Sort the given Array as per given conditions - GeeksforGeeks

07 Feb, 2022
Given an array arr[] consisting of N positive integers, the task is to sort the array such that – 
All even numbers must come before all odd numbers.
All even numbers that are divisible by 5 must come first than even numbers not divisible by 5.
If two even numbers are divisible by 5 then the number having a greater value will come first
If two even numbers were not divisible by 5 then the number having a greater index in the array will come first.
All odd numbers must come in relative order as they are present in the array.
Examples:
Input: arr[] = {5, 10, 30, 7}Output: 30 10 5 7Explanation: Even numbers = [10, 30]. Odd numbers = [5, 7]. After sorting of even numbers, even numbers = [30, 10] as both 10 and 30 divisible by 5 but 30 has a larger value so it will come before 10.After sorting A = [30, 10, 5, 7] as all even numbers must come before all odd numbers.
Approach: This problem can be solved by using sorting. Follow the steps below to solve the problem:
Create three vectors say v1, v2, v3 where v1 stores number which is even and divisible by 5, v2 stores number which is even but not divisible by 5, and v3 stores the numbers which are odd.
Iterate in the range [0, N-1] using the variable i and perform the following steps-If arr[i]%2 = 0 and arr[i]%5=0, then insert arr[i] in v1.If arr[i]%2 = 0 and arr[i]%5 != 0, then insert arr[i] in v2.If arr[i]%2 then insert arr[i] in v3.
If arr[i]%2 = 0 and arr[i]%5=0, then insert arr[i] in v1.
If arr[i]%2 = 0 and arr[i]%5 != 0, then insert arr[i] in v2.
If arr[i]%2 then insert arr[i] in v3.
Sort the vector v1 in descending order.
After performing the above steps, print vector v1 then print vector v2 and then v3 as the answer.
Below is the implementation of the above approach:
C++
Java
Python3
C#
Javascript
// C++ program for the above approach#include <bits/stdc++.h>using namespace std; // Function to sort array in the way// mentioned abovevoid AwesomeSort(vector<int> m, int n){    // Create three vectors    vector<int> v1, v2, v3;     int i;     // Traverse through the elements    // of the array    for (i = 0; i < n; i++) {         // If elements are even and        // divisible by 10        if (m[i] % 10 == 0)            v1.push_back(m[i]);         // If number is even but not divisible        // by 5        else if (m[i] % 2 == 0)            v2.push_back(m[i]);        else            // If number is odd            v3.push_back(m[i]);    }     // Sort  v1 in descending orde    sort(v1.begin(), v1.end(), greater<int>());     for (int i = 0; i < v1.size(); i++) {        cout << v1[i] << " ";    }    for (int i = v2.size()-1; i >= 0; i--) {        cout << v2[i] << " ";    }    for (int i = 0; i < v3.size(); i++) {        cout << v3[i] << " ";    }} // Driver Codeint main(){    // Given Input    vector<int> arr{ 5, 10, 30, 7 };    int N = arr.size();     // FunctionCall    AwesomeSort(arr, N);     return 0;}
// Java program for the above approachimport java.util.Collections;import java.util.Vector; public class GFG_JAVA {     // Function to sort array in the way    // mentioned above    static void AwesomeSort(Vector<Integer> m, int n)    {        // Create three vectors        Vector<Integer> v1 = new Vector<>();        Vector<Integer> v2 = new Vector<>();        Vector<Integer> v3 = new Vector<>();         int i;         // Traverse through the elements        // of the array        for (i = 0; i < n; i++) {             // If elements are even and            // divisible by 10            if (m.get(i) % 10 == 0)                v1.add(m.get(i));             // If number is even but not divisible            // by 5            else if (m.get(i) % 2 == 0)                v2.add(m.get(i));            else                // If number is odd                v3.add(m.get(i));        }         // Sort  v1 in descending orde        Collections.sort(v1, Collections.reverseOrder());         for (int ii = 0; ii < v1.size(); ii++) {            System.out.print(v1.get(ii) + " ");        }        for (int ii = v2.size()-1; ii >= 0; ii--) {            System.out.print(v2.get(ii) + " ");        }        for (int ii = 0; ii < v3.size(); ii++) {            System.out.print(v3.get(ii) + " ");        }    }     // Driver code    public static void main(String[] args)    {        // Given Input        Vector<Integer> arr = new Vector<>();        arr.add(5);        arr.add(10);        arr.add(30);        arr.add(7);         int N = arr.size();         // FunctionCall        AwesomeSort(arr, N);    }} // This code is contributed by abhinavjain194
