Ropes left after every removal of smallest

• Difficulty Level : Medium
• Last Updated : 06 Jul, 2021

Given an array of an integer of size, N. Array contains N ropes of length Ropes[i]. You have to perform a cut operation on ropes such that all of them are reduced by the length of the smallest rope. Display the number of ropes left after every cut. Perform operations till the length of each rope becomes zero.
Note: IF no ropes left after a single operation, in this case, we print 0.

Examples:

Input : Ropes[] = { 5, 1, 1, 2, 3, 5 }
Output : 4 3 2
Explanation : In first operation the minimum ropes is 1 so we reduce length 1 from all of them after reducing we left with 4 ropes and we do same for rest.

Input : Ropes[] = { 5, 1, 6, 9, 8, 11, 2, 2, 6, 5 }
Output : 9 7 5 3 2 1

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Simple solution is to we traverse a loop from [0…n-1], In each iterations first we find min length rope. After that, we reduce all ropes length by it and then count how many ropes are left whose length is greater than zero. this process is done until all ropes length is greater than zero. This solution work in O(n2) time.

Efficient solution works in O(nlog(n)). First we have to sort all Ropes in increasing order of there length. after that we have follow the step.

//initial cutting length "min rope"
CuttingLength = Ropes
Now Traverse a loop from left to right [1...n]
.During traverse we check that
is current ropes length is greater than zero or not
IF ( Ropes[i] - CuttingLength > 0 )
.... IF Yes then all ropes to it's right side also greater than 0
.... Print number of ropes remains (n - i)
....update Cutting Length by current rope length
...... CuttingLength = Ropes[i]
Do the same process for the rest.

Below is the implementation of above idea.

C++

 // C++ program to print how many// Ropes are Left After Every Cut#include using namespace std; // Function print how many Ropes are// Left AfterEvery Cutting operationvoid cuttringRopes(int Ropes[], int n){    // sort all Ropes in increase    // of there length    sort(Ropes, Ropes + n);     int singleOperation = 0;     // min length rope    int cuttingLenght = Ropes;     // now traverse through the given    // Ropes in increase order of length    for (int i = 1; i < n; i++)    {        // After cutting if current rope length        // is greater than '0' that mean all        // ropes to it's right side are also        // greater than 0        if (Ropes[i] - cuttingLenght > 0)        {            // print number of ropes remains            cout << (n - i) << " ";                         // now current rope become            // min length rope            cuttingLenght = Ropes[i];            singleOperation++;        }    }    if (singleOperation == 0)        cout << "0 ";}int main(){    int Ropes[] = { 5, 1, 1, 2, 3, 5 };    int n = sizeof(Ropes) / sizeof(Ropes);    cuttringRopes(Ropes, n);    return 0;}

Java

 // Java program to print how many// Ropes are Left After Every Cutimport java.util.*;import java.lang.*;import java.io.*; class GFG {         // function print how many Ropes are Left After    // Every Cutting operation    public static void cuttringRopes(int Ropes[], int n)    {        // sort all Ropes in increasing        // order of their length        Arrays.sort(Ropes);         int singleOperation = 0;         // min length rope        int cuttingLenght = Ropes;         // now traverse through the given Ropes in        // increase order of length        for (int i = 1; i < n; i++)        {            // After cutting if current rope length            // is greater than '0' that mean all            // ropes to it's right side are also            // greater than 0            if (Ropes[i] - cuttingLenght > 0)            {                System.out.print(n - i + " ");                                 // now current rope become                // min length rope                cuttingLenght = Ropes[i];                 singleOperation++;            }        }                 // after first operation all ropes        // length become zero        if (singleOperation == 0)            System.out.print("0");    }     public static void main(String[] arg)    {        int[] Ropes = { 5, 1, 1, 2, 3, 5 };        int n = Ropes.length;        cuttringRopes(Ropes, n);    }}

C#

 // C# program to print how many// Ropes are Left After Every Cutusing System; class GFG {         // function print how many Ropes are Left After    // Every Cutting operation    public static void cuttringRopes(int []Ropes, int n)    {        // sort all Ropes in increasing        // order of their length        Array.Sort(Ropes);         int singleOperation = 0;         // min length rope        int cuttingLenght = Ropes;         // now traverse through the given Ropes in        // increase order of length        for (int i = 1; i < n; i++)        {            // After cutting if current rope length            // is greater than '0' that mean all            // ropes to it's right side are also            // greater than 0            if (Ropes[i] - cuttingLenght > 0)            {                Console.Write(n - i + " ");                                 // now current rope become                // min length rope                cuttingLenght = Ropes[i];                 singleOperation++;            }        }                 // after first operation all ropes        // length become zero        if (singleOperation == 0)            Console.Write("0");    }     // Driver code    public static void Main()    {        int[] Ropes = { 5, 1, 1, 2, 3, 5 };        int n = Ropes.Length;        cuttringRopes(Ropes, n);    }} // This code is contributed by vt_m.

PHP

 0)        {            // print number of ropes remains            echo (\$n - \$i). " ";                         // now current rope become            // min length rope            \$cuttingLenght = \$Ropes[\$i];            \$singleOperation++;        }    }    if (\$singleOperation == 0)        echo "0 ";}     // Driver Code    \$Ropes = array(5, 1, 1, 2, 3, 5);    \$n = count(\$Ropes);    cuttringRopes(\$Ropes, \$n); // This code is contributed by Sam007?>

Javascript


4 3 2

Time Complexity : O(n long (n))
Space complexity : O(1)

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