Algorithm Puzzles: Jump Game

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Algorithm Puzzles everyday every week sometimes: Jump Game

Puzzle

Puzzle from leetcode:

You are given an integer array nums. You are initially positioned at the array’s first index, and each element in the array represents your maximum jump length at that position.

Return true if you can reach the last index, or false otherwise.

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

Looks like a typical DP puzzle, let dp[i] as largest index we can jump from index i, we can have:

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dp[0] = nums[0]
dp[1] = dp[0] > nums[1] + i ? dp[0] : nums[1] + i
dp[i] = dp[i-1] > nums[i] + i ? dp[i-1] : nums[i] + i

Base on this we have:

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class Solution {
  public:
    bool canJump(const std::vector<int>& nums) {
        dp.emplace_back(nums[0]);

        if (dp[0] == 0 && nums.size() > 1) {
            return false;
        }

        for (int i = 1; i < nums.size(); ++i) {
            int op = nums[i] + i;
            dp.emplace_back(dp[i - 1] > op ? dp[i - 1] : op);
            if (dp[i] == i) {
                return i == nums.size() - 1 ? true : false;
            }
            if (dp[i] >= nums.size() - 1) {
                return true;
            }
        }

        return true;
    }

  private:
    std::vector<int> dp = {};
};

We may use size-initialed vector to reduce runtime:

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class Solution {
  public:
    bool canJump(const std::vector<int>& nums) {
        const int length = nums.size();

        if (nums[0] == 0 && length > 1) {
            return false;
        }

        std::vector<int> dp(length);
        dp[0] = nums[0];

        for (int i = 1; i < length; ++i) {
            int op = nums[i] + i;
            op = dp[i - 1] > op ? dp[i - 1] : op;
            if (op == i) {
                return op == length - 1 ? true : false;
            }
            if (op >= nums.size() - 1) {
                return true;
            }
            dp[i] = op;
        }

        return true;
    }
};

Now we have much better runtime result:

In worst case T.C. should be O(n) and S.C. should be O(2n).