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: The approach to solving this problem is to sort the array and then find the number of minimum length sticks that are of the 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 iterationint 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 codeint 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 iterationsimport java.io.*;import java.util.*;class GFG{ // Function to calculate// sum of remaining sticks// after each iterationpublic 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 codepublic 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 |
Python3
# Python program to find the sum of# remaining sticks after each iterations# Function to calculate# sum of remaining sticks# after each iterationdef sum(arr, n): sum = 0 arr.sort() prev, count, s = 0, 1, n i = 1 while (i < n): if (arr[i] == arr[prev]): count += 1 else: prev = i sum += s - count s -= count count = 1 i += 1 return sum# Driver coden = 6ar = [ 5, 4, 4, 2, 2, 8 ]ans = sum(ar, n)print(ans)# This code is contributed by shinjanpatra |
C#
// C# program to find the sum of// remaining sticks after each iterationsusing System;using System.Collections.Generic; class GFG{ // Function to calculate // sum of remaining sticks // after each iteration public static int sum(int[] arr, int n) { int sum = 0; Array.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); Console.WriteLine(ans); }}// This code is contributed by sanjoy_62. |
Javascript
<script>// Java Script program to find the sum of// remaining sticks after each iterations // Function to calculate// sum of remaining sticks// after each iterationfunction 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> |
Output
7
Time Complexity: O(Nlog(N)) where N is the number of sticks.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
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 iterationint 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 codeint 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 program to find sum# of remaining sticks# Function to calculate # sum of remaining sticks# after each iterationdef 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 codedef 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 iterationfunction 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 codelet 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:
where N is the number of sticks
Auxiliary Space: O(N), where N is the number of sticks.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
