Given a list of integer each representing the length of each stick and an integer which tells how many times we can break a stick into half parts, we have to find maximum desired length sticks can be obtained from the group of sticks.
Note 1: When we break a stick it gets converted into two half parts for example for a stick of length 10 two sticks can be obtained of both 5 in length and for a stick of length 5 two sticks will be obtained of length 2 and 3 respectively.
Note 2: Discarded part can’t be used again for making sticks such that if a stick of length 11 is given we can break it into 5 and 6 of length pieces then we have to discard one of the pieces which can’t be used further.
Examples:
Input:
list = [2, 3, 4, 11]
n = 2
desired_length = 3
Output :
Maximum sticks of desired length that can be obtained are : 3
Explanation :
We already have one stick of length 3 and two more sticks can be obtained
by breaking stick of length 11 into [5, 3, 3] pieces therefore total sticks will be 3.Input:
list = [2, 1, 4, 5]
n = 2
desired_length = 4
Output :
Maximum sticks of desired length that can be obtained are : 1
Explanation :
We already have one stick of length 4 and no more sticks can be obtained
by breaking any stick therefore total sticks will be 1
Approach:
To solve the problem mentioned above we will first do a linear search operation to find all the sticks which have exact same length as of the desired stick length and count them. We will then store the count in the variable. Obviously, we have to discard all the sticks which have a length less than the desired length as with them we can’t make any desired length stick. Then pass the value of sticks that have a length more than the desired length to a function that calculates how many sticks can be obtained by breaking them. With the help of recursion find a number of ways in which sticks can be obtained.
Below is the implementation of the above approach:
C++
#include<bits/stdc++.h>using namespace std;// Method to find number of sticks by breaking them int sticks_break(int stick_length,int n, int desired_length){ // If stick cant be break any more if (n < 1) return 0; // Check if stick length became // smaller than the desired length if (stick_length < desired_length) return 0; // Check if stick length is even number if (stick_length % 2 == 0) // Check if half of stick // length is equal to desired length if (stick_length / 2 == desired_length) return 2; // Check if half of stick length // is greater than the desired length else if (stick_length / 2 > desired_length) return (sticks_break(stick_length / 2, n - 1, desired_length)); // Check if stick length is odd number if (stick_length % 2 != 0) // For odd number two halves will be // generated checking if first half // is equal to desired length if (stick_length / 2 == desired_length) return 1; // Checking if second half // is equal to desired length if (stick_length / 2 + 1 == desired_length) return 1; // Check if half of stick length // is greater than the desired length if (stick_length/2 > desired_length) return (max (sticks_break( stick_length / 2, n - 1, desired_length), sticks_break( stick_length / 2 + 1, n - 1, desired_length))); return 0;} // Method to find number of sticks int numberOfSticks(vector<int>list_length, int n, int desired_length){ int count = 0; for(auto stick_lenght : list_length) { // Check if desired length is found if (desired_length == stick_lenght) // Incrementing the count count = count + 1; // Check if stick length is // greater than desired length else if (stick_lenght> desired_length) // Incrementing count // after break the sticks count = count + sticks_break( stick_lenght, n, desired_length); } // Return count return count;} // Driver codeint main(){ // List of integers vector<int>list_length = { 1, 2, 3, 21 }; // Number of ways stick can be break int n = 3; // Desired length int desired_length = 3; int count = numberOfSticks(list_length, n, desired_length); // Print result cout << count << endl;}// This code is contributed by Stream_Cipher |
Java
import java.util.*; class GFG{ // Method to find number of sticks by breaking them static int sticks_break(int stick_length, int n, int desired_length){ // If stick cant be break any more if (n < 1) return 0; // Check if stick length became // smaller than the desired length if (stick_length < desired_length) return 0; // Check if stick length is even number if (stick_length % 2 == 0) // Check if half of stick // length is equal to desired length if (stick_length / 2 == desired_length) return 2; // Check if half of stick length // is greater than the desired length else if (stick_length / 2 > desired_length) return (sticks_break(stick_length / 2, n - 1, desired_length)); // Check if stick length is odd number if (stick_length % 2 != 0) // For odd number two halves will be // generated checking if first half // is equal to desired length if (stick_length / 2 == desired_length) return 1; // Checking if second half // is equal to desired length if (stick_length / 2 + 1 == desired_length) return 1; // Check if half of stick length // is greater than the desired length if (stick_length/2 > desired_length) return (Math.max (sticks_break( stick_length / 2, n - 1, desired_length), sticks_break( stick_length / 2 + 1, n - 1, desired_length))); return 0;} // Method to find number of sticks static int numberOfSticks(int list_length[], int n, int desired_length){ int count = 0; for(int i = 0; i < list_length.length; i++) { // Check if desired length is found if (desired_length == list_length[i]) // Incrementing the count count = count + 1; // Check if stick length is // greater than desired length else if (list_length[i]> desired_length) // Incrementing count // after break the sticks count = count + sticks_break(list_length[i], n, desired_length); } // Return count return count;}// Driver codepublic static void main(String args[]) { // List of integers int[] list_length = new int[]{ 1, 2, 3, 21 }; // Number of ways stick can be break int n = 3; // Desired length int desired_length = 3; int count = numberOfSticks(list_length, n, desired_length); // Print result System.out.println(count);}}// This code is contributed by Stream_Cipher |
