Algorithms‎ > ‎Dynamic Programming‎ > ‎

### Min jumps

Given an array of integers where each element represents the max number of steps that can be made forward from that element. Write a function to return the minimum number of jumps to reach the end of the array (starting from the first element). If an element is 0, then cannot move through that element.

Example:

```Input: arr[] = {1, 3, 5, 8, 9, 2, 6, 7, 6, 8, 9}
Output: 3 (1-> 3 -> 8 ->9)
```

First element is 1, so can only go to 3. Second element is 3, so can make at most 3 steps eg to 5 or 8 or 9.

Method 1 (Naive Recursive Approach)
A naive approach is to start from the first element and recursively call for all the elements reachable from first element. The minimum number of jumps to reach end from first can be calculated using minimum number of jumps needed to reach end from the elements reachable from first.

minJumps(start, end) = Min ( minJumps(k, end) ) for all k reachable from start

 `#include ``#include ` `// Returns minimum number of jumps to reach arr[h] from arr[l]``int` `minJumps(``int` `arr[], ``int` `l, ``int` `h)``{``   ``// Base case: when source and destination are same``   ``if` `(h == l)``     ``return` `0;` `   ``// When nothing is reachable from the given source``   ``if` `(arr[l] == 0)``     ``return` `INT_MAX;` `   ``// Traverse through all the points reachable from arr[l]. Recursively``   ``// get the minimum number of jumps needed to reach arr[h] from these``   ``// reachable points.``   ``int` `min = INT_MAX;``   ``for` `(``int` `i = l+1; i <= h && i <= l + arr[l]; i++)``   ``{``       ``int` `jumps = minJumps(arr, i, h);``       ``if``(jumps != INT_MAX && jumps + 1 < min)``           ``min = jumps + 1;``   ``}` `   ``return` `min;``}` `// Driver program to test above function``int` `main()``{``  ``int` `arr[] = {1, 3, 6, 3, 2, 3, 6, 8, 9, 5};``  ``int` `n = ``sizeof``(arr)/``sizeof``(arr[0]);``  ``printf``(``"Minimum number of jumps to reach end is %d "``, minJumps(arr, 0, n-1));``  ``return` `0;``}`

If we trace the execution of this method, we can see that there will be overlapping subproblems. For example, minJumps(3, 9) will be called two times as arr[3] is reachable from arr[1] and arr[2]. So this problem has both properties (optimal substructure and overlapping subproblems) of Dynamic Programming.

Method 2 (Dynamic Programming)
In this method, we build a jumps[] array from left to right such that jumps[i] indicates the minimum number of jumps needed to reach arr[i] from arr[0]. Finally, we return jumps[n-1].

 `#include ``#include ` `// Returns minimum number of jumps to reach arr[n-1] from arr[0]``int` `minJumps(``int` `arr[], ``int` `n)``{``    ``int` `*jumps = ``new` `int``[n];  ``// jumps[n-1] will hold the result``    ``int` `i, j;` `    ``if` `(n == 0 || arr[0] == 0)``        ``return` `INT_MAX;` `    ``jumps[0] = 0;` `    ``// Find the minimum number of jumps to reach arr[i]``    ``// from arr[0], and assign this value to jumps[i]``    ``for` `(i = 1; i < n; i++)``    ``{``        ``jumps[i] = INT_MAX;``        ``for` `(j = 0; j < i; j++)``        ``{``            ``if` `(i <= j + arr[j] && jumps[j] != INT_MAX)``            ``{``                 ``jumps[i] = jumps[j] + 1;``                 ``break``;``            ``}``        ``}``    ``}``    ``return` `jumps[n-1];``}` `// Driver program to test above function``int` `main()``{``    ``int` `arr[]= {1, 3, 6, 1, 0, 9};``    ``int` `size=``sizeof``(arr)/``sizeof``(``int``);``    ``printf``(``"Minimum number of jumps to reach end is %d "``, minJumps(arr,size));``    ``return` `0;``}`

Thanks to paras for suggesting this method.

Time Complexity: O(n^2)

Method 3 (Dynamic Programming)
In this method, we build jumps[] array from right to left such that jumps[i] indicates the minimum number of jumps needed to reach arr[n-1] from arr[i]. Finally, we return arr[0].

 `int` `minJumps(``int` `arr[], ``int` `n)``{``    ``int` `*jumps = ``new` `int``[n];  ``// jumps[0] will hold the result``    ``int` `min;` `    ``// Minimum number of jumps needed to reach last element``    ``// from last elements itself is always 0``    ``jumps[n-1] = 0;` `    ``int` `i, j;` `    ``// Start from the second element, move from right to left``    ``// and construct the jumps[] array where jumps[i] represents``    ``// minimum number of jumps needed to reach arr[m-1] from arr[i]``    ``for` `(i = n-2; i >=0; i--)``    ``{``        ``// If arr[i] is 0 then arr[n-1] can't be reached from here``        ``if` `(arr[i] == 0)``            ``jumps[i] = INT_MAX;` `        ``// If we can direcly reach to the end point from here then``        ``// jumps[i] is 1``        ``else` `if` `(arr[i] >= n - i - 1)``            ``jumps[i] = 1;` `        ``// Otherwise, to find out the minimum number of jumps needed``        ``// to reach arr[n-1], check all the points reachable from here``        ``// and jumps[] value for those points``        ``else``        ``{``            ``min = INT_MAX;  ``// initialize min value` `            ``// following loop checks with all reachable points and``            ``// takes the minimum``            ``for` `(j = i+1; j < n && j <= arr[i] + i; j++)``            ``{``                ``if` `(min > jumps[j])``                    ``min = jumps[j];``            ``}      ` `            ``// Handle overflow ``            ``if` `(min != INT_MAX)``              ``jumps[i] = min + 1;``            ``else``              ``jumps[i] = min; ``// or INT_MAX``        ``}``    ``}` `    ``return` `jumps[0];``}`

Time Complexity: O(n^2) in worst case.