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
View More Similar Problems
Compare two linked lists
You’re given the pointer to the head nodes of two linked lists. Compare the data in the nodes of the linked lists to check if they are equal. If all data attributes are equal and the lists are the same length, return 1. Otherwise, return 0. Example: list1=1->2->3->Null list2=1->2->3->4->Null The two lists have equal data attributes for the first 3 nodes. list2 is longer, though, so the lis
View Solution →Merge two sorted linked lists
This challenge is part of a tutorial track by MyCodeSchool Given pointers to the heads of two sorted linked lists, merge them into a single, sorted linked list. Either head pointer may be null meaning that the corresponding list is empty. Example headA refers to 1 -> 3 -> 7 -> NULL headB refers to 1 -> 2 -> NULL The new list is 1 -> 1 -> 2 -> 3 -> 7 -> NULL. Function Description C
View Solution →Get Node Value
This challenge is part of a tutorial track by MyCodeSchool Given a pointer to the head of a linked list and a specific position, determine the data value at that position. Count backwards from the tail node. The tail is at postion 0, its parent is at 1 and so on. Example head refers to 3 -> 2 -> 1 -> 0 -> NULL positionFromTail = 2 Each of the data values matches its distance from the t
View Solution →Delete duplicate-value nodes from a sorted linked list
This challenge is part of a tutorial track by MyCodeSchool You are given the pointer to the head node of a sorted linked list, where the data in the nodes is in ascending order. Delete nodes and return a sorted list with each distinct value in the original list. The given head pointer may be null indicating that the list is empty. Example head refers to the first node in the list 1 -> 2 -
View Solution →Cycle Detection
A linked list is said to contain a cycle if any node is visited more than once while traversing the list. Given a pointer to the head of a linked list, determine if it contains a cycle. If it does, return 1. Otherwise, return 0. Example head refers 1 -> 2 -> 3 -> NUL The numbers shown are the node numbers, not their data values. There is no cycle in this list so return 0. head refer
View Solution →Find Merge Point of Two Lists
This challenge is part of a tutorial track by MyCodeSchool Given pointers to the head nodes of 2 linked lists that merge together at some point, find the node where the two lists merge. The merge point is where both lists point to the same node, i.e. they reference the same memory location. It is guaranteed that the two head nodes will be different, and neither will be NULL. If the lists share
View Solution →