# Min Cost Path | DP-6

• Difficulty Level : Easy
• Last Updated : 28 Oct, 2022

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++

 `// A Naive recursive implementation``// of MCP(Minimum Cost Path) problem``#include ``using` `namespace` `std;` `#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 code``int` `main()``{``    ``int` `cost[R][C]``        ``= { { 1, 2, 3 }, { 4, 8, 2 }, { 1, 5, 3 } };` `    ``cout << minCost(cost, 2, 2) << endl;` `    ``return` `0;``}` `// This code is contributed by nikhilchhipa9`

## C

 `/* A Naive recursive implementation of MCP(Minimum Cost`` ``* Path) problem */``#include ``#include ``#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;``}`

## Java

 `/* 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`

## Python3

 `# A Naive recursive implementation of MCP(Minimum Cost Path) problem``import` `sys``R ``=` `3``C ``=` `3` `# 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`

## C#

 `/* 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.`

## PHP

 ``

## Javascript

 ``

Output

` 8`

Time Complexity: O( 3^(m*n))
Auxiliary Space: O(m + n), for recursive stack space.

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++

 `/* Dynamic Programming implementation of MCP problem */``#include ``#include ``#define R 3``#define C 3``using` `namespace` `std;``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 code*/``int` `main()``{``    ``int` `cost[R][C]``        ``= { { 1, 2, 3 }, { 4, 8, 2 }, { 1, 5, 3 } };``    ``cout << ``"  "` `<< minCost(cost, 2, 2);``    ``return` `0;``}` `// This code is contributed by shivanisinghss2110`

## C

 `/* Dynamic Programming implementation of MCP problem */``#include ``#include ``#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;``}`

## Java

 `/* 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`

## Python

 `# 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`

## C#

 `// 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.`

## PHP

 ``

## Javascript

 ``

Output

` 8`

Time Complexity: O(m * n), which is much better than Naive Recursive implementation.
Auxiliary Space: O(m * n)

Space Optimization: The idea is to use the same given array to store the solutions of subproblems.

## C++

 `#include ``using` `namespace` `std;` `#define row 3``#define 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;``}` `// This code is contributed by Aditya Kumar (adityakumar129)`

## C

 `#include ` `#define row 3``#define col 3` `// Find minimum between two numbers.``int` `min(``int` `num1, ``int` `num2)``{``    ``return` `(num1 > num2) ? num2 : num1;``}` `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` `cost[row][col]``        ``= { { 1, 2, 3 }, { 4, 8, 2 }, { 1, 5, 3 } };` `    ``printf``(``"%d \n"``, minCost(cost));``    ``return` `0;``}` `// This code is contributed by Aditya Kumar (adityakumar129)`

## Java

 `// 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`

## Python3

 `# Python3 program for the``# above approach`  `def` `minCost(cost, row, col):` `    ``# For 1st column``    ``for` `i ``in` `range``(``1``, row):``        ``cost[i][``0``] ``+``=` `cost[i ``-` `1``][``0``]` `    ``# For 1st row``    ``for` `j ``in` `range``(``1``, col):``        ``cost[``0``][j] ``+``=` `cost[``0``][j ``-` `1``]` `    ``# For rest of the 2d matrix``    ``for` `i ``in` `range``(``1``, row):``        ``for` `j ``in` `range``(``1``, col):``            ``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``]`  `# Driver code``if` `__name__ ``=``=` `'__main__'``:` `    ``row ``=` `3``    ``col ``=` `3` `    ``cost ``=` `[[``1``, ``2``, ``3``],``            ``[``4``, ``8``, ``2``],``            ``[``1``, ``5``, ``3``]]` `    ``print``(minCost(cost, row, col))` `# This code is contributed by Amit Katiyar`

## C#

 `// 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`

## Javascript

 ``

Output

`8`

Time Complexity: O(row * col)
Auxiliary Space: O(1), since no extra space has been taken.

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

## C++

 `/* Minimum Cost Path using Dijkstra’s shortest path``   ``algorithm with Min Heap by dinglizeng */``#include ``#include ``#include ``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, 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;``}`

## Java

 `/* Minimum Cost Path using Dijkstra’s shortest path``   ``algorithm with Min Heap by dinglizeng */``import` `java.util.*;` `public` `class` `GFG {``    ``/* define the number of rows and the number of columns``     ``*/``    ``static` `int` `R = ``4``;``    ``static` `int` `C = ``5``;` `    ``/* 8 possible moves */``    ``static` `int` `dx[] = { ``1``, -``1``, ``0``, ``0``, ``1``, ``1``, -``1``, -``1` `};``    ``static` `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. */``    ``static` `class` `Cell {``        ``int` `x;``        ``int` `y;``        ``int` `cost;``        ``Cell(``int` `x, ``int` `y, ``int` `z)``        ``{``            ``this``.x = x;``            ``this``.y = y;``            ``this``.cost = z;``        ``}``    ``}` `    ``/* To verify whether a move is within the boundary. */``    ``static` `boolean` `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.``     ``*/``    ``static` `int` `minCost(``int` `cost[][], ``int` `m, ``int` `n)``    ``{` `        ``/* the array to store the accumulated cost``           ``of path from top left corner */``        ``int``[][] dp = ``new` `int``[R][C];` `        ``/* the array to record whether a unit``           ``square has been visited */``        ``boolean``[][] visited = ``new` `boolean``[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] = Integer.MAX_VALUE;``                ``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.``         ``*/``        ``PriorityQueue pq``            ``= ``new` `PriorityQueue<>((Cell lhs, Cell rhs) -> {``                  ``return` `lhs.cost - rhs.cost;``              ``});` `        ``/* 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.add(``new` `Cell(``0``, ``0``, cost[``0``][``0``]));` `        ``while` `(!pq.isEmpty()) {` `            ``/* pop a move from the queue, ignore the units``               ``already visited */``            ``Cell cell = pq.peek();``            ``pq.remove();``            ``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] = Math.min(``                        ``dp[next_x][next_y],``                        ``dp[x][y] + cost[next_x][next_y]);``                    ``pq.add(``new` `Cell(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 */``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `cost[][] = { { ``1``, ``8``, ``8``, ``1``, ``5` `},``                         ``{ ``4``, ``1``, ``1``, ``8``, ``1` `},``                         ``{ ``4``, ``2``, ``8``, ``8``, ``1` `},``                         ``{ ``1``, ``5``, ``8``, ``8``, ``1` `} };``        ``System.out.println(minCost(cost, ``3``, ``4``));``    ``}``}``// This code is contributed by karandeep1234`

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 the priority queue in use, which is much better than Naive Recursive implementation.