Given a two-dimensional grid, each cell of which contains an integer cost which represents a cost to traverse through that cell, we need to find a path from the top left cell to the bottom right cell by which the total cost incurred is minimum.
Note: It is assumed that negative cost cycles do not exist in input matrix.
This problem is an extension of problem: Min Cost Path with right and bottom moves allowed.
In the previous problem only going right and the bottom was allowed but in this problem, we are allowed to go bottom, up, right and left i.e. in all 4 directions.
Examples:
A cost grid is given in below diagram, minimum cost to reach bottom right from top left is 327 (= 31 + 10 + 13 + 47 + 65 + 12 + 18 + 6 + 33 + 11 + 20 + 41 + 20) The chosen least cost path is shown in green.
It is not possible to solve this problem using dynamic programming similar to the previous problem because here current state depends not only on the right and bottom cells but also on the left and upper cells. We solve this problem using dijkstra’s algorithm. Each cell of the grid represents a vertex and neighbor cells adjacent vertices. We do not make an explicit graph from these cells instead we will use the matrix as it is in our Dijkstra’s algorithm.
In the below code, Dijkstra’s algorithm’s implementation is used. The code implemented below is changed to cope with matrix represented implicit graph. Please also see use of dx and dy arrays in the below code, these arrays are taken for simplifying the process of visiting neighbor vertices of each cell.
Below is the implementation of the above approach:
C++
// C++ program to get least cost path in a grid from// top-left to bottom-right#include <bits/stdc++.h>using namespace std;#define ROW 5#define COL 5// structure for information of each cellstruct cell { int x, y; int distance; cell(int x, int y, int distance) : x(x) , y(y) , distance(distance) { }};// Utility method for comparing two cellsbool operator<(const cell& a, const cell& b){ if (a.distance == b.distance) { if (a.x != b.x) return (a.x < b.x); else return (a.y < b.y); } return (a.distance < b.distance);}// Utility method to check whether a point is// inside the grid or notbool isInsideGrid(int i, int j){ return (i >= 0 && i < ROW && j >= 0 && j < COL);}// Method returns minimum cost to reach bottom// right from top leftint shortest(int grid[ROW][COL], int row, int col){ int dis[row][col]; // initializing distance array by INT_MAX for (int i = 0; i < row; i++) for (int j = 0; j < col; j++) dis[i][j] = INT_MAX; // direction arrays for simplification of getting // neighbour int dx[] = { -1, 0, 1, 0 }; int dy[] = { 0, 1, 0, -1 }; set<cell> st; // insert (0, 0) cell with 0 distance st.insert(cell(0, 0, 0)); // initialize distance of (0, 0) with its grid value dis[0][0] = grid[0][0]; // loop for standard dijkstra's algorithm while (!st.empty()) { // get the cell with minimum distance and delete // it from the set cell k = *st.begin(); st.erase(st.begin()); // looping through all neighbours for (int i = 0; i < 4; i++) { int x = k.x + dx[i]; int y = k.y + dy[i]; // if not inside boundary, ignore them if (!isInsideGrid(x, y)) continue; // If distance from current cell is smaller, // then update distance of neighbour cell if (dis[x][y] > dis[k.x][k.y] + grid[x][y]) { // If cell is already there in set, then // remove its previous entry if (dis[x][y] != INT_MAX) st.erase( st.find(cell(x, y, dis[x][y]))); // update the distance and insert new // updated cell in set dis[x][y] = dis[k.x][k.y] + grid[x][y]; st.insert(cell(x, y, dis[x][y])); } } } // uncomment below code to print distance // of each cell from (0, 0) /* for (int i = 0; i < row; i++, cout << endl) for (int j = 0; j < col; j++) cout << dis[i][j] << " "; */ // dis[row - 1][col - 1] will represent final // distance of bottom right cell from top left cell return dis[row - 1][col - 1];}// Driver code to test above methodsint main(){ int grid[ROW][COL] = { 31, 100, 65, 12, 18, 10, 13, 47, 157, 6, 100, 113, 174, 11, 33, 88, 124, 41, 20, 140, 99, 32, 111, 41, 20 }; cout << shortest(grid, ROW, COL) << endl; return 0;} |
Java
