通过将 N 给定的木材切割成至少 K 块来实现最大长度

给定一个大小为N的数组wood[] ，代表N块木头的长度和一个整数K ，至少需要从给定的木块中切割出K块相同长度的木头。任务是找到可以获得的这K 个木块的最大可能长度。

例子：

Input: wood[] = {5, 9, 7}, K = 3 
Output: 5 
Explanation: 
Cut arr[0] = 5 = 5 
Cut arr[1] = 9 = 5 + 4 
Cut arr[2] = 5 = 5 + 2 
Therefore, the maximum length that can be obtained by cutting the woods into 3 pieces is 5.

Input: wood[] = {5, 9, 7}, K = 4 
Output: 4 
Explanation: 
Cut arr[0] = 5 = 4 + 1 
Cut arr[1] = 9 = 2 * 4 + 1 
Cut arr[2] = 7 = 4 + 3 
Therefore, the maximum length that can be obtained by cutting the woods into 4 pieces is 4.

方法：该问题可以使用二分搜索来解决。请按照以下步骤解决问题：

• 从数组wood[] 中找到最大元素并将其存储在一个变量中，比如Max 。
• L的值必须位于范围[1, Max] 中。因此，在[1, Max]范围内应用二分搜索。
• 初始化两个变量，例如left = 1和right = Max来存储L值所在的范围。
• 检查是否可以将木材切割成K块，每块的长度等于(左+右)/ 2 。如果发现为真，则更新left = (left + right) / 2 。
• 否则，更新right = (left + right) / 2 。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to check if it is possible to cut
// woods into K pieces of length len
bool isValid(int wood[], int N, int len, int K)
{
 
    // Stores count of pieces
    // having length equal to K
    int count = 0;
 
    // Traverse wood[] array
    for (int i = 0; i < N; i++) {
 
        // Update count
        count += wood[i] / len;
    }
    return count >= K;
}
 
// Function to find the maximum value of L
int findMaxLen(int wood[], int N, int K)
{
 
    // Stores minimum possible of L
    int left = 1;
 
    // Stores maximum possible value of L
    int right = *max_element(wood,
                             wood + N);
 
    // Apply binary search over
    // the range [left, right]
    while (left <= right) {
 
        // Stores mid value of
        // left and right
        int mid = left + (right - left) / 2;
 
        // If it is possible to cut woods
        // into K pieces having length
        // of each piece equal to mid
        if (isValid(wood, N, mid,
                    K)) {
 
            // Update left
            left = mid + 1;
        }
        else {
 
            // Update right
            right = mid - 1;
        }
    }
    return right;
}
 
// Driver Code
int main()
{
    int wood[] = { 5, 9, 7 };
    int N = sizeof(wood) / sizeof(wood[0]);
    int K = 4;
    cout << findMaxLen(wood, N, K);
}

Java

// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
 
// Function to check if it is possible
// to cut woods into K pieces of
// length len
static boolean isValid(int wood[], int N,
                       int len, int K)
{
     
    // Stores count of pieces
    // having length equal to K
    int count = 0;
 
    // Traverse wood[] array
    for(int i = 0; i < N; i++)
    {
         
        // Update count
        count += wood[i] / len;
    }
    return count >= K;
}
 
// Function to find the maximum value of L
static int findMaxLen(int wood[], int N,
                      int K)
{
     
    // Stores minimum possible of L
    int left = 1;
 
    // Stores maximum possible value of L
    int right = Arrays.stream(wood).max().getAsInt();
 
    // Apply binary search over
    // the range [left, right]
    while (left <= right)
    {
         
        // Stores mid value of
        // left and right
        int mid = left + (right - left) / 2;
 
        // If it is possible to cut woods
        // into K pieces having length
        // of each piece equal to mid
        if (isValid(wood, N, mid, K))
        {
             
            // Update left
            left = mid + 1;
        }
        else
        {
             
            // Update right
            right = mid - 1;
        }
    }
    return right;
}
 
// Driver Code
public static void main(String[] args)
{
    int wood[] = { 5, 9, 7 };
    int N = wood.length;
    int K = 4;
     
