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Problem solution · Java

Minimum Time to Eat All Grains

Minimum Time to Eat All Grains: a Java solution using binary search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Binary search
Source
walkccc LeetCode Solutions
Length
50 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

Binary search

For Minimum Time to Eat All Grains, the implementation exploits a monotonic condition to discard half of the remaining search space after every check.

  1. Identify the ordered answer range or sorted search domain.
  2. Write a predicate whose truth changes only once.
  3. Move the appropriate boundary after each midpoint check and return the final feasible position.

Code notes

  • 50 lines of Java from the credited upstream file 2604.java.
  • The implementation visibly relies on sequence storage.
  • 2 loop blocks detected.

Complexity

Multiply the logarithmic number of midpoint checks by the cost of one predicate evaluation.

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 codeMinimum Time to Eat All Grains · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int minimumTime(int[] hens, int[] grains) {    Arrays.sort(hens);    Arrays.sort(grains);     final int maxPosition = Math.max(Arrays.stream(hens).max().getAsInt(), //                                     Arrays.stream(grains).max().getAsInt());    final int minPosition = Math.min(Arrays.stream(hens).min().getAsInt(), //                                     Arrays.stream(grains).min().getAsInt());    int l = 0;    int r = (int) (1.5 * (maxPosition - minPosition));     while (l < r) {      final int m = (int) ((l + (long) r) / 2);      if (canEat(hens, grains, m))        r = m;      else        l = m + 1;    }     return (int) l;  }   // Returns true if `hens` can eat all `grains` within `time`.  private boolean canEat(int[] hens, int[] grains, int time) {    int i = 0; // grains[i] := next grain to be ate    for (final int hen : hens) {      int rightMoves = time;      if (grains[i] < hen) {        // `hen` needs go back to eat `grains[i]`.        final int leftMoves = hen - grains[i];        if (leftMoves > time)          return false;        final int leftThenRight = time - 2 * leftMoves;        final int rightThenLeft = (time - leftMoves) / 2;        rightMoves = Math.max(0, Math.max(leftThenRight, rightThenLeft));      }      i = firstGreater(grains, hen + rightMoves);      if (i == grains.length)        return true;    }    return false;  }   private int firstGreater(int[] arr, int target) {    final int i = Arrays.binarySearch(arr, target + 1);    return i < 0 ? -i - 1 : i;  }}

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