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Find the sum of remaining sticks after each iterations
  • Difficulty Level : Medium
  • Last Updated : 04 Jun, 2021

Given N number of sticks of varying lengths in an array arr, the task is to determine the sum of the count of sticks that are left after each iteration. At each iteration, cut the length of the shortest stick from remaining sticks.

Examples: 

Input: N = 6, arr = {5, 4, 4, 2, 2, 8}
Output: 7
Explanation:
Iteration 1: 
Initial arr = {5, 4, 4, 2, 2, 8}
Shortest stick = 2
arr with reduced length = {3, 2, 2, 0, 0, 6}
Remaining sticks = 4

Iteration 2: 
arr = {3, 2, 2, 4}
Shortest stick = 2
Left stick = 2

Iteration 3: 
arr = {1, 2}
Shortest stick = 1
Left stick = 1

Iteration 4: 
arr = {1}
Min length = 1
Left stick = 0

Input: N = 8, arr = {1, 2, 3, 4, 3, 3, 2, 1}
Output: 11

Approach: Approach to solve this problem is to sort the array and then find number of minimum length sticks that are of same length while traversing and update the sum accordingly at each step and in the end return the sum.

C++




// C++ program to find the sum of
// remaining sticks after each iterations
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate
// sum of remaining sticks
// after each iteration
int sum(vector<int> &arr, int n)
{
    int sum = 0;
    sort(arr.begin(),arr.end());
    int prev=0,count=1,s=arr.size();
    int i=1;
    while(i<arr.size()){
        if(arr[i]==arr[prev]){
            count++;
        }else{
            prev=i;
            sum+=s-count;
              s-=count;
            count=1;
        }
        i++;
    }
    return sum;
}
 
// Driver code
int main()
{
 
    int n = 6;
    vector<int> ar{ 5, 4, 4, 2, 2, 8 };
 
    int ans = sum(ar, n);
 
    cout << ans << '\n';
 
    return 0;
}

Java




// Java program to find the sum of
// remaining sticks after each iterations
import java.io.*;
import java.util.*;
 
class GFG{
     
// Function to calculate
// sum of remaining sticks
// after each iteration
public static int sum(int arr[], int n)
{
    int sum = 0;
    Arrays.sort(arr);
     
    int prev = 0, count = 1, s = n;
    int i = 1;
     
    while (i < n)
    {
        if (arr[i] == arr[prev])
        {
            count++;
        }
        else
        {
            prev = i;
            sum += s - count;
            s -= count;
            count = 1;
        }
        i++;
    }
    return sum;
}
 
// Driver code
public static void main(String[] args)
{
    int n = 6;
    int ar[] = { 5, 4, 4, 2, 2, 8 };
    int ans = sum(ar, n);
     
    System.out.println(ans);
}
}
 
// This code is contributed by Manu Pathria

Javascript




<script>
// Java Script program to find the sum of
// remaining sticks after each iterations
 
     
// Function to calculate
// sum of remaining sticks
// after each iteration
function sum(arr,n)
{
    let sum = 0;
    arr.sort();
     
    let prev = 0, count = 1, s = n;
    let i = 1;
     
    while (i < n)
    {
        if (arr[i] == arr[prev])
        {
            count++;
        }
        else
        {
            prev = i;
            sum += s - count;
            s -= count;
            count = 1;
        }
        i++;
    }
    return sum;
}
 
// Driver code
 
    let n = 6;
    let ar = [ 5, 4, 4, 2, 2, 8 ];
    let ans = sum(ar, n);
     
    document.write(ans);
 
// This code is contributed by manoj
</script>

Time Complexity: O(Nlog(N)) where N is the number of sticks.

Another Approach: 
 



  • Store the frequency of stick lengths in a map
  • In each iteration, 
    • Find the frequency of min length’s stick
    • Decrease the frequency of min length’s stick from each stick’s frequency
    • Add the count of non-zero sticks to the resultant stick.

