[LeetCode] Jump Game II

Problem: You are given a 0-indexed array of integers nums of length n. You are initially positioned at nums[0].

Each element nums[i] represents the maximum length of a forward jump from index i. In other words, if you are at nums[i], you can jump to any nums[i + j] where:

• 0 <= j <= nums[i] and
• i + j < n

Return the minimum number of jumps to reach nums[n - 1]. The test cases are generated such that you can reach nums[n - 1].

Example:

Input: nums = [2,3,1,1,4]
Output: 2
Explanation: The minimum number of jumps to reach the last index is 2. Jump 1 step from index 0 to 1, then 3 steps to the last index.

Approach: My observation about the problem is if we can reach an index i and then we can definitely reach index i - 1. The proof is simple:

• If the length of jump is 1 then we are already at i - 1.
• Otherwise we can take length - 1 jump to reach i - 1.

In this way we can prove that we can reach any j where j = 0 to i - 1. Now we can maintain a table say jump where jump[i]  tells the farthest point it can reach. We can fill this table in following way:

• jump[0] = nums[0]
• jump[i] = MAX(jump[i - 1], i + jump[i]

Now we can simply use this table to get our answer. 

Implementation in C#:

    public int Jump(int[] nums)

    {

        int length = nums?.Length ?? 0;

        if(length <= 1)

        {

            return 0;

        }

        int[] jumps = new int[length];

        jumps[0] = nums[0];

        for (int i = 1; i < length; ++i)

        {

            jumps[i] = Math.Max(jumps[i - 1], i + nums[i]);

        }

        int numOfJumps = 0;

        for (int i = 0; i < length - 1; i = jumps[i])

        {

            ++numOfJumps;

        }

        return numOfJumps;

    }

Complexity: O(n)