48.Jump Game

Type I

Given an array of non-negative integers, you are initially positioned at the first index of the array. Each element in the array represents your maximum jump length at that position. Determine if you are able to reach the last index.

Example 1:

Input: nums = [2,3,1,1,4]
Output: true
Explanation: Jump 1 step from index 0 to 1, then 3 steps to the last index.

Example 2:
Input: nums = [3,2,1,0,4]
Output: false
Explanation: You will always arrive at index 3 no matter what. Its maximum jump length is 0, which makes it impossible to reach the last index.

Solution I: (Valley Peak Approach)

class Solution
{
public:
    bool canJump(vector<int> &nums)
    {

        int n = nums.size();

        int i = 0;
        int reach = 0;
        
        while (i < n)
        {
            if (i > reach)
            {
                return false;
            }
            reach = max(reach, nums[i] + i);
            i++;
        }
        return true;
    }
};

Time Complexity: O(n)

Type II

Given an array of non-negative integers, you are initially positioned at the first index of the array.Each element in the array represents your maximum jump length at that position.

Your goal is to reach the last index in the minimum number of jumps.

Example:

Input: [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.

Solution I: (Dynamic Programming)

class Solution
{
public:
    int jump(vector<int> &nums)
    {
        int n = nums.size();
        vector<int> v(n, INT_MAX);
        v[0] = 0;
        for (int i = 1; i < n; i++)
        {
            for (int j = 0; j < i; j++)
            {
                if (i <= j + nums[j])
                {
                    v[i] = min(v[i], v[j] + 1);
                }
            }
        }

        return v[n - 1];
    }
};

Time Complexity: O(n^2), Space Complexity: O(n)

Solution II: (Greedy Method)

class Solution
{
public:
    int jump(vector<int> &nums)
    {

        int maxEnd = 0;
        int curEnd = 0;
        int jump = 0;

        for (int i = 0; i < nums.size() - 1; i++)
        {

            maxEnd = max(maxEnd, nums[i] + i);
            if (curEnd == i)
            {
                jump++;
                curEnd = maxEnd;

                if (curEnd >= nums.size() - 1)
                {
                    break;
                }
            }
        }

        return jump;
    }
};

Time Complexity: O(n)

Solution III (Greedy)

Approach:- Check for a range and what is the max you can reach from that range.

 class Solution {
    public:
        int jump(vector<int>& nums) {
    
            int numJumps = 0;
    
            int l = 0;
            int r = 0;
    
            while (r < nums.size() - 1) {
    
                int farthest = -1;
                for (int i = l; i <= r; i++) {
                    farthest = max(farthest, i + nums[i]);
                }
    
                l = r + 1;
                r = farthest;
                numJumps++;
            }
    
            return numJumps;
        }
    };