// Java program to get least cost path // in a grid from top-left to bottom-rightimport java.io.*;import java.util.*;class GFG{ static int[] dx = { -1, 0, 1, 0 };static int[] dy = { 0, 1, 0, -1 };static int ROW = 5;static int COL = 5;// Custom class for representing// row-index, column-index &// distance of each cellstatic class Cell{ int x; int y; int distance; Cell(int x, int y, int distance) { this.x = x; this.y = y; this.distance = distance; }}// Custom comparator for inserting cells // into Priority Queuestatic class distanceComparator implements Comparator<Cell>{ public int compare(Cell a, Cell b) { if (a.distance < b.distance) { return -1; } else if (a.distance > b.distance) { return 1; } else {return 0;} }}// Utility method to check whether current// cell is inside grid or notstatic boolean isInsideGrid(int i, int j){ return (i >= 0 && i < ROW && j >= 0 && j < COL);}// Method to return shortest path from // top-corner to bottom-corner in 2D gridstatic int shortestPath(int[][] grid, int row, int col){ int[][] dist = new int[row][col]; // Initializing distance array by INT_MAX for(int i = 0; i < row; i++) { for(int j = 0; j < col; j++) { dist[i][j] = Integer.MAX_VALUE; } } // Initialized source distance as // initial grid position value dist[0][0] = grid[0][0]; PriorityQueue<Cell> pq = new PriorityQueue<Cell>( row * col, new distanceComparator()); // Insert source cell to priority queue pq.add(new Cell(0, 0, dist[0][0])); while (!pq.isEmpty()) { Cell curr = pq.poll(); for(int i = 0; i < 4; i++) { int rows = curr.x + dx[i]; int cols = curr.y + dy[i]; if (isInsideGrid(rows, cols)) { if (dist[rows][cols] > dist[curr.x][curr.y] + grid[rows][cols]) { // If Cell is already been reached once, // remove it from priority queue if (dist[rows][cols] != Integer.MAX_VALUE) { Cell adj = new Cell(rows, cols, dist[rows][cols]); pq.remove(adj); } // Insert cell with updated distance dist[rows][cols] = dist[curr.x][curr.y] + grid[rows][cols]; pq.add(new Cell(rows, cols, dist[rows][cols])); } } } } return dist[row - 1][col - 1];}// Driver codepublic static void main(String[] args) throws IOException{ int[][] grid = { { 31, 100, 65, 12, 18 }, { 10, 13, 47, 157, 6 }, { 100, 113, 174, 11, 33 }, { 88, 124, 41, 20, 140 }, { 99, 32, 111, 41, 20 } }; System.out.println(shortestPath(grid, ROW, COL));}}// This code is contributed by jigyansu |
Python3
# Python program to get least cost path in a grid from# top-left to bottom-rightfrom functools import cmp_to_keyROW = 5COL = 5def mycmp(a,b): if (a.distance == b.distance): if (a.x != b.x): return (a.x - b.x) else: return (a.y - b.y) return (a.distance - b.distance)# structure for information of each cellclass cell: def __init__(self,x, y, distance): self.x = x self.y = y self.distance = distance# Utility method to check whether a point is# inside the grid or notdef isInsideGrid(i, j): return (i >= 0 and i < ROW and j >= 0 and j < COL)# Method returns minimum cost to reach bottom# right from top leftdef shortest(grid, row, col): dis = [[0 for i in range(col)]for j in range(row)] # initializing distance array by INT_MAX for i in range(row): for j in range(col): dis[i][j] = 1000000000 # direction arrays for simplification of getting # neighbour dx = [-1, 0, 1, 0] dy = [0, 1, 0, -1] st = [] # insert (0, 0) cell with 0 distance st.append(cell(0, 0, 0)) # initialize distance of (0, 0) with its grid value dis[0][0] = grid[0][0] # loop for standard dijkstra's algorithm while (len(st)!=0): # get the cell with minimum distance and delete # it from the set k = st[0] st = st[1:] # looping through all neighbours for i in range(4): x = k.x + dx[i] y = k.y + dy[i] # if not inside boundary, ignore them if (isInsideGrid(x, y) == 0): continue # If distance from current cell is smaller, then # update distance of neighbour cell if (dis[x][y] > dis[k.x][k.y] + grid[x][y]): # update the distance and insert new updated # cell in set dis[x][y] = dis[k.x][k.y] + grid[x][y] st.append(cell(x, y, dis[x][y])) st.sort(key=cmp_to_key(mycmp)) # uncomment below code to print distance # of each cell from (0, 0) # for i in range(row): # for j in range(col): # print(dis[i][j] ,end= " ") # print() # dis[row - 1][col - 1] will represent final # distance of bottom right cell from top left cell return dis[row - 1][col - 1]# Driver code to test above methodsgrid = [[31, 100, 65, 12, 18], [10, 13, 47, 157, 6], [100, 113, 174, 11, 33], [88, 124, 41, 20, 140],[99, 32, 111, 41, 20]]print(shortest(grid, ROW, COL))# This code is contributed by shinjanpatra |
C#
using System;using System.Collections.Generic;class GFG { static int[] dx = { -1, 0, 1, 0 }; static int[] dy = { 0, 1, 0, -1 }; static int ROW = 5; static int COL = 5; // Custom class for representing // row-index, column-index & // distance of each cell class Cell { public int x; public int y; public int distance; public Cell(int x, int y, int distance) { this.x = x; this.y = y; this.distance = distance; } } // Custom comparator for sorting cells // in ascending order of distance class DistanceComparer : IComparer<Cell> { public int Compare(Cell a, Cell b) { if (a.distance < b.distance) { return -1; } else if (a.distance > b.distance) { return 1; } else { return 0; } } } // Utility method to check whether current // cell is inside grid or not static bool IsInsideGrid(int