# Find a permutation that causes worst case of Merge Sort in C++

Suppose we have a set of elements; we have to find which permutation of these elements would result in worst case of Merge Sort? As we know asymptotically, merge sort always consumes O (n log n) time, but some cases need more comparisons and consumes more time. Here we have to find a permutation of input elements that will require higher number of comparisons when sorted implementing a typical Merge Sort algorithm.

So, if the input is like [11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26] , then the output will be [11,19,15,23,13,21,17,25,12,20,16,24,14,22,18,26].

To solve this, we will follow these steps −

• Define a function merge(), this will take array arr, array left, array right, l_index, m_index, r_index,
• for initialize i := 0, when i <= m_index - l_index, update (increase i by 1), do −
• arr[i] := left[i]
• for initialize j := 0, when j < r_index - m_index, update (increase j by 1), do −
• arr[i + j] = right[j]
• Define a function divide(), this will take array arr, array left, array right, l_index, m_index, r_index,
• for initialize i := 0, when i <= m_index - l_index, update (increase i by 1), do −
• left[i] := arr[i * 2]
• for initialize i := 0, when i < r_index - m_index, update (increase i by 1), do −
• right[i] := arr[i * 2 + 1]
• Define a function gen_worst_seq(), this will take array arr[], l_index, r_index,
• if l_index < r_index, then −
• m_index := l_index + (r_index - l_index) / 2
• Define an array left of size: m_index-l_index+1.
• Define an array right of size: r_index-m_index.
• divide(arr, left, right, l_index, m_index, r_index)
• gen_worst_seq(left, l_index, m_index)
• gen_worst_seq(right, m_index + 1, r_index)
• merge(arr, left, right, l_index, m_index, r_index)

## Example

Let us see the following implementation to get better understanding −

Live Demo

#include <bits/stdc++.h>
using namespace std;
void display(int A[], int size) {
for (int i = 0; i < size; i++)
cout << A[i] << " ";
cout << endl;
}
int merge(int arr[], int left[], int right[],int l_index, int m_index, int r_index) {
int i;
for (i = 0; i <= m_index - l_index; i++)
arr[i] = left[i];
for (int j = 0; j < r_index - m_index; j++)
arr[i + j] = right[j];
}
int divide(int arr[], int left[], int right[], int l_index, int m_index, int r_index) {
for (int i = 0; i <= m_index - l_index; i++)
left[i] = arr[i * 2];
for (int i = 0; i < r_index - m_index; i++)
right[i] = arr[i * 2 + 1];
}
int gen_worst_seq(int arr[], int l_index, int r_index) {
if (l_index < r_index) {
int m_index = l_index + (r_index - l_index) / 2;
int left[m_index - l_index + 1];
int right[r_index - m_index];
divide(arr, left, right, l_index, m_index, r_index);
gen_worst_seq(left, l_index, m_index);
gen_worst_seq(right, m_index + 1, r_index);
merge(arr, left, right, l_index, m_index, r_index);
}
}
int main() {
int arr[] = {11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26};
int n = sizeof(arr) / sizeof(arr[0]);
gen_worst_seq(arr, 0, n - 1);
display(arr, n);
}

## Input

11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26

## Output

11 19 15 23 13 21 17 25 12 20 16 24 14 22 18 26

Updated on: 28-Aug-2020

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