# Python program for the above approach # Function to sort array in the way# mentioned abovedef AwesomeSort(m, n):       # Create three vectors    v1, v2, v3 = [],[],[]         # Traverse through the elements    # of the array    for i in range(n):               # If elements are even and        # divisible by 10        if (m[i] % 10 == 0):            v1.append(m[i])         # If number is even but not divisible        # by 5        elif (m[i] % 2 == 0):            v2.append(m[i])        else:            # If number is odd            v3.append(m[i])     # Sort  v1 in descending orde    v1 = sorted(v1)[::-1]     for i in range(len(v1)):        print(v1[i], end = " ")     for i in range(len(v2)):        print(v2[i], end = " ")     for i in range(len(v3)):        print (v3[i], end = " ") # Driver Codeif __name__ == '__main__':       # Given Input    arr = [5, 10, 30, 7]    N = len(arr)     # FunctionCall    AwesomeSort(arr, N)     # This code is contributed by mohit kumar 29.
// C# program for the above approachusing System;using System.Collections.Generic;class GFG {         // Function to sort array in the way    // mentioned above    static void AwesomeSort(List<int> m, int n)    {               // Create three vectors        List<int> v1 = new List<int>();        List<int> v2 = new List<int>();        List<int> v3 = new List<int>();          int i;          // Traverse through the elements        // of the array        for (i = 0; i < n; i++) {              // If elements are even and            // divisible by 10            if (m[i] % 10 == 0)                v1.Add(m[i]);              // If number is even but not divisible            // by 5            else if (m[i] % 2 == 0)                v2.Add(m[i]);            else                // If number is odd                v3.Add(m[i]);        }          // Sort  v1 in descending orde        v1.Sort();        v1.Reverse();          for (int ii = 0; ii < v1.Count; ii++) {            Console.Write(v1[ii] + " ");        }        for (int ii = 0; ii < v2.Count; ii++) {            Console.Write(v2[ii] + " ");        }        for (int ii = 0; ii < v3.Count; ii++) {            Console.Write(v3[ii] + " ");        }    }       static void Main()  {         // Given Input    List<int> arr = new List<int>();    arr.Add(5);    arr.Add(10);    arr.Add(30);    arr.Add(7);     int N = arr.Count;     // FunctionCall    AwesomeSort(arr, N);  }} // This code is contributed by divyeshrabadiya07.
<script> // JavaScript program for the above approach // Function to sort array in the way// mentioned abovefunction AwesomeSort(m, n){    // Create three vectors    let v1 = [];    let v2 = [];    let v3 = [];     let i;     // Traverse through the elements    // of the array    for (i = 0; i < n; i++) {         // If elements are even and        // divisible by 10        if (m[i] % 10 == 0)            v1.push(m[i]);         // If number is even but not divisible        // by 5        else if (m[i] % 2 == 0)            v2.push(m[i]);        else            // If number is odd            v3.push(m[i]);    }     // Sort  v1 in descending orde    v1.sort((a, b) => b - a);     for (let i = 0; i < v1.length; i++) {        document.write(v1[i] + " ");    }    for (let i = 0; i < v2.length; i++) {        document.write(v2[i] + " ");    }    for (let i = 0; i < v3.length; i++) {        document.write(v3[i] + " ");    }} // Driver Code // Given Inputlet arr = [5, 10, 30, 7];let N = arr.length; // FunctionCallAwesomeSort(arr, N); </script>
30 10 5 7 
Time Complexity: O(NlogN)Auxiliary Space: O(N)
abhinavjain194
mohit kumar 29
_saurabh_jaiswal
divyeshrabadiya07
pk218
Lowe's Company
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