Given an array arr[] consisting of heights of N chocolate bars, the task is to find the maximum height at which the horizontal cut is made to all the chocolates such that the sum remaining amount of chocolate is at least K.
Examples:
Input: K = 7, arr[] = [15, 20, 8, 17]
Output: 15
Explanation:
Suppose a cut is made at height 8:
-> chocolate taken from 1st chocolate bar = 15 – 8 =7
-> chocolate taken from 2nd chocolate bar = 20 – 8 =12
-> chocolate taken from 3rd chocolate bar = 8 – 8 = 0
-> chocolate taken from 4th chocolate bar = 17 – 8 = 9
=> Total chocolate wasted = (7 + 12 + 0 + 9) – K = 28 – 7 = 21Suppose a cut is made at height 15:
-> chocolate taken from 1st chocolate bar = 15 – 15 = 0
-> chocolate taken from 2nd chocolate bar = 20 – 15 = 5
-> 3rd chocolate bar wont be chosen as it is less than 15
-> chocolate taken from 4th chocolate bar = 17 – 15 = 2
=> Total chocolate wasted = (0 + 5 + 2) – K = 7 – 7 = 0Therefore when we take chocolate of height 15 then chocolate wasted is minimum. Therefore 15 is the answer.
Input: K = 12, arr[] = [30, 25, 22, 17, 20]
Output: 21
Explanation:
After a cut at height 18, the chocolate removed is 25 and chocolate wastage is (25 – 12) = 13 units. But if the cut is made at height 21 is made then 14 units of chocolate is removed and the wastage is (14 – 12) = 2 which is the least, hence 21 is the answer
Approach: The given problem can be solved based on Binary Search.
The idea is to perform the Binary Search over then range [0, max element of the array] and find that value in the range, say M, such that the sum of remaining chocolate after making the horizontal cut at M gives minimum difference with K.
Follow the steps below to solve the given problem:
- Initialize two variables, say low and high as 0 and the maximum array element respectively.
- Iterate until low <= high and perform the following steps:
- Find the value of mid as (low + high)/2.
- Find the sum of remaining chocolates after making the horizontal cut at height mid as M.
- If the value of M is less than K, then update the value of high as (mid – 1). Otherwise, update the value of low as (mid + 1).
- After performing the above steps, print the value as high as the resultant maximum height that must be cut.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the sum of remaining// chocolate after making the horizontal// cut at height midint cal(vector<int> arr, int mid){ // Stores the sum of chocolates int chocolate = 0; // Traverse the array arr[] for (auto i : arr) { // If the height is at least mid if (i >= mid) chocolate += i - mid; } // Return the possible sum return chocolate;}// Function to find the maximum horizontal// cut made to all the chocolates such that// the sum of the remaining element is// at least Kint maximumCut(vector<int> arr, int K){ // Ranges of Binary Search int low = 0; int high = *max_element(arr.begin(), arr.end()); // Perform the Binary Search while (low <= high) { int mid = (low + high) / 2; // Find the sum of removed after // making cut at height mid int chocolate = cal(arr, mid); // If the chocolate removed is // same as the chocolate needed // then return the height if (chocolate == K) return mid; // If the chocolate removed is // less than chocolate needed // then shift to the left range else if (chocolate < K) high = mid - 1; // Otherwise, shift to the right // range else { low = mid + 1; if (mid > high) high = mid; } } // Return the possible cut return high;}// Driver Codeint main(){ int N = 4; int K = 7; vector<int> arr{ 15, 20, 8, 17 }; cout << (maximumCut(arr, K));}// This code is contributed by ipg2016107. |
Java
// Java program for the above approachimport java.util.*;import java.util.Arrays;class GFG { // Function to find the sum of remaining // chocolate after making the horizontal // cut at height mid static int cal(int arr[], int mid) { // Stores the sum of chocolates int chocolate = 0; // Traverse the array arr[] for (int i = 0; i < arr.length; i++) { // If the height is at least mid if (arr[i] >= mid) chocolate += arr[i] - mid; } // Return the possible sum return chocolate; } // Function to find the maximum horizontal // cut made to all the chocolates such that // the sum of the remaining element is // at least K static int maximumCut(int arr[], int K) { // Ranges of Binary Search int low = 0; int high = Arrays.stream(arr).max().getAsInt(); // Perform the Binary Search while (low <= high) { int mid = (low + high) / 2; // Find the sum of removed after // making cut at height mid int chocolate = cal(arr, mid); // If the chocolate removed is // same as the chocolate needed // then return the height if (chocolate == K) return mid; // If the chocolate removed is // less than chocolate needed // then shift to the left range else if (chocolate < K) high = mid - 1; // Otherwise, shift to the right // range else { low = mid + 1; if (mid > high) high = mid; } } // Return the possible cut return high; } // Driver Code public static void main(String[] args) { int K = 7; int arr[] = { 15, 20, 8, 17 }; System.out.println(maximumCut(arr, K)); }}// This code is contributed by ukasp. |
