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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 ``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 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

 ``

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 ``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 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

 ``
Output:
`7`

Time Complexity: where N is the number of sticks

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