Min Cost Path | DP-6

 

Given a cost matrix cost[][] and a position (m, n) in cost[][], write a function that returns cost of minimum cost path to reach (m, n) from (0, 0). Each cell of the matrix represents a cost to traverse through that cell. The total cost of a path to reach (m, n) is the sum of all the costs on that path (including both source and destination). You can only traverse down, right and diagonally lower cells from a given cell, i.e., from a given cell (i, j), cells (i+1, j), (i, j+1), and (i+1, j+1) can be traversed. You may assume that all costs are positive integers.
For example, in the following figure, what is the minimum cost path to (2, 2)? 
 

The path with minimum cost is highlighted in the following figure. The path is (0, 0) –> (0, 1) –> (1, 2) –> (2, 2). The cost of the path is 8 (1 + 2 + 2 + 3).  

 



 

1) Optimal Substructure 
The path to reach (m, n) must be through one of the 3 cells: (m-1, n-1) or (m-1, n) or (m, n-1). So minimum cost to reach (m, n) can be written as “minimum of the 3 cells plus cost[m][n]”.
minCost(m, n) = min (minCost(m-1, n-1), minCost(m-1, n), minCost(m, n-1)) + cost[m][n]
2) Overlapping Subproblems 
Following is a simple recursive implementation of the MCP (Minimum Cost Path) problem. The implementation simply follows the recursive structure mentioned above. 
 

C

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/* A Naive recursive implementation of MCP(Minimum Cost Path) problem */
#include<stdio.h>
#include<limits.h>
#define R 3
#define C 3
 
int min(int x, int y, int z);
 
/* Returns cost of minimum cost path from (0,0) to (m, n) in mat[R][C]*/
int minCost(int cost[R][C], int m, int n)
{
   if (n < 0 || m < 0)
      return INT_MAX;
   else if (m == 0 && n == 0)
      return cost[m][n];
   else
      return cost[m][n] + min( minCost(cost, m-1, n-1),
                               minCost(cost, m-1, n),
                               minCost(cost, m, n-1) );
}
 
/* A utility function that returns minimum of 3 integers */
int min(int x, int y, int z)
{
   if (x < y)
      return (x < z)? x : z;
   else
      return (y < z)? y : z;
}
 
/* Driver program to test above functions */
int main()
{
   int cost[R][C] = { {1, 2, 3},
                      {4, 8, 2},
                      {1, 5, 3} };
   printf(" %d ", minCost(cost, 2, 2));
   return 0;
}

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Java

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/* A Naive recursive implementation of
MCP(Minimum Cost Path) problem */
public class GFG {
 
    /* A utility function that returns
    minimum of 3 integers */
    static int min(int x, int y, int z)
    {
        if (x < y)
            return (x < z) ? x : z;
        else
            return (y < z) ? y : z;
    }
     
    /* Returns cost of minimum cost path
    from (0,0) to (m, n) in mat[R][C]*/
    static int minCost(int cost[][], int m,
                                     int n)
    {
        if (n < 0 || m < 0)
            return Integer.MAX_VALUE;
        else if (m == 0 && n == 0)
            return cost[m][n];
        else
            return cost[m][n] +
                min( minCost(cost, m-1, n-1),
                     minCost(cost, m-1, n),
                     minCost(cost, m, n-1) );
    }
 
    // Driver code
    public static void main(String args[])
    {
         
        int cost[][] = { {1, 2, 3},
                         {4, 8, 2},
                         {1, 5, 3} };
                          
        System.out.print(minCost(cost, 2, 2));
    }
}
 
// This code is contributed by Sam007

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Python3

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# A Naive recursive implementation of MCP(Minimum Cost Path) problem
R = 3
C = 3
import sys
 
# Returns cost of minimum cost path from (0,0) to (m, n) in mat[R][C]
def minCost(cost, m, n):
    if (n < 0 or m < 0):
        return sys.maxsize
    elif (m == 0 and n == 0):
        return cost[m][n]
    else:
        return cost[m][n] + min( minCost(cost, m-1, n-1),
                                minCost(cost, m-1, n),
                                minCost(cost, m, n-1) )
 
#A utility function that returns minimum of 3 integers */
def min(x, y, z):
    if (x < y):
        return x if (x < z) else z
    else:
        return y if (y < z) else z
 
 
# Driver program to test above functions
cost= [ [1, 2, 3],
        [4, 8, 2],
        [1, 5, 3] ]
print(minCost(cost, 2, 2))
 
