Given an array arr[] of N elements. At any step, we can delete a number of a different parity from the just previous step, i.e., if, at the previous step, an odd number was deleted then delete an even number in the current step or vice versa.
It is allowed to start by deleting any number. Deletion is possible till we can delete numbers of different parity at every step. The task is to find the minimum possible sum of the elements left at the end.
Examples:
Input: arr[] = {1, 5, 7, 8, 2}
Output: 0
Delete elements in the order 1, 2, 5, 8 and finally 7.
There are multiple ways of deletion,
resulting in the same minimized sum.Input: arr[] = {2, 2, 2, 2}
Output: 6
Delete 2 in first step.
Cannot delete any number, since there are no odd numbers left.
Hence, the leftover elements sum is 6.
Approach: The following ways can be followed to solve the above problem:
- Count the number of odd and even elements and store in vectors v1 and v2.
- Check if the number of odd and even elements are the same or differ by 1, then we can perform N steps, resulting in leftover sum as 0.
- If the size differs by more than 1, then only there will be leftover elements.
- In order to minimize the leftover sum of elements, we select the larger elements first.
- Hence, the sum of the X smaller elements will be the answer, where X is v2.size() – v1.size() – 1 or v1.size() – v2.size() – 1 depending on the count of even and odd elements.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to find the minimized sumint MinimizeleftOverSum(int a[], int n){ vector<int> v1, v2; for (int i = 0; i < n; i++) { if (a[i] % 2) v1.push_back(a[i]); else v2.push_back(a[i]); } // If more odd elements if (v1.size() > v2.size()) { // Sort the elements sort(v1.begin(), v1.end()); sort(v2.begin(), v2.end()); // Left-over elements int x = v1.size() - v2.size() - 1; int sum = 0; int i = 0; // Find the sum of leftover elements while (i < x) { sum += v1[i++]; } // Return the sum return sum; } // If more even elements else if (v2.size() > v1.size()) { // Sort the elements sort(v1.begin(), v1.end()); sort(v2.begin(), v2.end()); // Left-over elements int x = v2.size() - v1.size() - 1; int sum = 0; int i = 0; // Find the sum of leftover elements while (i < x) { sum += v2[i++]; } // Return the sum return sum; } // If same elements else return 0;}// Driver codeint main(){ int a[] = { 2, 2, 2, 2 }; int n = sizeof(a) / sizeof(a[0]); cout << MinimizeleftOverSum(a, n); return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG{// Function to find the minimized sumstatic int MinimizeleftOverSum(int a[], int n){ Vector<Integer> v1 = new Vector<Integer>(), v2 = new Vector<Integer>(); for (int i = 0; i < n; i++) { if (a[i] % 2 == 1) v1.add(a[i]); else v2.add(a[i]); } // If more odd elements if (v1.size() > v2.size()) { // Sort the elements Collections.sort(v1); Collections.sort(v2); // Left-over elements int x = v1.size() - v2.size() - 1; int sum = 0; int i = 0; // Find the sum of leftover elements while (i < x) { sum += v1.get(i++); } // Return the sum return sum; } // If more even elements else if (v2.size() > v1.size()) { // Sort the elements Collections.sort(v1); Collections.sort(v2); // Left-over elements int x = v2.size() - v1.size() - 1; int sum = 0; int i = 0; // Find the sum of leftover elements while (i < x) { sum += v2.get(i++); } // Return the sum return sum; } // If same elements else return 0;}// Driver codepublic static void main(String[] args) { int a[] = { 2, 2, 2, 2 }; int n = a.length; System.out.println(MinimizeleftOverSum(a, n));}}// This code is contributed by Princi Singh |
Python3
# Python3 implementation of the approach # Function to find the minimized sum def MinimizeleftOverSum(a, n) : v1, v2 = [], []; for i in range(n) : if (a[i] % 2) : v1.append(a[i]); else : v2.append(a[i]); # If more odd elements if (len(v1) > len(v2)) : # Sort the elements v1.sort(); v2.sort(); # Left-over elements x = len(v1) - len(v2) - 1; sum = 0; i = 0; # Find the sum of leftover elements while (i < x) : sum += v1[i]; i += 1 # Return the sum return sum; # If more even elements elif (len(v2) > len(v1)) : # Sort the elements v1.sort(); v2.sort(); # Left-over elements x = len(v2) - len(v1) - 1; sum = 0; i = 0; # Find the sum of leftover elements while (i < x) : sum += v2[i]; i += 1 # Return the sum return sum; # If same elements else : return 0; # Driver code if __name__ == "__main__" : a = [ 2, 2, 2, 2 ]; n = len(a); print(MinimizeleftOverSum(a, n)); # This code is contributed by Ryuga |
C#
// C# implementation of the approachusing System;using System.Collections.Generic;class GFG{// Function to find the minimized sumstatic int MinimizeleftOverSum(int []a, int n){ List<int> v1 = new List<int>(), v2 = new List<int>(); for (int i = 0; i < n; i++) { if (a[i] % 2 == 1) v1.Add(a[i]); else v2.Add(a[i]); } // If more odd elements if (v1.Count > v2.Count) { // Sort the elements v1.Sort(); v2.Sort(); // Left-over elements int x = v1.Count - v2.Count - 1; int sum = 0; int i = 0; // Find the sum of leftover elements while (i < x) { sum += v1[i++]; } // Return the sum return sum; } // If more even elements else if (v2.Count > v1.Count) { // Sort the elements v1.Sort(); v2.Sort(); // Left-over elements int x = v2.Count - v1.Count - 1; int sum = 0; int i = 0; // Find the sum of leftover elements while (i < x) { sum += v2[i++]; } // Return the sum return sum; } // If same elements else return 0;}// Driver codepublic static void Main(String[] args) { int []a = { 2, 2, 2, 2 }; int n = a.Length; Console.WriteLine(MinimizeleftOverSum(a, n));}}// This code is contributed by PrinciRaj1992 |
Javascript
<script>// javascript implementation of the approach// Function to find the minimized sumfunction MinimizeleftOverSum(a , n){ var v1 = [], v2 =[]; for (i = 0; i < n; i++) { if (a[i] % 2 == 1) v1.push(a[i]); else v2.push(a[i]); } // If more odd elements if (v1.length > v2.length) { // Sort the elements v1.sort(); v2.sort(); // Left-over elements var x = v1.length - v2.length - 1; var sum = 0; var i = 0; // Find the sum of leftover elements while (i < x) { sum += v1[i++]; } // Return the sum return sum; } // If more even elements else if (v2.length > v1.length) { // Sort the elements v1.sort(); v2.sort(); // Left-over elements var x = v2.length - v1.length - 1; var sum = 0; var i = 0; // Find the sum of leftover elements while (i < x) { sum += v2[i++]; } // Return the sum return sum; } // If same elements else return 0;}// Driver code var a = [ 2, 2, 2, 2 ]; var n = a.length; document.write(MinimizeleftOverSum(a, n));// This code is contributed by Rajput-Ji </script> |
Output:
6
Time Complexity: O(n * log n), as in the worst case we will be using an inbuilt sort function to sort an array of size n. Where N is the number of elements in the array.
Auxiliary Space: O(n), as we are using extra space for the array v1 and v2. Where N is the number of elements in the array.
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