Python3
# Python program for the above approach# Function to find the sum of remaining# chocolate after making the horizontal# cut at height middef cal(arr, mid): # Stores the sum of chocolates chocolate = 0 # Traverse the array arr[] for i in arr: # If the height is at least mid if i >= mid: chocolate += i - mid # Return the possible sum return chocolate# Function to find the maximum horizontal# cut made to all the chocolates such that# the sum of the remaining element is# at least Kdef maximumCut(arr, K): # Ranges of Binary Search low = 0 high = max(arr) # Perform the Binary Search while low <= high: mid = (low + high) // 2 # Find the sum of removed after # making cut at height mid chocolate = cal(arr, mid) # If the chocolate removed is # same as the chocolate needed # then return the height if chocolate == K: return mid # If the chocolate removed is # less than chocolate needed # then shift to the left range elif chocolate < K: high = mid - 1 # Otherwise, shift to the right # range else: low = mid + 1 if mid > high: high = mid # Return the possible cut return high# Driver CodeN, K = 4, 7arr = [15, 20, 8, 17]print(maximumCut(arr, K)) |
C#
// C# program for the above approachusing System;using System.Collections.Generic;using System.Linq;class GFG { // Function to find the sum of remaining // chocolate after making the horizontal // cut at height mid static int cal(List<int> arr, int mid) { // Stores the sum of chocolates int chocolate = 0; // Traverse the array arr[] foreach(int i in arr) { // If the height is at least mid if (i >= mid) chocolate += i - mid; } // Return the possible sum return chocolate; } // Function to find the maximum horizontal // cut made to all the chocolates such that // the sum of the remaining element is // at least K static int maximumCut(List<int> arr, int K) { // Ranges of Binary Search int low = 0; int high = arr.Max(); // Perform the Binary Search while (low <= high) { int mid = (low + high) / 2; // Find the sum of removed after // making cut at height mid int chocolate = cal(arr, mid); // If the chocolate removed is // same as the chocolate needed // then return the height if (chocolate == K) return mid; // If the chocolate removed is // less than chocolate needed // then shift to the left range else if (chocolate < K) high = mid - 1; // Otherwise, shift to the right // range else { low = mid + 1; if (mid > high) high = mid; } } // Return the possible cut return high; } // Driver Code public static void Main() { int K = 7; List<int> arr = new List<int>() { 15, 20, 8, 17 }; Console.Write(maximumCut(arr, K)); }}// This code is contributed by SURENDRA_GANGWAR. |
Javascript
<script> // JavaScript Program to implement // the above approach // Function to find the sum of remaining // chocolate after making the horizontal // cut at height mid function cal(arr, mid) { // Stores the sum of chocolates let chocolate = 0 // Traverse the array arr[] for (let i = 0; i < arr.length; i++) { // If the height is at least mid if (arr[i] >= mid) chocolate += arr[i] - mid } // Return the possible sum return chocolate } // Function to find the maximum horizontal // cut made to all the chocolates such that // the sum of the remaining element is // at least K function maximumCut(arr, K) { // Ranges of Binary Search let low = 0 let high = arr[0]; for (let i = 1; i < arr.length; i++) { high = Math.max(high, arr[i]); } // Perform the Binary Search while (low <= high) { mid = Math.floor((low + high) / 2); // Find the sum of removed after // making cut at height mid chocolate = cal(arr, mid) // If the chocolate removed is // same as the chocolate needed // then return the height if (chocolate == K) { return mid } // If the chocolate removed is // less than chocolate needed // then shift to the left range else if (chocolate < K) { high = mid - 1 } // Otherwise, shift to the right // range else { low = mid + 1 if (mid > high) high = mid } } // Return the possible cut return high } // Driver Code let N = 4; let K = 7; let arr = [15, 20, 8, 17]; document.write(maximumCut(arr, K)) // This code is contributed by Potta Lokesh </script> |
Output
15
Time Complexity: O(N*log N)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