# This code is contributed by
# Smitha Dinesh Semwal

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C#

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/* A Naive recursive implementation of
MCP(Minimum Cost Path) problem */
using System;
 
class GFG
{
 
    /* A utility function that
    returns minimum of 3 integers */
    static int min(int x,
                   int y, int z)
    {
        if (x < y)
            return ((x < z) ? x : z);
        else
            return ((y < z) ? y : z);
    }
     
    /* Returns cost of minimum
    cost path from (0,0) to
    (m, n) in mat[R][C]*/
    static int minCost(int [,]cost,
                       int m , int n)
    {
        if (n < 0 || m < 0)
            return int.MaxValue;
        else if (m == 0 && n == 0)
            return cost[m, n];
        else
            return cost[m, n] +
                   min(minCost(cost, m - 1, n - 1),
                   minCost(cost, m - 1, n),
                   minCost(cost, m, n - 1) );
    }
 
    // Driver code
    public static void Main()
    {
         
        int [,]cost = {{1, 2, 3},
                       {4, 8, 2},
                       {1, 5, 3}};
                         
        Console.Write(minCost(cost, 2, 2));
    }
}
 
// This code is contributed
// by shiv_bhakt.

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PHP

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<?php
/* A Naive recursive implementation
of MCP(Minimum Cost Path) problem */
 
$R = 3;
$C = 3;
 
 
/* Returns cost of minimum 
cost path from (0,0) to
(m, n) in mat[R][C]*/
function minCost($cost, $m, $n)
{
global $R;
global $C;
if ($n < 0 || $m < 0)
    return PHP_INT_MAX;
else if ($m == 0 && $n == 0)
    return $cost[$m][$n];
else
    return $cost[$m][$n] + 
            min1(minCost($cost, $m - 1, $n - 1),
            minCost($cost, $m - 1, $n),
            minCost($cost, $m, $n - 1) );
}
 
/* A utility function that
returns minimum of 3 integers */
function min1($x, $y, $z)
{
if ($x < $y)
    return ($x < $z)? $x : $z;
else
    return ($y < $z)? $y : $z;
}
 
// Driver Code
$cost = array(array(1, 2, 3),
              array (4, 8, 2),
              array (1, 5, 3));
echo minCost($cost, 2, 2);
 
// This code is contributed by mits.
?>

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Output

 8 

It should be noted that the above function computes the same subproblems again and again. See the following recursion tree, there are many nodes which appear more than once. The time complexity of this naive recursive solution is exponential and it is terribly slow. 
 

mC refers to minCost()
                                    mC(2, 2)
                          /            |           \
                         /             |            \             
                 mC(1, 1)           mC(1, 2)             mC(2, 1)
              /     |     \       /     |     \           /     |     \ 
             /      |      \     /      |      \         /      |       \
       mC(0,0) mC(0,1) mC(1,0) mC(0,1) mC(0,2) mC(1,1) mC(1,0) mC(1,1) mC(2,0) 


So the MCP problem has both properties (see this and this) of a dynamic programming problem. Like other typical Dynamic Programming(DP) problems, recomputations of the same subproblems can be avoided by constructing a temporary array tc[][] in a bottom-up manner.
 

C++

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/* Dynamic Programming implementation of MCP problem */
#include<stdio.h>
#include<limits.h>
#define R 3
#define C 3
 
int min(int x, int y, int z);
 
int minCost(int cost[R][C], int m, int n)
{
     int i, j;
 
     // Instead of following line, we can use int tc[m+1][n+1] or
     // dynamically allocate memory to save space. The following line is
     // used to keep the program simple and make it working on all compilers.
     int tc[R][C]; 
 
     tc[0][0] = cost[0][0];
 
     /* Initialize first column of total cost(tc) array */
     for (i = 1; i <= m; i++)
        tc[i][0] = tc[i-1][0] + cost[i][0];
 
     /* Initialize first row of tc array */
     for (j = 1; j <= n; j++)
        tc[0][j] = tc[0][j-1] + cost[0][j];
 
     /* Construct rest of the tc array */
     for (i = 1; i <= m; i++)
        for (j = 1; j <= n; j++)
            tc[i][j] = min(tc[i-1][j-1],
                           tc[i-1][j],
                           tc[i][j-1]) + cost[i][j];
 
     return tc[m][n];
}
 
/* A utility function that returns minimum of 3 integers */
int min(int x, int y, int z)
{
   if (x < y)
      return (x < z)? x : z;
   else
      return (y < z)? y : z;
}
 
/* Driver program to test above functions */
int main()
{
   int cost[R][C] = { {1, 2, 3},
                      {4, 8, 2},
                      {1, 5, 3} };
   printf(" %d ", minCost(cost, 2, 2));
   return 0;
}