i, int j) { return (i >= 0 && i < ROW && j >= 0 && j < COL); } // Method to return shortest path from // top-corner to bottom-corner in 2D grid static int ShortestPath(int[][] grid, int row, int col) { int[][] dist = new int[row][]; for (int i = 0; i < row; i++) { dist[i] = new int[col]; for (int j = 0; j < col; j++) { dist[i][j] = int.MaxValue; } } // Initialized source distance as // initial grid position value dist[0][0] = grid[0][0]; List<Cell> pq = new List<Cell>(); pq.Add(new Cell(0, 0, dist[0][0])); pq.Sort(new DistanceComparer()); while (pq.Count > 0) { Cell curr = pq[0]; pq.RemoveAt(0); for (int i = 0; i < 4; i++) { int rows = curr.x + dx[i]; int cols = curr.y + dy[i]; if (IsInsideGrid(rows, cols)) { if (dist[rows][cols] > dist[curr.x][curr.y] + grid[rows][cols]) { // If Cell is already been reached // once, remove it from list if (dist[rows][cols] != int.MaxValue) { Cell adj = new Cell( rows, cols, dist[rows][cols]); pq.Remove(adj); } // Insert cell with updated distance dist[rows][cols] = dist[curr.x][curr.y] + grid[rows][cols]; pq.Add(new Cell(rows, cols, dist[rows][cols])); pq.Sort(new DistanceComparer()); } } } } return dist[row - 1][col - 1]; } // Driver code static void Main(string[] args) { int[][] grid = { new int[] { 31, 100, 65, 12, 18 }, new int[] { 10, 13, 47, 157, 6 }, new int[] { 100, 113, 174, 11, 33 }, new int[] { 88, 124, 41, 20, 140 }, new int[] { 99, 32, 111, 41, 20 } }; Console.WriteLine(ShortestPath(grid, ROW, COL)); }}// This code is contributed by phasing17 |
Javascript
<script> // Javascript program to get least cost path in a grid from// top-left to bottom-rightvar ROW = 5var COL = 5// structure for information of each cellclass cell{ constructor(x, y, distance) { this.x = x; this.y = y; this.distance = distance; }};// Utility method to check whether a point is// inside the grid or notfunction isInsideGrid(i, j){ return (i >= 0 && i < ROW && j >= 0 && j < COL);}// Method returns minimum cost to reach bottom// right from top leftfunction shortest(grid, row, col){ var dis = Array.from(Array(row), ()=>Array(col).fill(0)); // initializing distance array by INT_MAX for (var i = 0; i < row; i++) for (var j = 0; j < col; j++) dis[i][j] = 1000000000; // direction arrays for simplification of getting // neighbour var dx = [-1, 0, 1, 0]; var dy = [0, 1, 0, -1]; var st = []; // insert (0, 0) cell with 0 distance st.push(new cell(0, 0, 0)); // initialize distance of (0, 0) with its grid value dis[0][0] = grid[0][0]; // loop for standard dijkstra's algorithm while (st.length!=0) { // get the cell with minimum distance and delete // it from the set var k = st[0]; st.shift(); // looping through all neighbours for (var i = 0; i < 4; i++) { var x = k.x + dx[i]; var y = k.y + dy[i]; // if not inside boundary, ignore them if (!isInsideGrid(x, y)) continue; // If distance from current cell is smaller, then // update distance of neighbour cell if (dis[x][y] > dis[k.x][k.y] + grid[x][y]) { // update the distance and insert new updated // cell in set dis[x][y] = dis[k.x][k.y] + grid[x][y]; st.push(new cell(x, y, dis[x][y])); } } st.sort((a,b)=>{ if (a.distance == b.distance) { if (a.x != b.x) return (a.x - b.x); else return (a.y - b.y); } return (a.distance - b.distance); }); } // uncomment below code to print distance // of each cell from (0, 0) /* for (int i = 0; i < row; i++, cout << endl) for (int j = 0; j < col; j++) cout << dis[i][j] << " "; */ // dis[row - 1][col - 1] will represent final // distance of bottom right cell from top left cell return dis[row - 1][col - 1];}// Driver code to test above methodsvar grid =[ [31, 100, 65, 12, 18], [10, 13, 47, 157, 6], [100, 113, 174, 11, 33], [88, 124, 41, 20, 140], [99, 32, 111, 41, 20]];document.write(shortest(grid, ROW, COL));// This code is contributed by rutvik_56.</script> |
Output
327
Time Complexity: O(N2 log N), The Dijkstra’s algorithm used in the program has a time complexity of O(E log V) where E is the number of edges and V is the number of vertices. In this program, for each cell, we check its four neighbors, so the number of edges E in the worst case would be 4 times the number of cells (N^2) in the grid. Therefore, the time complexity of the program is O(4N^2logN^2), which simplifies to O(N^2 logN).
Auxiliary Space: O(N2), In this program, we are using a 2D array to store the distances from the top-left cell to each cell in the grid. The size of this array is N^2. We are also using a set to store the cells with their distances. In the worst case, all cells in the grid could be present in the set, so the size of the set could also be N^2. Therefore, the space complexity of the program is O(N^2).
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