Maximum of sum of length of rectangles and squares formed by given sticks
Given an array arr[] consisting of N integers, representing the length of the sticks, the task is to find the maximum sum possible of all lengths of the squares and rectangles constructed using these sticks.
Note A single side can be represented using only a single stick.
Examples:
Input: arr[] = {5, 3, 2, 3, 6, 3, 3}
Output: 12
Sum of length of Squares= 3 * 4 = 12
Sum of length of Rectangles = 0
Input: arr[] = {5, 3, 2, 3, 6, 4, 4, 4, 5, 5, 5 }
Output: 34
Sum of length of Squares = 5 * 4 = 20
Sum of length of Rectangles = 3 * 2 + 4 * 2 = 34
Approach: Follow the steps below to solve the problem:
- Traverse the array to count the frequencies of all the array elements, say freq.
- Count frequencies which are at least 2.
- Convert frequencies to nearest smaller even value.
- Convert the frequencies to single dimensional array in descending order.
- Now, sum elements upto indices which are multiples of 4.
- Print the sum all these elements
Below is the implementation of the above approach:
C++14
#include <bits/stdc++.h>using namespace std;// Function to find the maximum of sum// of lengths of rectangles and squares// formed using given set of sticksvoid findSumLength(vector<int> arr,int n){ // Stores the count of frequencies // of all the array elements map<int,int> freq; for(int i:arr) freq[i] += 1; // Stores frequencies which are at least 2 map<int,int> freq_2; for (auto i:freq) { if (freq[i.first] >= 2) freq_2[i.first] = freq[i.first]; } // Convert all frequencies to nearest // smaller even values. vector<int> arr1; for (auto i:freq_2) arr1.push_back((i.first) * (freq_2[(i.first)]/2)*2); sort(arr1.begin(), arr1.end()); // Sum of elements up to // index of multiples of 4 int summ = 0; for (int i:arr1) summ += i; // Print the sum cout << summ;}// Driver Codeint main(){ vector<int> arr = {5, 3, 2, 3, 6, 4, 4, 4, 5, 5, 5}; int n = arr.size(); findSumLength(arr, n); return 0;}// This code is contributed by mohit kumar 29. |
Python3
# Python3 implementation of# the above approach from collections import Counter# Function to find the maximum of sum# of lengths of rectangles and squares# formed using given set of sticks def findSumLength(arr, n) : # Stores the count of frequencies # of all the array elements freq = dict(Counter(arr)) # Stores frequencies which are at least 2 freq_2 = {} for i in freq: if freq[i]>= 2: freq_2[i] = freq[i] # Convert all frequencies to nearest # smaller even values. arr1 = [] for i in freq_2: arr1.extend([i]*(freq_2[i]//2)*2) # Sort the array in descending order arr1.sort(reverse = True) # Sum of elements up to # index of multiples of 4 summ = 0 for i in range((len(arr1)//4)*4): summ+= arr1[i] # Print the sum print(summ) # Driver Code if __name__ == "__main__" : arr = [5, 3, 2, 3, 6, 4, 4, 4, 5, 5, 5]; n = len(arr); findSumLength(arr, n); |
C#
using System;using System.Collections.Generic;class GFG{ // Function to find the maximum of sum // of lengths of rectangles and squares // formed using given set of sticks static void findSumLength(List<int> arr, int n) { // Stores the count of frequencies // of all the array elements Dictionary<int,int> freq = new Dictionary<int,int>(); foreach (int i in arr){ if(freq.ContainsKey(i)) freq[i] += 1; else freq[i] = 1; } // Stores frequencies which are at least 2 Dictionary<int,int> freq_2 = new Dictionary<int,int>(); foreach(KeyValuePair<int, int> entry in freq) { if (freq[entry.Key] >= 2) freq_2[entry.Key] = freq[entry.Key]; } // Convert all frequencies to nearest // smaller even values. List<int> arr1 = new List<int>(); foreach(KeyValuePair<int, int> entry in freq_2) arr1.Add(entry.Key * (freq_2[entry.Key]/2)*2); arr1.Sort(); // Sum of elements up to // index of multiples of 4 int summ = 0; foreach (int i in arr1){ summ += i; } // Print the sum Console.WriteLine(summ); } // Driver Code public static void Main() { List<int> arr = new List<int>(){5, 3, 2, 3, 6, 4, 4, 4, 5, 5, 5}; int n = arr.Count; findSumLength(arr, n); }}// THIS CODE IS CONTRIBUTED BY SURENDRA_GANGWAR. |
Javascript
<script>// Function to find the maximum of sum // of lengths of rectangles and squares // formed using given set of sticksfunction findSumLength(arr, n){ // Stores the count of frequencies // of all the array elements let freq = new Map(); for(let i = 0; i < arr.length; i++) { if(freq.has(arr[i])) freq.set(arr[i], freq.get(arr[i])+1); else freq.set(arr[i], 1); } // Stores frequencies which are at least 2 let freq_2 = new Map(); for(let [key, value] of freq.entries()) { if (freq.get(key) >= 2) freq_2.set(key,freq.get(key)); } // Convert all frequencies to nearest // smaller even values. let arr1 = []; for(let [key, value] of freq_2.entries()) arr1.push(key * Math.floor(freq_2.get(key)/2)*2); arr1.sort(function(a, b){return a - b;}); // Sum of elements up to // index of multiples of 4 let summ = 0; for(let i = 0; i < arr1.length; i++){ summ += arr1[i]; } // Print the sum document.write(summ);} // Driver Codelet arr=[5, 3, 2, 3, 6, 4, 4, 4, 5, 5, 5];let n = arr.Count;findSumLength(arr, n);// This code is contributed by unknown2108</script> |
Output:
34
Time Complexity: O(NlogN)
Auxiliary Space: O(1)
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