# PAT Advanced 1098 Insertion or Heap Sort: O(N) Time Without Simulation

## Overview

PAT Advanced 1098 Insertion or Heap Sort: O(N) Time Without Simulation, the challenge here is to find the next sorted sequences that differs from previous.

## PAT Advanced 1098 Insertion or Heap Sort

According to Wikipedia:

Insertion sort iterates, consuming one input element each repetition, and growing a sorted output list. Each iteration, insertion sort removes one element from the input data, finds the location it belongs within the sorted list, and inserts it there. It repeats until no input elements remain.

Heap sort divides its input into a sorted and an unsorted region, and it iteratively shrinks the unsorted region by extracting the largest element and moving that to the sorted region. it involves the use of a heap data structure rather than a linear-time search to find the maximum.

Now given the initial sequence of integers, together with a sequence which is a result of several iterations of some sorting method, can you tell which sorting method we are using?

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=100). Then in the next line, N integers are given as the initial sequence. The last line contains the partially sorted sequence of the N numbers. It is assumed that the target sequence is always ascending. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in the first line either “Insertion Sort” or “Heap Sort” to indicate the method used to obtain the partial result. Then run this method for one more iteration and output in the second line the resuling sequence. It is guaranteed that the answer is unique for each test case. All the numbers in a line must be separated by a space, and there must be no extra space at the end of the line.

Sample Input 1:

```10
3 1 2 8 7 5 9 4 6 0
1 2 3 7 8 5 9 4 6 0```

Sample Output 1:

```Insertion Sort
1 2 3 5 7 8 9 4 6 0```

Sample Input 2:

```10
3 1 2 8 7 5 9 4 6 0
6 4 5 1 0 3 2 7 8 9```

Sample Output 2:

```Heap Sort
5 4 3 1 0 2 6 7 8 9```

## Solution Analysis: O(N) Time Without Simulation

The easiest way to simulate the whole process of Insertion Sort and Heap Sort step by step and see if the partially sorted input is equal to any of the partial results from Insertion Sort or Heap Sort, if so, then continue perform the correct sort to output that result. This cost O(N^2) time. This could work because the data size is small.

But I won’t be satisfied about that approach, so I come up with the following O(N) solution. The key observation here is that:

1. Insertion Sort can be determined in linear time to check both the sorted portion and the unsorted portion
2. We can determine the correct position of the unsorted portion in linear time if it is determined as heap sort, and perform pop_heap opertion in O(logN) time

I became a bit lazy after coming up the solution, so I directly use the std::sort which internally adopts quick sort algorithm. But the truth is this sort could be implemented as insertion sort and it will run in O(N) time if the input is already sorted except the last element. So the following could runs indeed in O(N) time assume std::sort is implemented in Insertion Sort (which is absolutely possible).

```#include <vector>
#include <algorithm>
#include <unordered_set>
#include <cstdio>
#include <cassert>

using std::vector;
using std::make_heap;
using std::pop_heap;
using std::sort;

void printV(const vector<int> &v) {

printf("%d", v[0]);
for(int i = 1; i < v.size(); ++i) {
printf(" %d", v[i]);
}
printf("\n");

}

int checkIfInsertionSort(const vector<int> &origin, const vector<int> &partial) {

int i;
for (i = origin.size() - 1; i >=0; --i) {
if (origin[i] != partial[i]) break;
}

if (std::is_sorted(partial.begin(), partial.begin() + i + 1))
return i;
return -2;
}

int getIndex(const vector<int> &v) {
for (int i = v.size() - 1; i >= 0; --i) {
if (v[i] < v[0]) return i;
}
}

void performHeapSort(vector<int> &v) {
int i = getIndex(v);
pop_heap(v.begin(), v.begin() + i + 1);
}

int main() {
int n;
vector<int> original;
vector<int> partial;
scanf("%d", &n);

original.resize(n);
partial.resize(n);
for (int i = 0; i < n; ++i) scanf("%d", &original[i]);
for (int i = 0; i < n; ++i) scanf("%d", &partial[i]);

int r = checkIfInsertionSort(original, partial);

if (r != -2) {
printf("Insertion Sort\n") ;

int t = r + 2;
while (t <= partial.size()
&& partial[t -1] >= partial[t - 2])
++t;
sort(partial.begin(), partial.begin() + t);
printV(partial);
} else {
printf("Heap Sort\n") ;
performHeapSort(partial);
printV(partial);
}
}

```

In order to pass the PAT OJ for this 1098 Insertion or Heap Sort problem, just for you information:

1. There are 6 test cases, test case 0, 2 , 4 are insertion sort, test case 1, 3, 5 are heap sort

## Summary

PAT Advanced 1098 Insertion or Heap Sort: O(N) Time Without Simulation, the challenge here is to find the next sorted sequences that differs from previous.

Written on June 6, 2015