Python3
# method to find number of sticks by breaking them def sticks_break(stick_length, n, desired_length): # if stick cant be break any more if n < 1: return 0 # check if stick length became # smaller than the desired length if stick_length < desired_length: return 0 # check if stick length is even number if stick_length % 2 == 0: # check if half of stick # length is equal to desired length if stick_length / 2 == desired_length: return 2 # check if half of stick length # is greater than the desired length elif stick_length / 2 > desired_length: return sticks_break(stick_length / 2, n-1, desired_length) # check if stick length is odd number if stick_length % 2 != 0: # for odd number two halves will be generated # checking if first half is equal to desired length if stick_length // 2 == desired_length: return 1 # checking if second half # is equal to desired length if stick_length // 2 + 1 == desired_length: return 1 # check if half of stick length # is greater than the desired length if stick_length//2 > desired_length: return max (sticks_break(stick_length//2, n-1, desired_length), sticks_break(stick_length//2 + 1, n-1, desired_length)) return 0# method to find number of sticks def numberOfSticks(list_length, n, desired_length): count = 0 for stick_length in list_length: # check if desired length is found if desired_length == stick_length: # incrementing the count count = count + 1 # check if stick length is # greater than desired length elif stick_length > desired_length: # incrementing count # after break the sticks count = count + sticks_break(stick_length, n, desired_length) # return count return count # driver code if __name__ == "__main__": # list of integers list_length =[1, 2, 3, 21] # number of ways stick can be break n = 3 # desired length desired_length = 3 count = numberOfSticks(list_length, n, desired_length) # print result print(str(count)) |
C#
using System; class GFG{ // Method to find number of sticks by breaking them static int sticks_break(int stick_length, int n, int desired_length){ // If stick cant be break any more if (n < 1 ) return 0; // Check if stick length became // smaller than the desired length if (stick_length < desired_length) return 0; // Check if stick length is even number if (stick_length % 2 == 0) // Check if half of stick // length is equal to desired length if (stick_length / 2 == desired_length) return 2; // Check if half of stick length // is greater than the desired length else if (stick_length / 2 > desired_length) return (sticks_break(stick_length / 2, n - 1, desired_length)); // Check if stick length is odd number if (stick_length % 2 != 0) // For odd number two halves will be // generated checking if first half // is equal to desired length if (stick_length / 2 == desired_length) return 1; // Checking if second half // is equal to desired length if (stick_length / 2 + 1 == desired_length) return 1; // Check if half of stick length // is greater than the desired length if (stick_length/2 > desired_length) return (Math.Max(sticks_break( stick_length / 2, n - 1, desired_length), sticks_break( stick_length / 2 + 1, n - 1, desired_length))); return 0;} // Method to find number of sticks static int numberOfSticks(int []list_length, int n, int desired_length){ int count = 0; for(int i = 0; i < list_length.Length; i++) { // Check if desired length is found if (desired_length == list_length[i]) // Incrementing the count count = count + 1; // Check if stick length is // greater than desired length else if (list_length[i]> desired_length) // Incrementing count // after break the sticks count = count + sticks_break(list_length[i], n, desired_length); } // Return count return count;}// Driver codepublic static void Main() { // list of integers int []list_length = { 1, 2, 3, 21 }; // Number of ways stick can be break int n = 3; // Desired length int desired_length = 3; int count = numberOfSticks(list_length, n, desired_length); // Print result Console.WriteLine(count); }}// This code is contributed by Stream_Cipher |
Javascript
<script> // Method to find number of sticks by breaking them function sticks_break(stick_length, n, desired_length) { // If stick cant be break any more if (n < 1) return 0; // Check if stick length became // smaller than the desired length if (stick_length < desired_length) return 0; // Check if stick length is even number if (stick_length % 2 == 0) // Check if half of stick // length is equal to desired length if (parseInt(stick_length / 2, 10) == desired_length) return 2; // Check if half of stick length // is greater than the desired length else if (parseInt(stick_length / 2, 10) > desired_length) return (sticks_break(parseInt(stick_length / 2, 10), n - 1, desired_length)); // Check if stick length is odd number if (stick_length % 2 != 0) // For odd number two halves will be // generated checking if first half // is equal to desired length if (parseInt(stick_length / 2, 10) == desired_length) return 1; // Checking if second half // is equal to desired length if (parseInt(stick_length / 2, 10) + 1 == desired_length) return 1; // Check if half of stick length // is greater than the desired length if (parseInt(stick_length/2, 10) > desired_length) return (Math.max(sticks_break( parseInt(stick_length / 2, 10), n - 1, desired_length), sticks_break( parseInt(stick_length / 2, 10) + 1, n - 1, desired_length))); return 0; } // Method to find number of sticks function numberOfSticks(list_length, n, desired_length) { let count = 0; for(let i = 0; i < list_length.length; i++) { // Check if desired length is found if (desired_length == list_length[i]) // Incrementing the count count = count + 1; // Check if stick length is // greater than desired length else if (list_length[i]> desired_length) // Incrementing count // after break the sticks count = count + sticks_break(list_length[i], n, desired_length); } // Return count return count; } // List of integers let list_length = [ 1, 2, 3, 21 ]; // Number of ways stick can be break let n = 3; // Desired length let desired_length = 3; let count = numberOfSticks(list_length, n, desired_length); // Print result document.write(count); </script> |
Output:
3
Time Complexity: O(N^2)
Auxiliary Space: O(N)
