Given an array **wood[]** of size **N**, representing the length of **N** pieces of wood and an integer **K**, at least **K** pieces of the same length need to be cut from the given wooden pieces. The task is to find the maximum possible length of these **K** wooden pieces that can be obtained.

**Examples:**

Input:wood[] = {5, 9, 7}, K = 3Output:5Explanation:

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 = 4Output:4Explanation:

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.

**Approach:** The problem can be solved using a Binary search. Follow the steps below to solve the problem:

- Find the maximum element from the array
**wood[]**and store it in a variable, say**Max**. - The value of
**L**must lie in the raneg**[1, Max]**. Therefore, apply binary search over the range**[1, Max]**. - Initialize two variables say,
**left = 1**and**right = Max**to store the range in which the value of**L**lies. - Check if it is possible to cut the woods into
**K**pieces with length of each piece equal to**(left + right) / 2**or not. If found to be true, then update**left = (left + right) / 2**. - Otherwise, update
**right = (left + right) / 2**.

Below is the implementation of the above approach:

## C++

`// C++ program to implement` `// the above approach` `#include <bits/stdc++.h>` `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);` `}` |

*chevron_right*

*filter_none*

## 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` |

*chevron_right*

*filter_none*

## 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` |

*chevron_right*

*filter_none*

## 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` |

*chevron_right*

*filter_none*

**Output:**

4

**Time Complexity:** O(N * Log_{2}M), where M is the maximum element of the given array**Auxiliary Space:** O(1)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.