Value to be subtracted from array elements to make sum of all elements equals K

Given an integer K and an array, height[] where height[i] denotes the height of the i^th tree in a forest. The task is to make a cut of height X from the ground such that exactly K unit wood is collected. If it is not possible, then print -1 else print X.

Examples:

Input: height[] = {1, 2, 1, 2}, K = 2 
Output: 1 
Make a cut at height 1, the updated array will be {1, 1, 1, 1} and 
the collected wood will be {0, 1, 0, 1} i.e. 0 + 1 + 0 + 1 = 2.

Input: height = {1, 1, 2, 2}, K = 1 
Output: -1  

Approach: This problem can be solved using binary search.  

• Sort the heights of the trees.
• The lowest height to make the cut is 0 and the highest height is the maximum height among all the trees. So, set low = 0 and high = max(height[i]).
• Repeat the steps below while low ≤ high:
  1. Set mid = low + ((high – low) / 2).
  2. Count the amount of wood that can be collected if the cut is made at height mid and store it in a variable collected.
  3. If collected = K then mid is the answer.
  4. If collected > K then update low = mid + 1 as the cut needs to be made at a height higher than the current height.
  5. Else update high = mid – 1 as cuts need to be made at a lower height.
• Print -1 if no such value of mid is found.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
#include <iostream>
using namespace std;

// Function to return the amount of wood
// collected if the cut is made at height m
int woodCollected(int height[], int n, int m)
{
    int sum = 0;
    for (int i = n - 1; i >= 0; i--) {
        if (height[i] - m <= 0)
            break;
        sum += (height[i] - m);
    }

    return sum;
}

// Function that returns Height at
// which cut should be made
int collectKWood(int height[], int n, int k)
{
    // Sort the heights of the trees
    sort(height, height + n);

    // The minimum and the maximum
    // cut that can be made
    int low = 0, high = height[n - 1];

    // Binary search to find the answer
    while (low <= high) {
        int mid = low + ((high - low) / 2);

        // The amount of wood collected
        // when cut is made at the mid
        int collected = woodCollected(height, n, mid);

        // If the current collected wood is
        // equal to the required amount
        if (collected == k)
            return mid;

        // If it is more than the required amount
        // then the cut needs to be made at a
        // height higher than the current height
        if (collected > k)
            low = mid + 1;

        // Else made the cut at a lower height
        else
            high = mid - 1;
    }

    return -1;
}

// Driver code
int main()
{

    int height[] = { 1, 2, 1, 2 };
    int n = sizeof(height) / sizeof(height[0]);
    int k = 2;

    cout << collectKWood(height, n, k);

    return 0;
}

Java

// Java implementation of the approach
import java.util.Arrays;

class GFG
{
    static int[] height = new int[]{ 1, 2, 1, 2 };

    // Function to return the amount of wood
    // collected if the cut is made at height m
    public static int woodCollected(int n, int m)
    {
        int sum = 0;
        for (int i = n - 1; i >= 0; i--)
        {
            if (height[i] - m <= 0)
                break;
            sum += (height[i] - m);
        }
        return sum;
    }

    // Function that returns Height at
    // which cut should be made
    public static int collectKWood(int n, int k)
    {
        // Sort the heights of the trees
        Arrays.sort(height);

        // The minimum and the maximum
        // cut that can be made
        int low = 0, high = height[n - 1];

        // Binary search to find the answer
        while (low <= high)
        {
            int mid = low + ((high - low) / 2);

            // The amount of wood collected
            // when cut is made at the mid
            int collected = woodCollected(n, mid);

            // If the current collected wood is
            // equal to the required amount
            if (collected == k)
                return mid;

            // If it is more than the required amount
            // then the cut needs to be made at a
            // height higher than the current height
            if (collected > k)
                low = mid + 1;

            // Else made the cut at a lower height
            else
                high = mid - 1;
        }
        return -1;
    }

    // Driver code
    public static void main(String[] args)
    {
        int k = 2;
        int n = height.length;
        System.out.print(collectKWood(n,k));
    }
}

