# C Exercises: Return the minimum number of jumps to reach the end of the array

## C Array: Exercise-56 with Solution

Write a program in C to return the minimum number of jumps to reach the end of the array.

**Sample Solution:**

**C Code:**

```
#include <stdio.h>
#include <limits.h>
// Function to calculate the minimum number of jumps needed to reach the end
int noOfJumps(int arr1[], int low, int high) {
// If the start and end point are the same, no jump is needed
if (high == low)
return 0;
// If the current position is 0, it's impossible to move forward
if (arr1[low] == 0)
return INT_MAX;
int min = INT_MAX; // Initialize the minimum jumps to a maximum value
// Iterate through all possible steps from the current position
for (int i = low + 1; i <= high && i <= low + arr1[low]; i++) {
// Recursively find the minimum jumps needed from the next position
int jumps = noOfJumps(arr1, i, high);
// If it's possible to jump from the next position and it's minimum, update min
if (jumps != INT_MAX && jumps + 1 < min)
min = jumps + 1; // Update the minimum jumps
}
return min; // Return the minimum number of jumps
}
int main() {
int arr1[] = {1, 3, 5, 8, 9, 2, 6, 7, 6, 8, 9, 1, 1, 1};
int n = sizeof(arr1) / sizeof(arr1[0]);
int i;
//------------- print original array ------------------
printf("The given array is: ");
for (i = 0; i < n; i++) {
printf("%d ", arr1[i]);
}
printf("\n");
//------------------------------------------------------
// Calculate and display the minimum number of jumps needed to reach the end
printf("The minimum number of jumps required to reach the end is: %d\n", noOfJumps(arr1, 0, n - 1));
return 0;
}
```

Sample Output:

The given array is : 1 3 5 8 9 2 6 7 6 8 9 1 1 1 The minimum of number of jumps is required to reach the end is: 3

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