Shortest Absolute Value Distance - Google Top Interview Questions
Problem Statement :
You are given a two-dimensional list of integers matrix.
You can move up, left, right, down and each move from matrix[a][b] to matrix[c][d] costs abs(matrix[a][b] - matrix[c][d]).
Return the minimum cost to move from the top left corner to the bottom right corner.
Constraints
1 ≤ n * m ≤ 200,000 where n and m are the number of rows and columns in matrix
Example 1
Input
matrix = [
[1, 100, 1],
[2, 5, 3],
[1, 2, 3]
]
Output
4
Explanation
We can move from 1 -> 2 -> 1 -> 2 -> 3.
Solution :
Solution in C++ :
int solve(vector<vector<int>>& mat) {
int n = mat.size(), m = mat[0].size();
vector<vector<int>> dist(n, vector<int>(m, 1e9));
set<pair<int, pair<int, int>>> s;
s.insert({0, {0, 0}});
dist[0][0] = 0;
vector<int> dir{0, -1, 0, 1, 0};
while (!s.empty()) {
auto [dis, cor] = *s.begin();
s.erase(s.begin());
auto [i, j] = cor;
for (int k = 0; k < 4; k++) {
int x = i + dir[k], y = j + dir[k + 1];
if (x >= 0 and x < n and y >= 0 and y < m and
dist[x][y] > dist[i][j] + abs(mat[x][y] - mat[i][j])) {
if (s.find({dist[x][y], {x, y}}) != s.end()) {
s.erase(s.find({dist[x][y], {x, y}}));
}
dist[x][y] = dist[i][j] + abs(mat[x][y] - mat[i][j]);
s.insert({dist[x][y], {x, y}});
}
}
}
return dist[n - 1][m - 1];
}
Solution in Java :
import java.util.*;
class Solution {
class Node {
int r, c;
long dist;
Node(int r, int c, long dist) {
this.r = r;
this.c = c;
this.dist = dist;
}
}
boolean valid(int r, int c, int m, int n) {
if (r < 0 || c < 0 || r >= m || c >= n)
return false;
return true;
}
int[] dr = {0, 0, 1, -1};
int[] dc = {1, -1, 0, 0};
public int solve(int[][] matrix) {
int m = matrix.length, n = matrix[0].length;
PriorityQueue<Node> pq = new PriorityQueue<Node>((a, b) -> (Long.compare(a.dist, b.dist)));
boolean[][] vis = new boolean[m][n];
long[][] dists = new long[m][n];
for (int i = 0; i < m; i++) Arrays.fill(dists[i], 100000000);
vis[0][0] = false;
pq.add(new Node(0, 0, 0));
dists[0][0] = 0;
while (pq.size() > 0 && !vis[m - 1][n - 1]) {
Node curnode = pq.poll();
int r = curnode.r;
int c = curnode.c;
long d = curnode.dist;
if (vis[r][c])
continue;
vis[r][c] = true;
for (int z = 0; z < 4; z++) {
int nr = r + dr[z];
int nc = c + dc[z];
if (!valid(nr, nc, m, n) || vis[nr][nc])
continue;
if (d + Math.abs(matrix[r][c] - matrix[nr][nc]) < dists[nr][nc]) {
dists[nr][nc] = d + Math.abs(matrix[r][c] - matrix[nr][nc]);
pq.add(new Node(nr, nc, dists[nr][nc]));
}
}
}
return (int) dists[m - 1][n - 1];
}
}
Solution in Python :
class Solution:
def solve(self, matrix):
m, n = len(matrix), len(matrix[0])
def neighbours(i, j):
dr = ((i + 1, j), (i - 1, j), (i, j - 1), (i, j + 1))
return [(x, y) for x, y in dr if 0 <= x < m and 0 <= y < n]
visited = [[False for j in range(n)] for i in range(m)]
distance = [[float("inf") for j in range(n)] for i in range(m)]
distance[0][0] = 0
pq = []
heapq.heapify(pq)
heapq.heappush(pq, (0, 0, 0)) # (distance,i index, j index)
while not visited[m - 1][n - 1]:
dist, i, j = heapq.heappop(pq)
if visited[i][j]:
continue
visited[i][j] = True
if i == m - 1 and j == n - 1:
break
for x, y in neighbours(i, j):
dist = distance[i][j] + abs(matrix[i][j] - matrix[x][y])
if not visited[x][y] and dist < distance[x][y]:
distance[x][y] = dist
heapq.heappush(pq, (dist, x, y))
return dist
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