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Java

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/* Java program for Dynamic Programming implementation
   of Min Cost Path problem */
import java.util.*;
 
class MinimumCostPath
{
    /* A utility function that returns minimum of 3 integers */
    private static int min(int x, int y, int z)
    {
        if (x < y)
            return (x < z)? x : z;
        else
            return (y < z)? y : z;
    }
 
    private static int minCost(int cost[][], int m, int n)
    {
        int i, j;
        int tc[][]=new int[m+1][n+1];
 
        tc[0][0] = cost[0][0];
 
        /* Initialize first column of total cost(tc) array */
        for (i = 1; i <= m; i++)
            tc[i][0] = tc[i-1][0] + cost[i][0];
 
        /* Initialize first row of tc array */
        for (j = 1; j <= n; j++)
            tc[0][j] = tc[0][j-1] + cost[0][j];
 
        /* Construct rest of the tc array */
        for (i = 1; i <= m; i++)
            for (j = 1; j <= n; j++)
                tc[i][j] = min(tc[i-1][j-1],
                               tc[i-1][j],
                               tc[i][j-1]) + cost[i][j];
 
        return tc[m][n];
    }
 
    /* Driver program to test above functions */
    public static void main(String args[])
    {
        int cost[][]= {{1, 2, 3},
                       {4, 8, 2},
                       {1, 5, 3}};
        System.out.println(minCost(cost,2,2));
    }
}
// This code is contributed by Pankaj Kumar

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Python

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# Dynamic Programming Python implementation of Min Cost Path
# problem
R = 3
C = 3
 
def minCost(cost, m, n):
 
    # Instead of following line, we can use int tc[m+1][n+1] or
    # dynamically allocate memoery to save space. The following
    # line is used to keep te program simple and make it working
    # on all compilers.
    tc = [[0 for x in range(C)] for x in range(R)]
 
    tc[0][0] = cost[0][0]
 
    # Initialize first column of total cost(tc) array
    for i in range(1, m+1):
        tc[i][0] = tc[i-1][0] + cost[i][0]
 
    # Initialize first row of tc array
    for j in range(1, n+1):
        tc[0][j] = tc[0][j-1] + cost[0][j]
 
    # Construct rest of the tc array
    for i in range(1, m+1):
        for j in range(1, n+1):
            tc[i][j] = min(tc[i-1][j-1], tc[i-1][j], tc[i][j-1]) + cost[i][j]
 
    return tc[m][n]
 
# Driver program to test above functions
cost = [[1, 2, 3],
        [4, 8, 2],
        [1, 5, 3]]
print(minCost(cost, 2, 2))
 
# This code is contributed by Bhavya Jain

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C#

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// C# program for Dynamic Programming implementation
// of Min Cost Path problem
using System;
 
class GFG
{
    // A utility function that
    // returns minimum of 3 integers
    private static int min(int x, int y, int z)
    {
        if (x < y)
            return (x < z)? x : z;
        else
            return (y < z)? y : z;
    }
 
    private static int minCost(int [,]cost, int m, int n)
    {
        int i, j;
        int [,]tc=new int[m+1,n+1];
 
        tc[0,0] = cost[0,0];
 
        /* Initialize first column of total cost(tc) array */
        for (i = 1; i <= m; i++)
            tc[i, 0] = tc[i - 1, 0] + cost[i, 0];
 
        /* Initialize first row of tc array */
        for (j = 1; j <= n; j++)
            tc[0, j] = tc[0, j - 1] + cost[0, j];
 
        /* Construct rest of the tc array */
        for (i = 1; i <= m; i++)
            for (j = 1; j <= n; j++)
                tc[i, j] = min(tc[i - 1, j - 1],
                            tc[i - 1, j],
                            tc[i, j - 1]) + cost[i, j];
 
        return tc[m, n];
    }
 
    // Driver program
    public static void Main()
    {
        int [,]cost= {{1, 2, 3},
                    {4, 8, 2},
                    {1, 5, 3}};
        Console.Write(minCost(cost,2,2));
    }
}
 
// This code is contributed by Sam007.

