I’m pretty sure the answer is true
Given an array of integers (both odd and even), sort them in such a way that the first part of the array contains odd numbers sorted in descending order, rest portion contains even numbers sorted in ascending order.
Explanation:
- Partition the input array such that all odd elements are moved to left and all even elements on right. This step takes O(n).
- Once the array is partitioned, sort left and right parts individually. This step takes O(n Log n).
#include <bits/stdc++.h>
using namespace std;
void twoWaySort(int arr[], int n)
{
int l = 0, r = n - 1;
int k = 0;
while (l < r) {
while (arr[l] % 2 != 0) {
l++;
k++;
}
while (arr[r] % 2 == 0 && l < r)
r--;
if (l < r)
swap(arr[l], arr[r]);
}
sort(arr, arr + k, greater<int>());
sort(arr + k, arr + n);
}
int main()
{
int arr[] = { 1, 3, 2, 7, 5, 4 };
int n = sizeof(arr) / sizeof(int);
twoWaySort(arr, n);
for (int i = 0; i < n; i++)
cout << arr[i] << " ";
return 0;
}
