Find the sum of remaining sticks after each iterations

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:

  • 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++

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// 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;
}

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Java

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// 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

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Python 3

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# 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()

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C#

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// 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

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Output:

7

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

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Improved By : kanugargng, 29AjayKumar