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 ith 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 is the maximum height among all the trees. So, set low = 0 and high = max(height[i]).
  • Repeat the below steps 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 varible collected.
    3. If collected = K then mid is the answer.
    4. If collecetd > 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 made at a lower height.
  • Print -1 if no such value of mid is found.

Below is the implementation of the above approach:

C++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

1

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