# Minimum jumps to reach last building in a matrix

Given a matrix containing an integer value, In which each cell of the matrix represents height of building. Find minimum jumps needed reach from First building (0, 0) to last (n-1, m-1). Jump from a cell to next cell is absolute difference between two building heights.
Examples :

```Input :  int height[][] = {{ 5, 4, 2 },
{ 9, 2, 1 },
{ 2, 5, 9 },
{ 1, 3, 11}};
Output : 12
The minimum jump path is 5 -> 2 -> 5
-> 11. Total jumps is 3 + 3 + 6 = 12.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Recursive Solution:
The above problem can be solve easily by using recursion. 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 jump to reach (m, n) can be written as “minimum jump of the 3 cells plus current jump.

Below is the implementation of the approach.

## C++

 `// Recursive CPP program to find minimum jumps ` `// to reach last building from first. ` `#include ` `using` `namespace` `std; ` ` `  `# define R 4 ` `# define C 3 ` ` `  `bool` `isSafe(``int` `x, ``int` `y) ` `{ ` `    ``return` `(x < R && y < C); ` `} ` ` `  `/* Returns minimum jump path from (0, 0) to  ` `  ``(m, n) in hight[R][C]*/` `int` `minJump(``int` `height[R][C], ``int` `x, ``int` `y) ` `{ ` `    ``// base case ` `    ``if` `(x == R - 1 && y == C - 1) ` `        ``return` `0; ` ` `  `    ``// Find minimum jumps if we go through diagonal ` `    ``int` `diag = INT_MAX; ` `    ``if` `(isSafe(x + 1, y + 1)) ` `        ``diag = minJump(height, x + 1, y + 1) + ` `           ``abs``(height[x][y] - height[x + 1][y + 1]); ` ` `  `    ``// Find minimum jumps if we go through down ` `    ``int` `down = INT_MAX; ` `    ``if` `(isSafe(x + 1, y)) ` `        ``down = minJump(height, x + 1, y) +  ` `             ``abs``(height[x][y] - height[x + 1][y]); ` ` `  `    ``// Find minimum jumps if we go through right ` `    ``int` `right = INT_MAX; ` `    ``if` `(isSafe(x, y + 1)) ` `        ``right = minJump(height, x, y + 1) +  ` `              ``abs``(height[x][y] - height[x][y + 1]); ` ` `  `    ``// return minimum jumps ` `    ``return` `min({down, right, diag}); ` `} ` ` `  `/* Driver program to test above functions */` `int` `main() ` `{ ` `    ``int` `height[][C] = { { 5, 4, 2 }, ` `                       ``{ 9, 2, 1 }, ` `                       ``{ 2, 5, 9 }, ` `                       ``{ 1, 3, 11 } }; ` ` `  `    ``cout << minJump(height, 0, 0); ` `    ``return` `0; ` `} `

## Java

 `// Recursive Java program ` `// to find minimum jumps ` `// to reach last building  ` `// from first. ` `class` `GFG { ` `     `  `    ``static` `boolean` `isSafe(``int` `x, ``int` `y) ` `    ``{ ` `        ``return` `(x < ``4` `&& y < ``3``); ` `    ``} ` `     `  `    ``// Returns minimum jump ` `    ``// path from (0, 0) to  ` `    ``// (m, n) in hight[R][C] ` `    ``static` `int` `minJump(``int` `height[][], ``int` `x, ` `                                       ``int` `y) ` `    ``{ ` `        ``// base case ` `        ``if` `(x == ``4` `- ``1` `&& y == ``3` `- ``1``) ` `            ``return` `0``; ` `     `  `        ``// Find minimum jumps  ` `        ``// if we go through  ` `        ``// diagonal ` `        ``int` `diag = Integer.MAX_VALUE; ` `         `  `        ``if` `(isSafe(x + ``1``, y + ``1``)) ` `            ``diag = minJump(height, x + ``1``, y + ``1``) + ` `                   ``Math.abs(height[x][y] - height[x + ``1``][y + ``1``]); ` `     `  `        ``// Find minimum jumps ` `        ``// if we go through ` `        ``// down ` `        ``int` `down = Integer.MAX_VALUE; ` `         `  `        ``if` `(isSafe(x + ``1``, y)) ` `            ``down = minJump(height, x + ``1``, y) + ` `                   ``Math.abs(height[x][y] - height[x + ``1``][y]); ` `     `  `        ``// Find minimum jumps ` `        ``// if we go through right ` `        ``int` `right = Integer.MAX_VALUE; ` `         `  `        ``if` `(isSafe(x, y + ``1``)) ` `            ``right = minJump(height, x, y + ``1``) + ` `                    ``Math.abs(height[x][y] - height[x][y + ``1``]); ` `     `  `        ``// return minimum jumps ` `        ``return` `Math.min(down, Math.min(right, diag)); ` `    ``} ` `     `  `    ``// Driver program  ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `height[][] = { { ``5``, ``4``, ``2` `}, ` `                           ``{ ``9``, ``2``, ``1` `}, ` `                           ``{ ``2``, ``5``, ``9` `}, ` `                           ``{ ``1``, ``3``, ``11``} }; ` `     `  `        ``System.out.println(minJump(height, ``0``, ``0``)); ` `    ``} ` `} ` ` `  `// This article is contributed by Prerna Saini. `

## Python3

# Recursive Python3 program to find minimum jumps
# to reach last building from first.
R = 4
C = 3

def isSafe(x, y):
return (x < R and y < C) # Returns minimum jump path from # (0, 0) to (m, n) in hight[R][C] def minJump(height, x, y): # base case if (x == R - 1 and y == C - 1): return 0 # Find minimum jumps if we go # through diagonal diag = 10**9 if (isSafe(x + 1, y + 1)): diag = (minJump(height, x + 1, y + 1) + abs(height[x][y] - height[x + 1][y + 1])) # Find minimum jumps if we go through down down = 10**9 if (isSafe(x + 1, y)): down = (minJump(height, x + 1, y) + abs(height[x][y] - height[x + 1][y])) # Find minimum jumps if we go through right right = 10**9 if (isSafe(x, y + 1)): right = (minJump(height, x, y + 1) + abs(height[x][y] - height[x][y + 1])) # return minimum jumps return min([down, right, diag]) # Driver Code height = [ [ 5, 4, 2 ], [ 9, 2, 1 ], [ 2, 5, 9 ], [ 1, 3, 11 ] ] print(minJump(height, 0, 0)) # This code is contributed by mohit kumar [tabby title="C#"]

 `// Recursive C# program ` `// to find minimum jumps ` `// to reach last building  ` `// from first. ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``static` `bool` `isSafe(``int` `x, ``int` `y) ` `    ``{ ` `        ``return` `(x < 4 && y < 3); ` `    ``} ` `     `  `    ``// Returns minimum jump ` `    ``// path from (0, 0) to  ` `    ``// (m, n) in hight[R][C] ` `    ``static` `int` `minJump(``int` `[,]height,  ` `                       ``int` `x, ``int` `y) ` `    ``{ ` `         `  `        ``// base case ` `        ``if` `(x == 4 - 1 && y == 3 - 1) ` `            ``return` `0; ` `     `  `        ``// Find minimum jumps  ` `        ``// if we go through  ` `        ``// diagonal ` `        ``int` `diag = ``int``.MaxValue; ` `         `  `        ``if` `(isSafe(x + 1, y + 1)) ` `            ``diag = minJump(height, x + 1, y + 1) + ` `                   ``Math.Abs(height[x,y] -  ` `                   ``height[x + 1,y + 1]); ` `     `  `        ``// Find minimum