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PHP

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<?php
// DP implementation
// of MCP problem
$R = 3;
$C = 3;
 
function minCost($cost, $m, $n)
{
    global $R;
    global $C;
    // Instead of following line,
    // we can use int tc[m+1][n+1]
    // or dynamically allocate
    // memory to save space. The
    // following line is used to keep
    // the program simple and make
    // it working on all compilers.
    $tc;
    for ($i = 0; $i <= $R; $i++)
    for ($j = 0; $j <= $C; $j++)
    $tc[$i][$j] = 0;
 
    $tc[0][0] = $cost[0][0];
 
    /* Initialize first column of
       total cost(tc) array */
    for ($i = 1; $i <= $m; $i++)
        $tc[$i][0] = $tc[$i - 1][0] +
                     $cost[$i][0];
 
    /* Initialize first
       row of tc array */
    for ($j = 1; $j <= $n; $j++)
        $tc[0][$j] = $tc[0][$j - 1] +
                     $cost[0][$j];
 
    /* Construct rest of
       the tc array */
    for ($i = 1; $i <= $m; $i++)
        for ($j = 1; $j <= $n; $j++)
         
            // returns minimum of 3 integers
            $tc[$i][$j] = min($tc[$i - 1][$j - 1],
                              $tc[$i - 1][$j],
                              $tc[$i][$j - 1]) +
                              $cost[$i][$j];
 
    return $tc[$m][$n];
}
 
 
// Driver Code
$cost = array(array(1, 2, 3),
              array(4, 8, 2),
              array(1, 5, 3));
echo minCost($cost, 2, 2);
 
// This code is contributed by mits
?>

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Output

 8 

Time Complexity of the DP implementation is O(mn) which is much better than Naive Recursive implementation.
Space Optimization: The idea is to use the same given array to store the solutions of subproblems.
 

C++

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#include <bits/stdc++.h>
#define endl "\n"
using namespace std;
 
const int row = 3;
const int col = 3;
 
int minCost(int cost[row][col]) {
 
    // for 1st column
    for (int i=1 ; i<row ; i++){
        cost[i][0] += cost[i-1][0];
    }
 
    // for 1st row
    for (int j=1 ; j<col ; j++){
        cost[0][j] += cost[0][j-1];
    }
 
    // for rest of the 2d matrix
    for (int i=1 ; i<row ; i++) {
        for (int j=1 ; j<col ; j++) {
            cost[i][j] += min(cost[i-1][j-1], min(cost[i-1][j], cost[i][j-1]));
        }
    }
 
    // returning the value in last cell
    return cost[row-1][col-1];
}
int main(int argc, char const *argv[])
{   
    int cost[row][col] = {  {1, 2, 3},
                              {4, 8, 2},
                              {1, 5, 3} };
 
    cout << minCost(cost) << endl;
    return 0;
}
 
// Contributed by Mansimar-Anand TU_2022 p_e_k_k_a Task @ Codechef/Codeforces

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Java

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// Java program for the
// above approach
import java.util.*;
class GFG{
 
static int row = 3;
static int col = 3;
 
static int minCost(int cost[][])
{
  // for 1st column
  for (int i = 1; i < row; i++)
  {
    cost[i][0] += cost[i - 1][0];
  }
 
  // for 1st row
  for (int j = 1; j < col; j++)
  {
    cost[0][j] += cost[0][j - 1];
  }
 
  // for rest of the 2d matrix
  for (int i = 1; i < row; i++)
  {
    for (int j = 1; j < col; j++)
    {
      cost[i][j] += Math.min(cost[i - 1][j - 1],
                    Math.min(cost[i - 1][j],
                             cost[i][j - 1]));
    }
  }
 
  // Returning the value in
  // last cell
  return cost[row - 1][col - 1];
}
   
// Driver code 
public static void  main(String[] args)
{    
  int cost[][] = {{1, 2, 3},
                  {4, 8, 2},
                  {1, 5, 3} };
  System.out.print(minCost(cost) + "\n");
}
}
 
// This code is contributed by Amit Katiyar

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C#

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// C# program for the
// above approach
using System;
class GFG{
 
static int row = 3;
static int col = 3;
 
static int minCost(int [,]cost)
{
  // for 1st column
  for (int i = 1; i < row; i++)
  {
    cost[i, 0] += cost[i - 1, 0];
  }
 
  // for 1st row
  for (int j = 1; j < col; j++)
  {
    cost[0, j] += cost[0, j - 1];
  }
 
  // for rest of the 2d matrix
  for (int i = 1; i < row; i++)
  {
    for (int j = 1; j < col; j++)
    {
      cost[i,j] += Math.Min(cost[i - 1,
                                 j - 1],
                   Math.Min(cost[i - 1, j],
                            cost[i, j - 1]));
    }
  }
 