// This code is contributed by Sanjit_Prasad

Python3

# Python3 implementation of the approach

# Function to return the amount of wood
# collected if the cut is made at height m
def woodCollected(height, n, m):
    sum = 0
    for i in range(n - 1, -1, -1):
        if (height[i] - m <= 0):
            break
        sum += (height[i] - m)

    return sum

# Function that returns Height at
# which cut should be made
def collectKWood(height, n, k):

    # Sort the heights of the trees
    height = sorted(height)

    # The minimum and the maximum
    # cut that can be made
    low = 0
    high = height[n - 1]

    # Binary search to find the answer
    while (low <= high):
        mid = low + ((high - low) // 2)

        # The amount of wood collected
        # when cut is made at the mid
        collected = woodCollected(height, n, mid)

        # If the current collected wood is
        # equal to the required amount
        if (collected == k):
            return mid

        # If it is more than the required amount
        # then the cut needs to be made at a
        # height higher than the current height
        if (collected > k):
            low = mid + 1

        # Else made the cut at a lower height
        else:
            high = mid - 1

    return -1

# Driver code
height = [1, 2, 1, 2]
n = len(height)
k = 2

print(collectKWood(height, n, k))

# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach
using System;
using System.Collections;

class GFG
{
    static int[] height = { 1, 2, 1, 2 };

    // Function to return the amount of wood
    // collected if the cut is made at height m
    public static int woodCollected(int n, int m)
    {
        int sum = 0;
        for (int i = n - 1; i >= 0; i--)
        {
            if (height[i] - m <= 0)
                break;
            sum += (height[i] - m);
        }
        return sum;
    }

    // Function that returns Height at
    // which cut should be made
    public static int collectKWood(int n, int k)
    {
        // Sort the heights of the trees
        Array.Sort(height);

        // The minimum and the maximum
        // cut that can be made
        int low = 0, high = height[n - 1];

        // Binary search to find the answer
        while (low <= high)
        {
            int mid = low + ((high - low) / 2);

            // The amount of wood collected
            // when cut is made at the mid
            int collected = woodCollected(n, mid);

            // If the current collected wood is
            // equal to the required amount
            if (collected == k)
                return mid;

            // If it is more than the required amount
            // then the cut needs to be made at a
            // height higher than the current height
            if (collected > k)
                low = mid + 1;

            // Else made the cut at a lower height
            else
                high = mid - 1;
        }
        return -1;
    }

    // Driver code
    public static void Main()
    {
        int k = 2;
        int n = height.Length;
        Console.WriteLine(collectKWood(n,k));
    }
}

// This code is contributed by AnkitRai01

Javascript

<script>
// Javascript implementation of the approach

// Function to return the amount of wood
// collected if the cut is made at height m
function woodCollected(height, n, m) {
    let sum = 0;
    for (let i = n - 1; i >= 0; i--) {
        if (height[i] - m <= 0)
            break;
        sum += (height[i] - m);
    }

    return sum;
}

// Function that returns Height at
// which cut should be made
function collectKWood(height, n, k) {
    // Sort the heights of the trees
    height.sort((a, b) => a - b);

    // The minimum and the maximum
    // cut that can be made
    let low = 0, high = height[n - 1];

    // Binary search to find the answer
    while (low <= high) {
        let mid = low + ((high - low) / 2);

        // The amount of wood collected
        // when cut is made at the mid
        let collected = woodCollected(height, n, mid);

        // If the current collected wood is
        // equal to the required amount
        if (collected == k)
            return mid;

        // If it is more than the required amount
        // then the cut needs to be made at a
        // height higher than the current height
        if (collected > k)
            low = mid + 1;

        // Else made the cut at a lower height
        else
            high = mid - 1;
    }

    return -1;
}

// Driver code


let height = [ 1, 2, 1, 2 ];
let n = height.length;
let k = 2;

document.write(collectKWood(height, n, k));
</script>

Output:

1

Time Complexity: O(nlog(max_element in the array))

Auxiliary Space: O(1)