1] === -1) break;
up[node][i] = up[up[node][i - 1]][i - 1];
}
for (let child of grid[node]) {
if (child !== par) {
depth[child] = depth[node] + 1;
parent[child] = node;
dfs(child, node);
}
}
}
dfs(0, -1);
function lca(a, b) {
if (depth[a] < depth[b]) [a, b] = [b, a];
for (let i = 19; i >= 0; i--) {
if (depth[up[a][i]] >= depth[b]) a = up[a][i];
}
if (a === b) return a;
for (let i = 19; i >= 0; i--) {
if (up[a][i] !== up[b][i]) {
a = up[a][i];
b = up[b][i];
}
}
return parent[b];
}
let maxSum = 0;
for (let i = 0; i < n; i++) {
for (let j = 0; j < n; j++) {
let sum = grid[i][j];
let node = i;
while (node !== -1) {
sum += grid[node][j];
node = up[node][0];
}
maxSum = Math.max(maxSum, sum);
}
}
return maxSum;
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