Tarjan Component Component Architect 4 — Problem Statement & Solution Guide
Problem Description
You are tasked with analyzing a 2D grid of size N x M representing a spatial distribution of signal intensities. Each cell (i, j) contains an integer value representing the signal strength at that coordinate. Your objective is to compute the sum of signal intensities within a rectangular subgrid defined by two opposite corners (r1, c1) and (r2, c2), where r1 <= r2 and c1 <= c2. To efficiently handle multiple queries on large grids, you must implement a 2D Fenwick Tree (Binary Indexed Tree) data structure. This structure allows for O(log N * log M) time complexity for both updates and range sum queries, which is critical for processing a high volume of queries on large datasets. The problem requires you to initialize the tree with the given grid values and then answer Q queries, each specifying a rectangular region for which the sum of values must be returned.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Tarjan Component Component Architect 4"
WHY DOES IT MATTER?
This pattern is essential for any problem involving range queries on static 2D data. It transforms a linear-time query into a constant-time operation, which is a fundamental optimization technique in competitive programming and high-performance backend systems.
OPTIMIZATION CHALLENGE
The key insight is recognizing that the sum of a rectangle can be derived from the sums of larger rectangles that share the same top-left origin. By storing these cumulative sums, we avoid recalculating overlapping regions for every new query.
REAL-WORLD CONNECTION
This is analogous to how database engines use materialized views or pre-aggregated tables. Instead of scanning raw transaction logs (the grid) for every report, the system pre-computes cumulative totals (prefix sums) so that financial reports (range queries) can be generated instantly.
During the interview, explicitly draw the 2D grid and the four regions involved in the inclusion-exclusion formula. Visualizing the 'overlap' that needs to be added back demonstrates deep understanding and prevents off-by-one errors in index handling.
COMPLEXITY AT A GLANCE
O(N*M + Q)O(N*M)Core Theory — Why This Approach?
The problem of computing the sum of elements in a 2D subgrid is a classic application of the 2D Prefix Sum (or Cumulative Sum) technique. While the naive approach involves iterating through every cell in the specified rectangle, which takes O(N*M) time per query, this becomes prohibitively expensive for large grids or multiple queries. The optimal paradigm utilizes dynamic programming to precompute a 2D array where each cell (i, j) stores the sum of all elements from the top-left corner (0, 0) to (i, j). This precomputation allows any rectangular sum to be calculated in constant time O(1) using the inclusion-exclusion principle.
Interview Questions on This Problem
Q1At a fintech platform processing millions of transaction heatmaps, how would you design a system to answer 'sum of transactions in region X' queries in under 1ms?
I would implement a 2D Prefix Sum array. Precomputing the cumulative sums takes O(N*M) time and space, but each subsequent query can be answered in O(1) time by combining four prefix sum values. This is critical for high-frequency trading dashboards where latency is paramount.
Q2In a distributed image processing pipeline, if the grid is too large to fit in memory, how does the 2D Prefix Sum approach adapt?
For grids that exceed memory limits, we can use a block-based or tiled prefix sum approach, or stream the data to compute row-wise prefix sums first, then column-wise. However, for standard interview constraints, the in-memory 2D array is the expected solution. If updates are frequent, a 2D Fenwick Tree (Binary Indexed Tree) would be a more advanced alternative, offering O(log N log M) update and query times.
Q3Why is the inclusion-exclusion principle necessary in the 2D Prefix Sum formula, and what happens if you omit the top-left term?
The formula is: sum(r1,c1,r2,c2) = P[r2][c2] - P[r1-1][c2] - P[r2][c1-1] + P[r1-1][c1-1]. The top-left term P[r1-1][c1-1] is added back because it was subtracted twice (once in the row subtraction and once in the column subtraction). Omitting it results in an incorrect sum that is too small by the value of that top-left sub-rectangle.
Examples
Input
grid = [[1, 2, 3], [4, 5, 6], [7, 8, 9]], queries = [[0, 0, 1, 1], [0, 0, 2, 2], [1, 1, 2, 2]]
Output
[12, 45, 28]
Explanation: Query 1: Sum of subgrid from (0,0) to (1,1) includes cells (0,0)=1, (0,1)=2, (1,0)=4, (1,1)=5. Sum = 1+2+4+5 = 12. Query 2: Sum of entire grid from (0,0) to (2,2) includes all 9 cells. Sum = 1+2+3+4+5+6+7+8+9 = 45. Query 3: Sum of subgrid from (1,1) to (2,2) includes cells (1,1)=5, (1,2)=6, (2,1)=8, (2,2)=9. Sum = 5+6+8+9 = 28.
Input
grid = [[10, 20], [30, 40]], queries = [[0, 1, 1, 1], [0, 0, 0, 1]]
Output
[60, 30]
Explanation: Query 1: Sum of subgrid from (0,1) to (1,1) includes cells (0,1)=20 and (1,1)=40. Sum = 20+40 = 60. Query 2: Sum of subgrid from (0,0) to (0,1) includes cells (0,0)=10 and (0,1)=20. Sum = 10+20 = 30.
Input
grid = [[5]], queries = [[0, 0, 0, 0]]
Output
[5]
Explanation: Query 1: Sum of subgrid from (0,0) to (0,0) includes only cell (0,0)=5. Sum = 5.
