# Minimum jumps to reach last building in a matrix

• Difficulty Level : Medium
• Last Updated : 27 Jul, 2022

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

NaÃ¯ve Recursive Solution: The above problem can be solved 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 â€œminimum 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 height[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 height[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 height[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`

## 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 height[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

 ``

## Javascript

 ``

Output

`12`

The time complexity of this solution is exponential.
Auxiliary Space: O(1)

Dynamic Programming Solution: If we draw the recursion tree of the 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.

Implementation:

## 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 height[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 height[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 height[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`

## 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 height[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

 ``

## Javascript

 ``

Output

`12`

Time complexity: (R*C)
Auxiliary Space: O(1)

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