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

```#include <stdio.h>
#include <limits.h>

// 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].

C / C++

```#include <stdio.h>
#include <limits.h>

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;
}
```

Java

```// JAVA Code for Minimum number of jumps to reach end
class GFG{

private static 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 Integer.MAX_VALUE;  // if first element is 0,
// end cannot be reached

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] = Integer.MAX_VALUE;
for (j = 0; j < i; j++)
{
if (i <= j + arr[j] && jumps[j] != Integer.MAX_VALUE)
{
jumps[i] = Math.min(jumps[i], jumps[j] + 1);
break;
}
}
}
return jumps[n-1];
}

// driver program to test above function
public static void main(String[] args) {
int arr[] = {1, 3, 6, 1, 0, 9};

System.out.println("Minimum number of jumps to reach end is : "+
minJumps(arr,arr.length));
}
}

// This code is contributed by Arnav Kr. Mandal.
```

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

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.

Minimum number of jumps to reach end | Set 2 (O(n) solution)

Thanks to Ashish for suggesting this solution.

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