Minimum number of jumps to reach the end in JAVA

We start from an array of integers, where each number means how many steps can be made from that position. First, we are on index 0 of the array, and we start jumping maximum the amount that we find at that index.

Input: int[] inputNumbers = {1, 4, 2, 7, 9, 3, 5, 8, 4, 7, 8}; Output: 3 Explanation: The jumps are: 1 -> 4 -> 9 -> 8. We start from index 0 where we have value 1. So, we jump one step and we arrive at value 4. This means our next options to choose from are: 2, 7, 9, 3. I chose 9 and reach the end of the array, to value 8. The same result is achieved by selecting 7.

Solution:

1. private static int minimumJumps(int[] inputNumbers) { 2. int length = inputNumbers.length; 3. int[] jumps = new int[length]; 4. 5. if (length == 0 || inputNumbers[0] == 0) { 6. return -1; 7. } 8. 9. for (int i = 1; i < length; i++) { 10. jumps[i] = Integer.MAX_VALUE; 11. for (int j = 0; j < i; j++) { 12. if (i <= j + inputNumbers[j] && jumps[j] != Integer.MAX_VALUE) { 13. jumps[i] = Math.min(jumps[i], jumps[j] + 1); 14. break; 15. } 16. } 17. } 18. return jumps[length - 1]; 19. } 20.