Pick Up Gold in Two Locations - Amazon Top Interview Questions
Problem Statement :
You are given a two-dimensional list of integers matrix and integers row, col, erow0, ecol0, erow1, and ecol1. You are currently at matrix[row][col] and want to pick up gold that is at matrix[erow0][ecol0] and matrix[erow1][ecol1].
You can move up, down, left, and right but when you are at a cell (r, c), you incur cost matrix[r][c], although if you land at a cell more than once, you don't incur that cost again. Return the minimum cost to pick up gold at both locations.
Constraints
n, m ≤ 250 where n and m are the number of rows and columns in matrix
Example 1
Input
matrix = [
[1, 1, 1, 1],
[1, 9, 9, 9],
[1, 1, 1, 9]
]
row = 0
col = 0
erow0 = 0
ecol0 = 3
erow1 = 2
ecol1 = 2
Output
8
Explanation
We're currently at (0, 0) and want to pick up gold at both (0, 3) and (2, 2).
First we move from (0, 0) to (0, 3) by moving right 3 steps. Then we come back to (0, 0). And then
follow the 1s to (2, 2).
Example 2
Input
matrix = [
[1, 1, 1]
]
row = 0
col = 0
erow0 = 0
ecol0 = 1
erow1 = 0
ecol1 = 2
Output
3
Explanation
We go right 2 steps. Note that we still incur the cost at the starting square.
Solution :
Solution in C++ :
int r, c;
int dx[4]{-1, 0, 1, 0};
int dy[4]{0, 1, 0, -1};
void dijkstra(vector<vector<int>>& dp, vector<vector<int>>& g, int sx, int sy) {
dp[sx][sy] = g[sx][sy];
priority_queue<pair<int, pair<int, int>>> q;
q.emplace(-dp[sx][sy], make_pair(sx, sy));
while (q.size()) {
tie(sx, sy) = q.top().second;
if (dp[sx][sy] != -q.top().first) {
q.pop();
continue;
}
q.pop();
for (int k = 0; k < 4; k++) {
int nx = sx + dx[k];
int ny = sy + dy[k];
if (nx < 0 || nx >= r || ny < 0 || ny >= c) {
continue;
}
if (dp[nx][ny] > g[nx][ny] + dp[sx][sy]) {
dp[nx][ny] = g[nx][ny] + dp[sx][sy];
q.emplace(-dp[nx][ny], make_pair(nx, ny));
}
}
}
}
void init(vector<vector<int>>& dp) {
for (int i = 0; i < r; i++) {
vector<int> v(c);
fill(v.begin(), v.end(), 1e9);
dp.push_back(v);
}
}
int solve(vector<vector<int>>& matrix, int row, int col, int erow0, int ecol0, int erow1,
int ecol1) {
r = matrix.size();
c = matrix[0].size();
vector<vector<int>> adp, bdp, cdp;
init(adp);
init(bdp);
init(cdp);
long long ret = 1e18;
dijkstra(adp, matrix, row, col);
dijkstra(bdp, matrix, erow0, ecol0);
dijkstra(cdp, matrix, erow1, ecol1);
for (int i = 0; i < r; i++) {
for (int j = 0; j < c; j++) {
ret = min(ret, adp[i][j] + bdp[i][j] + cdp[i][j] - 2LL * matrix[i][j]);
}
}
return ret;
}
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[][] matrix, int row, int col, int erow0, int ecol0, int erow1, int ecol1) {
// Run dijkstra's algorithm from each end point
int[][] d1 = dijkstra(matrix, row, col);
int[][] d2 = dijkstra(matrix, erow0, ecol0);
int[][] d3 = dijkstra(matrix, erow1, ecol1);
int ans = Integer.MAX_VALUE;
int N = matrix.length;
int M = matrix[0].length;
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
// cell (i,j) is the "meeting point" of the 3 paths
int cost = d1[i][j] + d2[i][j] + d3[i][j] - 2 * matrix[i][j];
ans = Math.min(ans, cost);
}
}
return ans;
}
public int[][] dijkstra(int[][] m, int row, int col) {
int N = m.length;
int M = m[0].length;
int[][] dist = new int[N][M];
for (int i = 0; i < N; i++) {
Arrays.fill(dist[i], Integer.MAX_VALUE);
}
dist[row][col] = m[row][col];
PriorityQueue<int[]> pq = new PriorityQueue<int[]>(new Comparator<int[]>() {
@Override
public int compare(int[] arr1, int[] arr2) {
return arr1[2] - arr2[2];
}
});
int[][] dirs = {{-1, 0}, {1, 0}, {0, 1}, {0, -1}};
pq.add(new int[] {row, col, dist[row][col]});
while (!pq.isEmpty()) {
int[] cell = pq.poll();
for (int[] dir : dirs) {
int newR = cell[0] + dir[0];
int newC = cell[1] + dir[1];
if (newR < 0 || newR >= N || newC < 0 || newC >= M)
continue;
int d = dist[cell[0]][cell[1]];
if (dist[newR][newC] > d + m[newR][newC]) {
dist[newR][newC] = d + m[newR][newC];
pq.add(new int[] {newR, newC, dist[newR][newC]});
}
}
}
return dist;
}
}
Solution in Python :
class Solution:
def solve(self, matrix, row, col, erow0, ecol0, erow1, ecol1):
def is_valid(x, y):
return x >= 0 and y >= 0 and x < len(matrix) and y < len(matrix[0])
def min_cost(sx, sy):
heap = [(matrix[sx][sy], sx, sy)]
dists = [[math.inf] * len(matrix[0]) for _ in range(len(matrix))]
dists[sx][sy] = matrix[sx][sy]
while heap:
cost, x, y = heapq.heappop(heap)
for nx, ny in [(x, y - 1), (x + 1, y), (x - 1, y), (x, y + 1)]:
if is_valid(nx, ny) and matrix[nx][ny] + cost < dists[nx][ny]:
edge = matrix[nx][ny]
dists[nx][ny] = edge + cost
heapq.heappush(heap, (edge + cost, nx, ny))
return dists
res = math.inf
a, b, c = min_cost(row, col), min_cost(erow0, ecol0), min_cost(erow1, ecol1)
for i in range(len(matrix)):
for j in range(len(matrix[0])):
res = min(res, a[i][j] + b[i][j] + c[i][j] - 2 * matrix[i][j])
return res
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