    System.out.print(findMaxLen(wood, N, K));
}
}
 
// This code is contributed by Princi Singh

Python3

# Python3 program to implement
# the above approach
 
# Function to check if it is possible to
# cut woods into K pieces of length len
def isValid(wood, N, len, K):
 
    # Stores count of pieces
    # having length equal to K
    count = 0
 
    # Traverse wood[] array
    for i in range(N):
 
        # Update count
        count += wood[i] // len
         
    return (count >= K)
 
# Function to find the maximum value of L
def findMaxLen(wood, N, K):
 
    # Stores minimum possible of L
    left = 1
 
    # Stores maximum possible value of L
    right = max(wood)
 
    # Apply binary search over
    # the range [left, right]
    while (left <= right):
         
        # Stores mid value of
        # left and right
        mid = left + (right - left) // 2
 
        # If it is possible to cut woods
        # into K pieces having length
        # of each piece equal to mid
        if (isValid(wood, N, mid, K)):
 
            # Update left
            left = mid + 1
        else:
             
            # Update right
            right = mid - 1
             
    return right
 
# Driver Code
if __name__ == '__main__':
 
    wood = [ 5, 9, 7 ]
    N = len(wood)
    K = 4
     
    print(findMaxLen(wood, N, K))
 
# This code is contributed by mohit kumar 29

C#

// C# program to implement
// the above approach
using System;
using System.Linq;
class GFG{
 
// Function to check if it is possible
// to cut woods into K pieces of
// length len
static bool isValid(int []wood, int N,
                    int len, int K)
{   
  // Stores count of pieces
  // having length equal to K
  int count = 0;
 
  // Traverse wood[] array
  for(int i = 0; i < N; i++)
  {
    // Update count
    count += wood[i] / len;
  }
  return count >= K;
}
 
// Function to find the maximum
// value of L
static int findMaxLen(int []wood,
                      int N, int K)
{   
  // Stores minimum possible
  // of L
  int left = 1;
 
  // Stores maximum possible
  // value of L
  int right = wood.Max();
 
  // Apply binary search over
  // the range [left, right]
  while (left <= right)
  {
    // Stores mid value of
    // left and right
    int mid = left +
              (right - left) / 2;
 
    // If it is possible to cut woods
    // into K pieces having length
    // of each piece equal to mid
    if (isValid(wood, N,
                mid, K))
    {
      // Update left
      left = mid + 1;
    }
    else
    {
      // Update right
      right = mid - 1;
    }
  }
  return right;
}
 
// Driver Code
public static void Main(String[] args)
{
  int []wood = {5, 9, 7};
  int N = wood.Length;
  int K = 4;
  Console.Write(findMaxLen(wood,
                           N, K));
}
}
 
// This code is contributed by shikhasingrajput

Javascript

 
// Javascript program to implement
// the above approach
 
    // Function to check if it is possible
    // to cut woods into K pieces of
    // length len
    function isValid(wood , N , len , K) {
 
        // Stores count of pieces
        // having length equal to K
        var count = 0;
 
        // Traverse wood array
        for (i = 0; i < N; i++) {
 
            // Update count
            count += parseInt(wood[i] / len);
        }
        return count >= K;
    }
 
    // Function to find the maximum value of L
    function findMaxLen(wood , N , K) {
 
        // Stores minimum possible of L
        var left = 1;
 
        // Stores maximum possible value of L
        var right = Math.max.apply(Math,wood);
 
        // Apply binary search over
        // the range [left, right]
        while (left <= right) {
 
            // Stores mid value of
            // left and right
            var mid = left + (right - left) / 2;
 
            // If it is possible to cut woods
            // into K pieces having length
            // of each piece equal to mid
            if (isValid(wood, N, mid, K)) {
 
                // Update left
                left = mid + 1;
            } else {
 
                // Update right
                right = mid - 1;
            }
        }
        return right;
    }
 
    // Driver Code
     
        var wood = [ 5, 9, 7 ];
        var N = wood.length;
        var K = 4;
 
        document.write(findMaxLen(wood, N, K));
 
// This code is contributed by todaysgaurav
 

输出：

4

时间复杂度： O(N * Log ₂ M)，其中 M 是给定数组的最大元素
辅助空间： O(1)

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