Leetcode 2498: Frog Jump II
A frog starts on the first stone in a river and wants to jump to the last stone and then return to the first stone. The frog can jump to any stone at most once. The cost of a jump is defined as the absolute difference between the positions of two stones. The frog must choose a path that minimizes the maximum jump length during the entire journey. Return the minimum cost of the path where the cost is defined as the maximum length of any jump.
Problem
Approach
Steps
Complexity
Input: You are given an integer array stones where each element represents the position of a stone, sorted in strictly increasing order. The frog will start at the first stone and needs to travel to the last stone and back.
Example: Input: stones = [0, 2, 5, 7, 10]
Constraints:
• 2 <= stones.length <= 10^5
• 0 <= stones[i] <= 10^9
• stones[0] == 0
• stones is sorted in strictly increasing order
Output: Return the minimum possible cost of the path where the cost is the maximum length of any jump the frog makes.
Example: Output: 5
Constraints:
• The minimum cost is determined by the path with the smallest maximum jump length.
Goal: The goal is to find the minimum cost of a path where the maximum jump length is minimized during the frog's journey.
Steps:
• 1. Iterate through the stones array and calculate the maximum jump length if the frog were to travel from the first stone to the last and back.
• 2. For each possible jump, calculate the jump length and track the largest jump.
• 3. The solution is the smallest maximum jump length among all possible paths.
Goal: Ensure that the input is valid, with stones sorted in strictly increasing order, and there are at least two stones.
Steps:
• The number of stones is between 2 and 100,000, and the position of each stone is between 0 and 1 billion.
Assumptions:
• The frog always starts at the first stone (position 0) and ends at the last stone, completing the round trip.
• Input: Input: stones = [0, 2, 5, 7, 10]
• Explanation: The frog starts at position 0 and needs to jump to 2, then to 5, 7, and finally 10. The maximum jump length is from 5 to 10, which is 5. Hence, the minimum cost is 5.
• Input: Input: stones = [0, 3, 9]
• Explanation: The frog starts at 0, jumps to 9, and returns to 0. The maximum jump length is 9. Hence, the minimum cost is 9.
Approach: We can solve this problem by considering the maximum difference between every second stone (representing the farthest distance the frog might jump at a time), which ensures that the maximum jump length is minimized.
Observations:
• The frog's journey consists of multiple jumps, and the goal is to minimize the maximum jump distance.
• The optimal approach would involve looking at the maximum differences between stones as the frog hops between them.
Steps:
• 1. Iterate through the list of stones and compute the maximum jump length between each stone and the stone that is two steps ahead.
• 2. The frog's path will be optimal when the largest jump between any two stones is minimized.
• 3. Track the maximum jump during the round trip and return it as the minimum cost.
Empty Inputs:
• There will always be at least two stones, so this case is not applicable.
Large Inputs:
• Ensure the solution is efficient for large inputs, where the number of stones is near the upper constraint of 100,000.
Special Values:
• Handle the case where the frog needs to make large jumps between stones located far apart.
Constraints:
• The solution should be optimal and run efficiently within time limits for large inputs.
int maxJump(vector<int>& A) {
int res = A[1] - A[0];
int n = A.size();
for(int i = 2; i < n; i++)
res = max(res, A[i] - A[i - 2]);
return res;
}
1 : Function Definition
int maxJump(vector<int>& A) {
Defines the function `maxJump` that takes a vector of integers `A` and returns an integer representing the maximum jump between elements with one element skipped in between.
2 : Mathematical Operations
int res = A[1] - A[0];
Initializes `res` with the difference between the second and first elements of the array, as the first possible jump.
3 : Array Operations
int n = A.size();
Gets the size of the array `A` and stores it in the variable `n`, which is used to control the loop iteration.
4 : Looping
for(int i = 2; i < n; i++)
Starts a loop from the third element (index 2) of the array and iterates through the rest of the array to compute the jump differences.
5 : Mathematical Operations
res = max(res, A[i] - A[i - 2]);
In each iteration, the difference between the current element `A[i]` and the element two positions earlier `A[i-2]` is calculated. The maximum of this difference and the previous `res` is stored back in `res`.
6 : Return Statement
return res;
Returns the maximum jump found after iterating through the array.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is linear (O(n)) since we only need to iterate through the stones list once.
Best Case: O(1)
Worst Case: O(1)
Description: The space complexity is constant (O(1)) since we only need a few extra variables to track the maximum jump.
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