Question

2604. Minimum Time to Eat All Grains

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Description

There are n hens and m grains on a line. You are given the initial positions of the hens and the grains in two integer arrays hens and grains of size n and m respectively.

Any hen can eat a grain if they are on the same position. The time taken for this is negligible. One hen can also eat multiple grains.

In 1 second, a hen can move right or left by 1 unit. The hens can move simultaneously and independently of each other.

Return _the minimum time to eat all grains if the hens act optimally._

 

Example 1:

Input: hens = [3,6,7], grains = [2,4,7,9]
Output: 2
Explanation: 
One of the ways hens eat all grains in 2 seconds is described below:
- The first hen eats the grain at position 2 in 1 second. 
- The second hen eats the grain at position 4 in 2 seconds. 
- The third hen eats the grains at positions 7 and 9 in 2 seconds. 
So, the maximum time needed is 2.
It can be proven that the hens cannot eat all grains before 2 seconds.

Example 2:

Input: hens = [4,6,109,111,213,215], grains = [5,110,214]
Output: 1
Explanation: 
One of the ways hens eat all grains in 1 second is described below:
- The first hen eats the grain at position 5 in 1 second. 
- The fourth hen eats the grain at position 110 in 1 second.
- The sixth hen eats the grain at position 214 in 1 second. 
- The other hens do not move. 
So, the maximum time needed is 1.

 

Constraints:

• 1 <= hens.length, grains.length <= 2*104
• 0 <= hens[i], grains[j] <= 109

Solutions

Solution 1: Sorting + Binary Search

First, sort the chickens and grains by their position from left to right. Then enumerate the time $t$ using binary search to find the smallest $t$ such that all the grains can be eaten up in $t$ seconds.

For each chicken, we use the pointer $j$ to point to the leftmost grain that has not been eaten, and the current position of the chicken is $x$ and the position of the grain is $y$. There are the following cases:

• If $y \leq x$, we note that $d = x - y$. If $d \gt t$, the current grain cannot be eaten, so directly return false. Otherwise, move the pointer $j$ to the right until $j=m$ or $grains[j] \gt x$. At this point, we need to check whether the chicken can eat the grain pointed to by $j$. If it can, continue to move the pointer $j$ to the right until $j=m$ or $min(d, grains[j] - x) + grains[j] - y \gt t$.
• If $y \lt x$, move the pointer $j$ to the right until $j=m$ or $grains[j] - x \gt t$.

If $j=m$, it means that all the grains have been eaten, return true, otherwise return false.

Time complexity $O(n \times \log n + m \times \log m + (m + n) \times \log U)$, space complexity $O(\log m + \log n)$. $n$ and $m$ are the number of chickens and grains respectively, and $U$ is the maximum value of all the chicken and grain positions.

class Solution:
  def minimumTime(self, hens: List[int], grains: List[int]) -> int:
    def check(t):
      j = 0
      for x in hens:
        if j == m:
          return True
        y = grains[j]
        if y <= x:
          d = x - y
          if d > t:
            return False
          while j < m and grains[j] <= x:
            j += 1
          while j < m and min(d, grains[j] - x) + grains[j] - y <= t:
            j += 1
        else:
          while j < m and grains[j] - x <= t:
            j += 1
      return j == m

    hens.sort()
    grains.sort()
    m = len(grains)
    r = abs(hens[0] - grains[0]) + grains[-1] - grains[0] + 1
    return bisect_left(range(r), True, key=check)

class Solution {
  private int[] hens;
  private int[] grains;
  private int m;

  public int minimumTime(int[] hens, int[] grains) {
    m = grains.length;
    this.hens = hens;
    this.grains = grains;
    Arrays.sort(hens);
    Arrays.sort(grains);
    int l = 0;
    int r = Math.abs(hens[0] - grains[0]) + grains[m - 1] - grains[0];
    while (l < r) {
      int mid = (l + r) >> 1;
      if (check(mid)) {
        r = mid;
      } else {
        l = mid + 1;
      }
    }
    return l;
  }

  private boolean check(int t) {
    int j = 0;
    for (int x : hens) {
      if (j == m) {
        return true;
      }
      int y = grains[j];
      if (y <= x) {
        int d = x - y;
        if (d > t) {
          return false;
        }
        while (j < m && grains[j] <= x) {
          ++j;
        }
        while (j < m && Math.min(d, grains[j] - x) + grains[j] - y <= t) {
          ++j;
        }
      } else {
        while (j < m && grains[j] - x <= t) {
          ++j;
        }
      }
    }
    return j == m;
  }
}

class Solution {
public:
  int minimumTime(vector<int>& hens, vector<int>& grains) {
    int m = grains.size();
    sort(hens.begin(), hens.end());
    sort(grains.begin(), grains.end());
    int l = 0;
    int r = abs(hens[0] - grains[0]) + grains[m - 1] - grains[0];
    auto check = [&](int t) -> bool {
      int j = 0;
      for (int x : hens) {
        if (j == m) {
          return true;
        }
        int y = grains[j];
        if (y <= x) {
          int d = x - y;
          if (d > t) {
            return false;
          }
          while (j < m && grains[j] <= x) {
            ++j;
          }
          while (j < m && min(d, grains[j] - x) + grains[j] - y <= t) {
            ++j;
          }
        } else {
          while (j < m && grains[j] - x <= t) {
            ++j;
          }
        }
      }
      return j == m;
    };
    while (l < r) {
      int mid = (l + r) >> 1;
      if (check(mid)) {
        r = mid;
      } else {
        l = mid + 1;
      }
    }
    return l;
  }
};

func minimumTime(hens []int, grains []int) int {
  sort.Ints(hens)
  sort.Ints(grains)
  m := len(grains)
  l, r := 0, abs(hens[0]-grains[0])+grains[m-1]-grains[0]
  check := func(t int) bool {
    j := 0
    for _, x := range hens {
      if j == m {
        return true
      }
      y := grains[j]
      if y <= x {
        d := x - y
        if d > t {
          return false
        }
        for j < m && grains[j] <= x {
          j++
        }
        for j < m && min(d, grains[j]-x)+grains[j]-y <= t {
          j++
        }
      } else {
        for j < m && grains[j]-x <= t {
          j++
        }
      }
    }
    return j == m
  }
  for l < r {
    mid := (l + r) >> 1
    if check(mid) {
      r = mid
    } else {
      l = mid + 1
    }
  }
  return l
}

func abs(x int) int {
  if x < 0 {
    return -x
  }
  return x
}

function minimumTime(hens: number[], grains: number[]): number {
  hens.sort((a, b) => a - b);
  grains.sort((a, b) => a - b);
  const m = grains.length;
  let l = 0;
  let r = Math.abs(hens[0] - grains[0]) + grains[m - 1] - grains[0] + 1;

  const check = (t: number): boolean => {
    let j = 0;
    for (const x of hens) {
      if (j === m) {
        return true;
      }
      const y = grains[j];
      if (y <= x) {
        const d = x - y;
        if (d > t) {
          return false;
        }
        while (j < m && grains[j] <= x) {
          ++j;
        }
        while (j < m && Math.min(d, grains[j] - x) + grains[j] - y <= t) {
          ++j;
        }
      } else {
        while (j < m && grains[j] - x <= t) {
          ++j;
        }
      }
    }
    return j === m;
  };

  while (l < r) {
    const mid = (l + r) >> 1;
    if (check(mid)) {
      r = mid;
    } else {
      l = mid + 1;
    }
  }
  return l;
}