jumps ` `        ``// if we go through ` `        ``// down ` `        ``int` `down = ``int``.MaxValue; ` `         `  `        ``if` `(isSafe(x + 1, y)) ` `            ``down = minJump(height, x + 1, y) + ` `                   ``Math.Abs(height[x,y] -  ` `                   ``height[x + 1,y]); ` `     `  `        ``// Find minimum jumps ` `        ``// if we go through right ` `        ``int` `right = ``int``.MaxValue; ` `         `  `        ``if` `(isSafe(x, y + 1)) ` `            ``right = minJump(height, x, y + 1) + ` `                    ``Math.Abs(height[x,y] -  ` `                    ``height[x,y + 1]); ` `     `  `        ``// return minimum jumps ` `        ``return` `Math.Min(down, Math.Min(right, diag)); ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `[,]height = {{5, 4, 2}, ` `                        ``{9, 2, 1}, ` `                        ``{2, 5, 9}, ` `                        ``{1, 3, 11}}; ` `     `  `        ``Console.Write(minJump(height, 0, 0)); ` `    ``} ` `} ` ` `  `// This code is contributed by nitin mittal `

## PHP

 ` `

Output :

```12
```

Time complexity of this solution is exponential.

Dynamic Programming Solution:
If we draw recursion tree of above recursive solution, we can observe overlapping subproblems. Since the problem has overlapping subproblems, we can solve it efficiently using Dynamic Programming. Below is Dynamic Programming based solution.

 `// A Dynamic Programming based CPP program to find ` `// minimum jumps to reach last building from first. ` `#include ` `using` `namespace` `std; ` ` `  `#define R 4 ` `#define C 3 ` ` `  `bool` `isSafe(``int` `x, ``int` `y) ` `{ ` `    ``return` `(x < R && y < C); ` `} ` ` `  `// Lookup table used for memoization. ` `int` `dp[R][C]; ` ` `  `/* Returns minimum jump path from (0, 0) to (m, n) ` `   ``in hight[R][C]*/` `int` `minJump(``int` `height[R][C], ``int` `x, ``int` `y) ` `{ ` `    ``// if we visited it before ` `    ``if` `(dp[x][y] != -1) ` `        ``return` `dp[x][y]; ` ` `  `    ``if` `(x == R - 1 && y == C - 1) ` `        ``return` `(dp[x][y] = 0); ` ` `  `    ``// Find minimum jumps if we go through diagonal ` `    ``int` `diag = INT_MAX; ` `    ``if` `(isSafe(x + 1, y + 1)) ` `        ``diag = minJump(height, x + 1, y + 1) + ` `           ``abs``(height[x][y] - height[x + 1][y + 1]); ` ` `  `    ``// Find minimum jumps if we go through down ` `    ``int` `down = INT_MAX; ` `    ``if` `(isSafe(x + 1, y)) ` `        ``down = minJump(height, x + 1, y) +  ` `             ``abs``(height[x][y] - height[x + 1][y]); ` ` `  `    ``// Find minimum jumps if we go through right ` `    ``int` `right = INT_MAX; ` `    ``if` `(isSafe(x, y + 1)) ` `        ``right = minJump(height, x, y + 1) +  ` `              ``abs``(height[x][y] - height[x][y + 1]); ` ` `  `    ``// return minimum jump ` `    ``dp[x][y] = min({down, right, diag}); ` `    ``return` `dp[x][y]; ` `} ` ` `  `/* Driver program to test above functions */` `int` `main() ` `{ ` `    ``int` `height[][C] = { { 5, 4, 2 }, ` `                       ``{ 9, 2, 1 }, ` `                       ``{ 2, 5, 9 }, ` `                       ``{ 1, 3, 11 } }; ` `    ``memset``(dp, -1, ``sizeof``(dp)); ` `    ``cout << minJump(height, 0, 0); ` `    ``return` `0; ` `} `

Output:

```12
```

Time complexity: (R*C)