  // Returning the value in
  // last cell
  return cost[row - 1, col - 1];
}
   
// Driver code 
public static void  Main(String[] args)
{    
  int [,]cost = {{1, 2, 3},
                 {4, 8, 2},
                 {1, 5, 3} };
  Console.Write(minCost(cost) + "\n");
}
}
 
// This code is contributed by Rajput-Ji

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Output

8

Alternate Solution 
We can also use the Dijkstra’s shortest path algorithm. Below is the implementation of the approach: 
 

C++

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/* Minimum Cost Path using Dijkstra’s shortest path
   algorithm with Min Heap by dinglizeng */
#include <stdio.h>
#include <queue>
#include <limits.h>
using namespace std;
 
/* define the number of rows and the number of columns */
#define R 4
#define C 5
 
/* 8 possible moves */
int dx[] = {1,-1, 0, 0, 1, 1,-1,-1};
int dy[] = {0, 0, 1,-1, 1,-1, 1,-1};
 
/* The data structure to store the coordinates of \\
  the unit square and the cost of path from the top left. */
struct Cell{
    int x;
    int y;
    int cost;
};
 
/* The compare class to be used by a Min Heap.
 * The greater than condition is used as this
   is for a Min Heap based on priority_queue.
 */
class mycomparison
{
public:
  bool operator() (const Cell &lhs, const Cell &rhs) const
  {
    return (lhs.cost > rhs.cost);
  }
};
 
/* To verify whether a move is within the boundary. */
bool isSafe(int x, int y){
    return x >= 0 && x < R && y >= 0 && y < C;
}
 
/* This solution is based on Dijkstra’s shortest path algorithm
 * For each unit square being visited, we examine all
    possible next moves in 8 directions,
 *    calculate the accumulated cost of path for each
     next move, adjust the cost of path of the adjacent
     units to the minimum as needed.
 *    then add the valid next moves into a Min Heap.
 * The Min Heap pops out the next move with the minimum
   accumulated cost of path.
 * Once the iteration reaches the last unit at the lower
   right corner, the minimum cost path will be returned.
 */
int minCost(int cost[R][C], int m, int n) {
 
    /* the array to store the accumulated cost
       of path from top left corner */
    int dp[R][C];
 
    /* the array to record whether a unit
       square has been visited */
    bool visited[R][C];
    
    /* Initialize these two arrays, set path cost
      to maximum integer value, each unit as not visited */
    for(int i = 0; i < R; i++) {
        for(int j = 0; j < C; j++) {
            dp[i][j] = INT_MAX;
            visited[i][j] = false;
        }
    }
     
    /* Define a reverse priority queue.
     * Priority queue is a heap based implementation.
     * The default behavior of a priority queue is
        to have the maximum element at the top.
     * The compare class is used in the definition of the Min Heap.
     */
    priority_queue<Cell, vector<Cell>, mycomparison> pq;
     
    /* initialize the starting top left unit with the
      cost and add it to the queue as the first move. */
    dp[0][0] = cost[0][0];
    pq.push({0, 0, cost[0][0]});
     
    while(!pq.empty()) {
 
        /* pop a move from the queue, ignore the units
           already visited */
        Cell cell = pq.top();
        pq.pop();
        int x = cell.x;
        int y = cell.y;
        if(visited[x][y]) continue;
 
        /* mark the current unit as visited */
        visited[x][y] = true;
 
        /* examine all non-visited adjacent units in 8 directions 
         * calculate the accumulated cost of path for
           each next move from this unit,
         * adjust the cost of path for each next adjacent
           units to the minimum if possible.
         */
        for(int i = 0; i < 8; i++) {
            int next_x = x + dx[i];
            int next_y = y + dy[i];
            if(isSafe(next_x, next_y) && !visited[next_x][next_y]) {
                dp[next_x][next_y] = min(dp[next_x][next_y],
                   dp[x][y] + cost[next_x][next_y]);
                pq.push({next_x, next_y, dp[next_x][next_y]});
            }
        }
    }
 
    /* return the minimum cost path at the lower
       right corner */
    return dp[m][n];       
}
 
/* Driver program to test above functions */
int main()
{
   int cost[R][C] = { {1, 8, 8, 1, 5},
                      {4, 1, 1, 8, 1},
                      {4, 2, 8, 8, 1},
                      {1, 5, 8, 8, 1} };
   printf(" %d ", minCost(cost, 3, 4));
   return 0;
}

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Output

 7 

Using a reverse priority queue in this solution can reduce the time complexity compared with a full scan looking for the node with minimum path cost. The overall Time Complexity of the DP implementation is O(mn) without consideration of priority queue in use, which is much better than Naive Recursive implementation.
 

Asked in: Amazon

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