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Problem solution · Java

Maximum Fruits Harvested After at Most K Steps

Maximum Fruits Harvested After at Most K Steps: a Java solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Sliding window or two pointers
Source
walkccc LeetCode Solutions
Length
37 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Sliding window or two pointers

For Maximum Fruits Harvested After at Most K Steps, the implementation maintains a moving interval and updates only the information that enters or leaves the window.

  1. Choose the invariant that makes a window valid or useful.
  2. Advance the right boundary and add the new element.
  3. Move the left boundary only as needed while maintaining the invariant and updating the answer.

Code notes

  • 37 lines of Java from the credited upstream file 2106.java.
  • The implementation visibly relies on sequence storage.
  • 4 loop blocks detected.

Complexity

Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure operations.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMaximum Fruits Harvested After at Most K Steps · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int maxTotalFruits(int[][] fruits, int startPos, int k) {    final int maxRight = Math.max(startPos, fruits[fruits.length - 1][0]);    int ans = 0;    int[] amounts = new int[1 + maxRight];    int[] prefix = new int[2 + maxRight];     for (int[] f : fruits)      amounts[f[0]] = f[1];     for (int i = 0; i + 1 < prefix.length; ++i)      prefix[i + 1] = prefix[i] + amounts[i];     // Go right first.    final int maxRightSteps = Math.min(maxRight - startPos, k);    for (int rightSteps = 0; rightSteps <= maxRightSteps; ++rightSteps) {      final int leftSteps = Math.max(0, k - 2 * rightSteps); // Turn left      ans = Math.max(ans, getFruits(startPos, maxRight, leftSteps, rightSteps, prefix));    }     // Go left first.    final int maxLeftSteps = Math.min(startPos, k);    for (int leftSteps = 0; leftSteps <= maxLeftSteps; ++leftSteps) {      final int rightSteps = Math.max(0, k - 2 * leftSteps); // Turn right      ans = Math.max(ans, getFruits(startPos, maxRight, leftSteps, rightSteps, prefix));    }     return ans;  }   private int getFruits(int startPos, int maxRight, int leftSteps, int rightSteps, int[] prefix) {    final int l = Math.max(0, startPos - leftSteps);    final int r = Math.min(maxRight, startPos + rightSteps);    return prefix[r + 1] - prefix[l];  }}

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