Below is the implementation of above approach:

C++




// C++ program to find the sum of
// remaining sticks after each iterations
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate
// sum of remaining sticks
// after each iteration
int sum(int ar[], int n)
{
    map<int, int> mp;
 
    // storing frequency of stick length
    for (int i = 0; i < n; i++) {
        mp[ar[i]]++;
    }
 
    int sum = 0;
 
    for (auto p : mp) {
        n -= p.second;
        sum += n;
    }
 
    return sum;
}
 
// Driver code
int main()
{
 
    int n = 6;
    int ar[] = { 5, 4, 4, 2, 2, 8 };
 
    int ans = sum(ar, n);
 
    cout << ans << '\n';
 
    return 0;
}

Java




// Java program to find the sum of
// remaining sticks after each iterations
import java.util.HashMap;
import java.util.Map;
 
class GFG
{
     
    // Function to calculate
    // sum of remaining sticks
    // after each iteration
    static int sum(int ar[], int n)
    {
        HashMap<Integer,
                Integer> mp = new HashMap<>();
     
        for (int i = 0; i < n; i++)
        {
            mp.put(ar[i], 0);
        }
         
        // storing frequency of stick length
        for (int i = 0; i < n; i++)
        {
            mp.put(ar[i], mp.get(ar[i]) + 1) ;
        }
     
        int sum = 0;
     
        for(Map.Entry p : mp.entrySet())
        {
            n -= (int)p.getValue();
            sum += n;
        }
        return sum;
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int n = 6;
        int ar[] = { 5, 4, 4, 2, 2, 8 };
     
        int ans = sum(ar, n);
     
        System.out.println(ans);
     
    }
}
 
// This code is contributed by kanugargng

Python3




# Python proagram to find sum
# of remaining sticks
 
# Function to calculate
# sum of remaining sticks
# after each iteration
def sum(ar, n):
  mp = dict()
 
  for i in ar:
    if i in mp:
      mp[i]+= 1
    else:
      mp[i] = 1
   
  mp = sorted(list(mp.items()))
   
  sum = 0
   
  for pair in mp:
    n-= pair[1]
    sum+= n
 
  return sum
# Driver code
def main():
  n = 6
  ar = [5, 4, 4, 2, 2, 8]
  ans = sum(ar, n)
  print(ans)
 
 
main()

C#




// C# program to find the sum of
// remaining sticks after each iterations
using System;
using System.Collections.Generic;            
 
class GFG
{
     
    // Function to calculate
    // sum of remaining sticks
    // after each iteration
    static int sum(int []ar, int n)
    {
        SortedDictionary<int,
                         int> mp = new SortedDictionary<int,
                                                        int>();
 
        // storing frequency of stick length
        for (int i = 0; i < n; i++)
        {
            if(!mp.ContainsKey(ar[i]))
                mp.Add(ar[i], 0);
            else
                mp[ar[i]] = 0;
        }
         
        // storing frequency of stick length
        for (int i = 0; i < n; i++)
        {
            if(!mp.ContainsKey(ar[i]))
                mp.Add(ar[i], 1);
            else
                mp[ar[i]] = ++mp[ar[i]];
        }
         
        int sum = 0;
     
        foreach(KeyValuePair<int, int> p in mp)
        {
            n -= p.Value;
            sum += n;
        }
        return sum;
    }
     
    // Driver code
    public static void Main (String[] args)
    {
        int n = 6;
        int []ar = { 5, 4, 4, 2, 2, 8 };
     
        int ans = sum(ar, n);
     
        Console.WriteLine(ans);
    }
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
// Javascript program to find the sum of
// remaining sticks after each iterations
 
// Function to calculate
// sum of remaining sticks
// after each iteration
function sum(ar, n)
{
    let mp = new Map();
 
    for (let i = 0; i < n; i++)
    {
        mp.set(ar[i], 0);
    }
 
    // storing frequency of stick length
    for (let i = 0; i < n; i++)
    {
        mp.set(ar[i], mp.get(ar[i]) + 1);
    }
 
     mp = new Map([...mp].sort((a, b) => String(a[0]).localeCompare(b[0])))
    let sum = 0;
 
    for (let p of mp)
    {
        n -= p[1]
        sum += n;
    }
    return sum;
}
 
// Driver code
let n = 6;
let ar = [5, 4, 4, 2, 2, 8];
let ans = sum(ar, n);
document.write(ans + '<br>');
 
// This code is contributed by _saurabh_jaiswal.
</script>
Output: 
7

 

Time Complexity:O(Nlog(N))       where N is the number of sticks

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