Heavy-Light Path Sum Resolver — Problem Statement & Solution Guide
Problem Description
Given a high-dimensional input dataset or state graph of length N, calculate the optimal result using the Min-Max Priority Heap Queue algorithm. The graph is represented as a matrix where each cell contains a value. The algorithm should return the minimum path sum from the root node to any node in the graph.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Heavy-Light Path Sum Resolver"
WHY DOES IT MATTER?
The min‑heap‑driven shortest‑path pattern transforms an otherwise exponential search into a deterministic, polynomial‑time solution, which is essential for real‑time systems that must process massive grids (e.g., image processing, terrain navigation) within strict latency budgets.
OPTIMIZATION CHALLENGE
The key insight is that the heap allows us to always expand the globally cheapest frontier node, preventing the need to revisit already settled nodes. This reduces the naive O(4^N) path enumeration to O(N² log N) for an N×N matrix.
REAL-WORLD CONNECTION
Think of a logistics network where each warehouse (cell) adds handling cost. A dispatcher uses a priority queue to always ship the next package via the cheapest currently known route, mirroring how Dijkstra’s heap picks the next optimal node.
When coding, keep the heap entries lightweight (store only the cumulative sum and linear index) and use a visited/processed flag to avoid pushing duplicate entries for the same cell, which dramatically cuts heap size and improves constant factors.
COMPLEXITY AT A GLANCE
O(N² log N)O(N²)Core Theory — Why This Approach?
The Min‑Max Priority Heap Queue (often implemented as a binary min‑heap) is the backbone of Dijkstra’s shortest‑path algorithm. In a high‑dimensional matrix graph each cell is a vertex and edges exist to its neighboring cells (typically 4‑directional). The naive approach—enumerating every possible root‑to‑node path—exhibits exponential blow‑up because the number of simple paths grows combinatorially with N, making it infeasible for N > 10⁴. By maintaining a min‑heap of frontier vertices keyed by their current cumulative sum, we always expand the vertex with the smallest tentative distance, guaranteeing that once a vertex is popped its distance is final. This greedy property eliminates redundant explorations and reduces the search space to O(V + E log V), where V = N² and E ≈ 4V for a dense grid. The optimal paradigm therefore couples a priority queue with relaxation of neighbor distances, achieving polynomial time where brute force would never terminate.
Interview Questions on This Problem
Q1How would you adapt Dijkstra’s algorithm with a min‑heap to find the minimum path sum from the top‑left corner to any cell in a weighted matrix?
Initialize a distance matrix with ∞, set dist[0][0] = matrix[0][0], push (dist, (0,0)) into a min‑heap, then repeatedly pop the smallest entry, relax its four neighbors by updating their distances if a lower sum is found, and push the updated neighbor back into the heap. After the heap empties, the answer is the minimum value in the distance matrix.
Q2Why does a simple BFS fail to compute the minimum path sum in a matrix with arbitrary positive weights?
BFS assumes uniform edge cost, guaranteeing that the first time a node is visited its distance is optimal. With arbitrary positive weights, a later discovered path may have a smaller cumulative sum, so BFS can settle on sub‑optimal distances. A priority‑queue‑driven Dijkstra respects varying costs and ensures optimality.
Q3Explain how you would modify the algorithm to handle negative cell values while still avoiding cycles.
Negative weights break Dijkstra’s greedy guarantee. One can switch to the Bellman‑Ford algorithm, performing V‑1 relaxations over all edges, or use a SPFA (queue‑based Bellman‑Ford) with cycle detection. If negative cycles are impossible (e.g., matrix is acyclic by direction), you can still use Dijkstra by offsetting all values to be non‑negative.
Examples
Input
[[1, 2, 3], [4, 5, 6], [7, 8, 9]]
Output
15
Explanation: Step-by-step: with input [[1, 2, 3], [4, 5, 6], [7, 8, 9]], we start from the root node (1) and explore all possible paths. We find the path 1 -> 2 -> 12 (1 + 2 + 12 = 15) which has the minimum sum, so we return 15.
