713994. Frogs and Jumps
Frogs and Jumps
Summary
In this problem, we are given an array of integers representing the positions of frogs on a log. Each frog can jump at most k units to the right or left. Our task is to find the maximum number of consecutive jumps each frog can make. This problem involves arrays and dynamic programming.
Detailed Explanation
To solve this problem, we need to iterate through the array and for each position, calculate the maximum number of consecutive jumps that a frog at that position can make. We use a variable maxJumps to keep track of the maximum number of consecutive jumps. For each position, we update maxJumps based on the positions that are within the jump distance (k) from the current position.
Here's the step-by-step breakdown:
- Initialize
maxJumpsto 0. - Iterate through the array starting from the second element (since the first element can only make one jump).
- For each position, calculate the maximum number of consecutive jumps that a frog at that position can make by considering the positions within the jump distance (
k) to the left and right. - Update
maxJumpsbased on the calculation in step 3. - Return the maximum value of
maxJumps.
Time complexity: O(n), where n is the length of the array. Space complexity: O(1), since we only use a constant amount of space.
Optimized Solutions
Java
public int maxConsecutiveJumps(int[] frogs, int k) {
int maxJumps = 0;
for (int i = 1; i < frogs.length; i++) {
int jumps = 0;
for (int j = Math.max(0, i - k); j <= Math.min(frogs.length - 1, i + k); j++) {
if (Math.abs(i - j) > k) break;
jumps++;
}
maxJumps = Math.max(maxJumps, jumps);
}
return maxJumps;
}