Trail to Minimize Effort - Google Top Interview Questions
Problem Statement :
You are given a two-dimensional list of integers matrix where element represents the height of a hill.
You are currently on the top left cell and want to go to the bottom right cell.
In each move, you can go up, down, left, or right.
A path's cost is defined to the largest absolute difference of heights between any two consecutive cells in the path.
Return the minimum cost of any path.
Constraints
1 ≤ n, m ≤ 250 where n and m are the number of rows and columns in matrix
Example 1
\
Input
matrix = [
[1, 5, 3],
[2, 4, 3],
[3, 5, 3]
]
Output
2
Explanation
We can take the following path [1, 2, 4, 5, 3]. The largest absolute difference of heights between any
two consecutive cells is between 2 and 4.
Solution :
Solution in C++ :
int solve(vector<vector<int>>& matrix) {
int n = matrix.size(), m = matrix[0].size();
vector<int> dir{0, -1, 0, 1, 0};
vector<vector<int>> dist(n, vector<int>(m, INT_MAX));
set<pair<int, pair<int, int>>> s;
// setting the base case
dist[0][0] = 0;
s.insert({0, {0, 0}});
while (!s.empty()) {
auto [distance, cord] = *s.begin();
auto [xcord, ycord] = cord;
s.erase(s.begin());
for (int k = 0; k < 4; k++) {
int x = xcord + dir[k];
int y = ycord + dir[k + 1];
if (x >= 0 and x < n and y >= 0 and y < m) {
int maxd = max(distance, abs(matrix[x][y] - matrix[xcord][ycord]));
if (dist[x][y] > maxd) {
if (s.find({dist[x][y], {x, y}}) != s.end()) {
s.erase(s.find({dist[x][y], {x, y}}));
}
dist[x][y] = maxd;
s.insert({dist[x][y], {x, y}});
}
}
}
}
return dist[n - 1][m - 1];
}
Solution in Java :
import java.util.*;
class Solution {
static int M;
public int solve(int[][] matrix) {
int N = matrix.length;
M = matrix[0].length;
DisjointSetUnion dsu = new DisjointSetUnion(N * M);
ArrayList<Edge> edges = new ArrayList<Edge>();
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
if (i < N - 1)
edges.add(new Edge(i, j, 'D', Math.abs(matrix[i][j] - matrix[i + 1][j])));
if (j < M - 1)
edges.add(new Edge(i, j, 'R', Math.abs(matrix[i][j] - matrix[i][j + 1])));
}
}
Collections.sort(edges);
for (Edge e : edges) {
dsu.connect(e.a, e.b);
if (dsu.connected(0, N * M - 1))
return e.w;
}
return 0;
}
static class Edge implements Comparable<Edge> {
int a;
int b;
int w;
public Edge(int i, int j, char d, int w) {
a = i * M + j;
if (d == 'D')
b = (i + 1) * M + j;
else
b = i * M + (j + 1);
this.w = w;
}
public int compareTo(Edge e) {
return w - e.w;
}
}
static class DisjointSetUnion {
public int[] parent;
public int[] weight;
public int count;
public DisjointSetUnion(int N) {
count = N;
parent = new int[N];
for (int i = 0; i < N; i++) parent[i] = i;
weight = new int[N];
Arrays.fill(weight, 1);
}
//"find"
public int root(int p) {
if (p == parent[p])
return p;
return parent[p] = root(parent[p]);
}
//"union"
public void connect(int p, int q) {
p = root(p);
q = root(q);
if (p == q)
return;
if (weight[p] < weight[q]) {
parent[p] = q;
weight[q] += weight[p];
} else {
parent[q] = p;
weight[p] += weight[q];
}
count--;
}
public boolean connected(int p, int q) {
return root(p) == root(q);
}
}
}
Solution in Python :
class Solution:
def solve(self, A):
R, C = len(A), len(A[0])
DIRS = [[1, 0], [0, 1], [-1, 0], [0, -1]]
def bfs(cost):
q = [[0, 0]]
seen = set(tuple([0, 0]))
while q:
i, j = q.pop()
if i == R - 1 and j == C - 1:
return True
for di, dj in DIRS:
ni, nj = di + i, dj + j
if (
0 <= ni < R
and 0 <= nj < C
and abs(A[ni][nj] - A[i][j]) <= cost
and tuple([ni, nj]) not in seen
):
seen.add(tuple([ni, nj]))
q.append([ni, nj])
return False
l = 0
r = max([max(x) for x in A])
while l < r:
m = l + r >> 1
if bfs(m):
r = m
else:
l = m + 1
return l
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