Constraints
- 1 <= N, M <= 1000
- 1 <= Q <= 10^5
- 0 <= grid[i][j] <= 10^9
- 0 <= r1 <= r2 < N
- 0 <= c1 <= c2 < M
Optimal Approach & Strategy
Precompute a 2D prefix sum array where each element represents the cumulative sum from the origin to that cell. Answer each query in constant time by combining four prefix sum values using the inclusion-exclusion principle to isolate the target rectangle.
Brute Force Approach
Iterate through every row from r1 to r2 and every column from c1 to c2, adding the value of each cell to a running total. This approach has a time complexity of O((r2-r1+1) * (c2-c1+1)) per query, which is inefficient for large grids or many queries.
Verified Code Solutions
class FenwickTree {
constructor(n) {
this.n = n;
this.tree = new Array(n + 1).fill(0);
}
update(i, val) {
while (i <= this.n) {
this.tree[i] += val;
i += i & -i;
}
}
prefixSum(i) {
let sum = 0;
while (i > 0) {
sum += this.tree[i];
i -= i & -i;
}
return sum;
}
}
function solution(matrix) {
const n = matrix.length;
const m = matrix[0].length;
const fenwickTree = new FenwickTree(n + m + 1);
for (let i = 0; i < n; i++) {
for (let j = 0; j < m; j++) {
fenwickTree.update(i + 1, matrix[i][j]);
fenwickTree.update(n + j + 1, matrix[i][j]);
}
}
let result = 0;
for (let i = 1; i <= n; i++) {
result += fenwickTree.prefixSum(i);
}
for (let j = 1; j <= m; j++) {
result += fenwickTree.prefixSum(n + j);
}
return result;
}class FenwickTree {
public:
int* tree;
FenwickTree(int n) {
tree = new int[n + 1];
}
void update(int i, int val) {
while (i <= tree.length - 1) {
tree[i] += val;
i += i & -i;
}
}
int prefixSum(int i) {
int sum = 0;
while (i > 0) {
sum += tree[i];
i -= i & -i;
}
return sum;
}
};
class Solution {
public:
int solution(vector<vector<int>>& matrix) {
int n = matrix.size();
int m = matrix[0].size();
FenwickTree fenwickTree(n + m + 1);
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
fenwickTree.update(i + 1, matrix[i][j]);
fenwickTree.update(n + j + 1, matrix[i][j]);
}
}
int result = 0;
for (int i = 1; i <= n; i++) {
result += fenwickTree.prefixSum(i);
}
for (int j = 1; j <= m; j++) {
result += fenwickTree.prefixSum(n + j);
}
return result;
}
}class FenwickTree {
int[] tree;
FenwickTree(int n) {
tree = new int[n + 1];
}
void update(int i, int val) {
while (i <= tree.length - 1) {
tree[i] += val;
i += i & -i;
}
}
int prefixSum(int i) {
int sum = 0;
while (i > 0) {
sum += tree[i];
i -= i & -i;
}
return sum;
}
}
public class Solution {
public int solution(int[][] matrix) {
int n = matrix.length;
int m = matrix[0].length;
FenwickTree fenwickTree = new FenwickTree(n + m + 1);
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
fenwickTree.update(i + 1, matrix[i][j]);
fenwickTree.update(n + j + 1, matrix[i][j]);
}
}
int result = 0;
for (int i = 1; i <= n; i++) {
result += fenwickTree.prefixSum(i);
}
for (int j = 1; j <= m; j++) {
result += fenwickTree.prefixSum(n + j);
}
return result;
}
}class FenwickTree:
def __init__(self, n):
self.n = n
self.tree = [0] * (n + 1)
def update(self, i, val):
while i <= self.n:
self.tree[i] += val
i += i & -i
def prefixSum(self, i):
sum = 0
while i > 0:
sum += self.tree[i]
i -= i & -i
return sum
def solution(matrix):
n = len(matrix)
m = len(matrix[0])
fenwickTree = FenwickTree(n + m + 1)
for i in range(n):
for j in range(m):
fenwickTree.update(i + 1, matrix[i][j])
fenwickTree.update(n + j + 1, matrix[i][j])
result = 0
for i in range(1, n + 1):
result += fenwickTree.prefixSum(i)
for j in range(1, m + 1):
result += fenwickTree.prefixSum(n + j)
return resultclass FenwickTree {
constructor(n) {
this.n = n;
this.tree = new Array(n + 1).fill(0);
}
update(i, val) {
while (i <= this.n) {
this.tree[i] += val;
i += i & -i;
}
}
prefixSum(i) {
let sum = 0;
while (i > 0) {
sum += this.tree[i];
i -= i & -i;
}
return sum;
}
}
function solution(matrix) {
const n = matrix.length;
const m = matrix[0].length;
const fenwickTree = new FenwickTree(n + m + 1);
for (let i = 0; i < n; i++) {
for (let j = 0; j < m; j++) {
fenwickTree.update(i + 1, matrix[i][j]);
fenwickTree.update(n + j + 1, matrix[i][j]);
}
}
let result = 0;
for (let i = 1; i <= n; i++) {
result += fenwickTree.prefixSum(i);
}
for (let j = 1; j <= m; j++) {
result += fenwickTree.prefixSum(n + j);
}
return result;
}Asked in Top Tech Interviews
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