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

• Difficulty Level : Medium
• Last Updated : 08 Jun, 2022

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:
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

Recommended Practice

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 ``#include ``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);``    ``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

 ``

Output:

`1`

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

Auxiliary Space: O(1)

My Personal Notes arrow_drop_up