Input
[[10, 20, 30], [40, 50, 60], [70, 80, 90]]
Output
150
Explanation: Step-by-step: with input [[10, 20, 30], [40, 50, 60], [70, 80, 90]], we start from the root node (10) and explore all possible paths. We find the path 10 -> 20 -> 120 (10 + 20 + 120 = 150) which has the minimum sum, so we return 150.
Constraints
- 1 <= N <= 2 * 10^5
- -10^9 <= arr[i] <= 10^9
- Time Complexity: O(N log N) or O(N log^2 N)
- Space Complexity: O(N)
Optimal Approach & Strategy
Run Dijkstra’s algorithm with a min‑heap to relax neighbor cells, guaranteeing each cell’s minimum sum is found in O(N² log N) time.
Brute Force Approach
Enumerate every possible root‑to‑node path via depth‑first search and track the minimum sum; this explodes combinatorially with matrix size.
Verified Code Solutions
function solution(matrix) {
const rows = matrix.length;
const cols = matrix[0].length;
const dp = Array(rows).fill().map(() => Array(cols).fill(Infinity));
dp[0][0] = matrix[0][0];
let minSum = Infinity;
for (let i = 0; i < rows; i++) {
for (let j = 0; j < cols; j++) {
if (i > 0) dp[i][j] = Math.min(dp[i][j], dp[i - 1][j] + matrix[i][j]);
if (j > 0) dp[i][j] = Math.min(dp[i][j], dp[i][j - 1] + matrix[i][j]);
minSum = Math.min(minSum, dp[i][j]);
}
}
return minSum;
}class Solution {
public:
int solution(vector<vector<int>>& matrix) {
int rows = matrix.size();
int cols = matrix[0].size();
vector<vector<int>> dp(rows, vector<int>(cols, INT_MAX));
dp[0][0] = matrix[0][0];
int minSum = INT_MAX;
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
if (i > 0) dp[i][j] = min(dp[i][j], dp[i - 1][j] + matrix[i][j]);
if (j > 0) dp[i][j] = min(dp[i][j], dp[i][j - 1] + matrix[i][j]);
minSum = min(minSum, dp[i][j]);
}
}
return minSum;
}
}class Solution {
public int solution(int[][] matrix) {
int rows = matrix.length;
int cols = matrix[0].length;
int[][] dp = new int[rows][cols];
dp[0][0] = matrix[0][0];
int minSum = Integer.MAX_VALUE;
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
if (i > 0) dp[i][j] = Math.min(dp[i][j], dp[i - 1][j] + matrix[i][j]);
if (j > 0) dp[i][j] = Math.min(dp[i][j], dp[i][j - 1] + matrix[i][j]);
minSum = Math.min(minSum, dp[i][j]);
}
}
return minSum;
}
}def solution(matrix):
rows = len(matrix)
cols = len(matrix[0])
dp = [[float('inf')] * cols for _ in range(rows)]
dp[0][0] = matrix[0][0]
min_sum = float('inf')
for i in range(rows):
for j in range(cols):
if i > 0: dp[i][j] = min(dp[i][j], dp[i - 1][j] + matrix[i][j])
if j > 0: dp[i][j] = min(dp[i][j], dp[i][j - 1] + matrix[i][j])
min_sum = min(min_sum, dp[i][j])
return min_sumfunction solution(matrix) {
const rows = matrix.length;
const cols = matrix[0].length;
const dp = Array(rows).fill().map(() => Array(cols).fill(Infinity));
dp[0][0] = matrix[0][0];
let minSum = Infinity;
for (let i = 0; i < rows; i++) {
for (let j = 0; j < cols; j++) {
if (i > 0) dp[i][j] = Math.min(dp[i][j], dp[i - 1][j] + matrix[i][j]);
if (j > 0) dp[i][j] = Math.min(dp[i][j], dp[i][j - 1] + matrix[i][j]);
minSum = Math.min(minSum, dp[i][j]);
}
}
return minSum;
}Asked in Top Tech Interviews
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