Skip to content
Related Articles

Related Articles

Value to be subtracted from array elements to make sum of all elements equals K
  • Difficulty Level : Medium
  • Last Updated : 23 Jul, 2020

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 variable 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++




// 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
Output:
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.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer Geeks Classes Live and Geeks Classes Live USA

My Personal Notes arrow_drop_up
Recommended Articles
Page :