# C Program for Minimum number of jumps to reach end

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.

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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

## C

 `#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; ` `} `

Output:

```Minimum number of jumps to reach end is 4
```

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].

## C / C++

 `#include ` `#include ` ` `  `int` `min(``int` `x, ``int` `y) { ``return` `(x < y) ? x : y; } ` ` `  `// 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] = min(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; ` `} `

Output:

```Minimum number of jumps to reach end is 3
```

Please refer complete article on Minimum number of jumps to reach end for more details!

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