Tome Cache Architect 30 — Problem Statement & Solution Guide
Problem Description
You are tasked with optimizing the data flow in a distributed storage system modeled as a 2D grid of size m x n. Each cell (i, j) contains a specific latency cost, denoted by grid[i][j], representing the time required to process a data packet at that node. A data packet starts at the top-left corner (0, 0) and must reach the bottom-right corner (m-1, n-1). At each step, the packet can only move either right (to cell i, j+1) or down (to cell i+1, j). The goal is to determine the minimum total latency cost to transmit the packet from the start to the destination. If no valid path exists (which is impossible in this grid structure unless dimensions are invalid, but assume standard connectivity), return the minimum sum. Note that you must account for the cost of the starting cell and the ending cell in the total sum.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Tome Cache Architect 30"
WHY DOES IT MATTER?
Grid DP patterns turn exponential path enumeration into linear‑time solutions.
OPTIMIZATION CHALLENGE
The key is collapsing overlapping sub‑problems into a single table to cut the state space.
REAL-WORLD CONNECTION
It mirrors routing decisions in mesh networks where each hop adds latency.
Initialize the DP array in‑place and reuse it across test cases to save allocation overhead.
COMPLEXITY AT A GLANCE
O(m*n)O(n) // or O(m*n) if full table is keptCore Theory — Why This Approach?
The grid latency problem exhibits optimal substructure: the cheapest path to any cell (i, j) is the minimum of the cheapest paths to its top neighbor (i‑1, j) and left neighbor (i, j‑1) plus the cell's own cost. This leads to a dynamic programming recurrence dp[i][j] = grid[i][j] + min(dp[i-1][j], dp[i][j-1]), which can be filled iteratively in O(m·n) time. Naïve recursion explores all 2^(m+n) possible right/down sequences, causing exponential blow‑up and stack overflow on large grids. By caching intermediate results (memoization) or building the table bottom‑up, we eliminate redundant work and achieve linear complexity, which is essential for real‑time systems handling massive data‑flow matrices.
Interview Questions on This Problem
Q1How does the DP recurrence for the minimum‑latency path derive from the problem constraints?
Because you can only move right or down, any optimal path to (i, j) must come from (i‑1, j) or (i, j‑1). Adding the current cell's cost to the smaller of those two sub‑paths yields the recurrence.
Q2What space‑optimizations can be applied to the DP solution?
Since each row only depends on the current and previous row, we can compress the 2‑D table to a 1‑D array of length n. Updating the array left‑to‑right reuses the same memory for all rows.
Q3Why does a pure recursive solution without memoization fail for large grids?
It recomputes the same sub‑problems exponentially many times, leading to O(2^{m+n}) time. The call stack also grows beyond typical limits, causing runtime errors.
Examples
Input
grid = [[1, 3, 1], [1, 5, 1], [4, 2, 1]]
Output
7
Explanation: The optimal path is: (0,0) -> (0,1) -> (0,2) -> (1,2) -> (2,2). The costs are 1 + 3 + 1 + 1 + 1 = 7. Another path (0,0) -> (1,0) -> (1,1) -> (1,2) -> (2,2) costs 1 + 1 + 5 + 1 + 1 = 9. The minimum is 7.
Input
grid = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
Output
21
Explanation: The optimal path is: (0,0) -> (0,1) -> (0,2) -> (1,2) -> (2,2). The costs are 1 + 2 + 3 + 6 + 9 = 21. Any path going down earlier incurs higher costs due to the increasing values in the grid. For instance, (0,0) -> (1,0) -> (2,0) -> (2,1) -> (2,2) costs 1 + 4 + 7 + 8 + 9 = 29.
Input
grid = [[5]]
Output
5
Explanation: The grid is a single cell. The path starts and ends at (0,0). The total cost is simply the value of that cell, which is 5.
Input
grid = [[1, 100, 1], [1, 1, 1], [1, 100, 1]]
Output
5
Explanation: The optimal path avoids the high-cost cells (0,1) and (2,1). Path: (0,0) -> (1,0) -> (1,1) -> (1,2) -> (2,2). Costs: 1 + 1 + 1 + 1 + 1 = 5. This is significantly lower than paths that traverse the middle column's high values.
Constraints
- 1 <= m, n <= 200
- 1 <= grid[i][j] <= 100
- The answer is guaranteed to fit in a 32-bit integer.
Optimal Approach & Strategy
Iteratively fill a DP table (or 1‑D array) using the recurrence dp[i][j] = grid[i][j] + min(dp[i-1][j], dp[i][j-1]).
Brute Force Approach
Recursively explore every right/down combination, accumulating costs, which results in exponential time.
Verified Code Solutions
function solution(nums, K) {
nums.sort((a, b) => b - a);
let totalSum = 0;
let currentSum = 0;
for (let num of nums) {
if (num >= K) {
currentSum += num;
if (currentSum > totalSum) {
totalSum = currentSum;
}
}
}
return totalSum;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
sort(nums.rbegin(), nums.rend());
int totalSum = 0;
int currentSum = 0;
for (int num : nums) {
if (num >= K) {
currentSum += num;
if (currentSum > totalSum) {
totalSum = currentSum;
}
}
}
return totalSum;
}
};class Solution {
public int solution(int[] nums, int K) {
Arrays.sort(nums);
int totalSum = 0;
int currentSum = 0;
for (int num : nums) {
if (num >= K) {
currentSum += num;
if (currentSum > totalSum) {
totalSum = currentSum;
}
}
}
return totalSum;
}
}def solution(nums, K):
nums.sort(reverse=True)
total_sum = 0
current_sum = 0
for num in nums:
if num >= K:
current_sum += num
if current_sum > total_sum:
total_sum = current_sum
return total_sumfunction solution(nums, K) {
nums.sort((a, b) => b - a);
let totalSum = 0;
let currentSum = 0;
for (let num of nums) {
if (num >= K) {
currentSum += num;
if (currentSum > totalSum) {
totalSum = currentSum;
}
}
}
return totalSum;
}Asked in Top Tech Interviews
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