Given an array of positive integers, remove all the occurrences of element to get the maximum sum of the remaining array
Examples:
Input : arr = {1, 1, 3}
Output : 3
On removing 1 from array we get {3} total value is 3Input : arr = {1, 1, 3, 3, 2, 2, 1, 1, 1}
Output : 11
On removing 2 from array we get {1, 1, 3, 3, 1, 1, 1} total value is 11
Brute Force solution is to first find the sum of array after that find all the frequencies of element in the array. Find the value contributed by them to array sum. Select the minimum value among them. To get maximum sum of array after removing is equal difference of total value of sum and minimum value contributed by individual element total frequent value.
Time complexity: O(n2)
A better approach We first find the total sum of array and then sort the array, count the individual frequencies while traversing the array and get maximum value. After sorting, we can frequencies of all elements in O(n) tine,
Time complexity of this approach is O(n Log n)
An Efficient Approach is to use hashing to count the frequencies of elements while traversing the array.Find the minimum value using the frequencies stored in array
C++
#include <bits/stdc++.h> using namespace std; int maxSumArray(int arr[], int n) { // Find total sum and frequencies of elements int sum = 0; unordered_map<int, int> mp; for (int i = 0; i < n; i++) { sum += arr[i]; mp[arr[i]]++; } // Find minimum value to be subtracted. int minimum = INT_MAX; for (auto x : mp) minimum = min(minimum, x.second * x.first); // Find maximum sum after removal return (sum - minimum); } // Drivers code int main() { int arr[] = { 1, 1, 3, 3, 2, 2, 1, 1, 1 }; int n = sizeof(arr) / sizeof(int); cout << maxSumArray(arr, n); return 0; } |
Java
// Java program to convert fractional decimal // to binary number import java.util.*; class GFG { static int maxSumArray(int arr[], int n) { // Find total sum and frequencies of elements int sum = 0; Map<Integer,Integer> m = new HashMap<>(); for (int i = 0 ; i < n; i++) { sum += arr[i]; if(m.containsKey(arr[i])) { m.put(arr[i], m.get(arr[i])+1); } else { m.put(arr[i], 1); } } // Find minimum value to be subtracted. int minimum = Integer.MAX_VALUE; for (Map.Entry<Integer,Integer> x : m.entrySet()) minimum = Math.min(minimum, x.getValue() * x.getKey()); // Find maximum sum after removal return (sum - minimum); } // Drivers code public static void main(String[] args) { int arr[] = { 1, 1, 3, 3, 2, 2, 1, 1, 1 }; int n = arr.length; System.out.println(maxSumArray(arr, n)); } } // This code contributed by Rajput-Ji |
Python3
# Python3 program to convert # fractional decimal to binary number from sys import maxsize def maxSumArray(arr, n): # Find total sum and frequencies of elements sum1 = 0 mp = {i:0 for i in range(4)} for i in range(n): sum1 += arr[i] mp[arr[i]] += 1 # Find minimum value to be subtracted. minimum = maxsize for key, value in mp.items(): if(key == 0): continue minimum = min(minimum, value * key) # Find maximum sum after removal return (sum1 - minimum) # Driver Code if __name__ =='__main__': arr = [1, 1, 3, 3, 2, 2, 1, 1, 1] n = len(arr) print(maxSumArray(arr, n)) # This code is contributed by # Surendra_Gangwar |
C#
// C# program to convert fractional decimal // to binary number using System; using System.Collections.Generic; class GFG { static int maxSumArray(int []arr, int n) { // Find total sum and frequencies of elements int sum = 0; Dictionary<int,int> m = new Dictionary<int,int>(); for (int i = 0 ; i < n; i++) { sum += arr[i]; if(m.ContainsKey(arr[i])) { var val = m[arr[i]]; m.Remove(arr[i]); m.Add(arr[i], val + 1); } else { m.Add(arr[i], 1); } } // Find minimum value to be subtracted. int minimum = int.MaxValue; foreach(KeyValuePair<int, int> x in m) minimum = Math.Min(minimum, (x.Value * x.Key)); // Find maximum sum after removal return (sum - minimum); } // Driver code public static void Main(String[] args) { int []arr = { 1, 1, 3, 3, 2, 2, 1, 1, 1 }; int n = arr.Length; Console.WriteLine(maxSumArray(arr, n)); } } // This code is contributed by 29AjayKumar |
11
Time complexity: O(n)
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
