Almost Sorted – HackerRank Solution

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In this post, we will solve Almost Sorted HackerRank Solution. This problem (Almost Sorted) is a part of HackerRank Problem Solving series.

Task

Given an array of integers, determine whether the array can be sorted in ascending order using only one of the following operations one time.

  1. Swap two elements.
  2. Reverse one sub-segment.

Determine whether one, both or neither of the operations will complete the task. Output is as follows.

  1. If the array is already sorted, output yes on the first line. You do not need to output anything else.
  2. If you can sort this array using one single operation (from the two permitted operations) then output yes on the first line and then:
    • If elements can only be swapped, d[l] and d[r], output swap l r in the second line. l and r are the indices of the elements to be swapped, assuming that the array is indexed from 1 to n.
    • If elements can only be reversed, for the segment d[l . . . r], output reverse l r in the second line. l and r are the indices of the first and last elements of the subarray to be reversed, assuming that the array is indexed from 1 to n. Here d[l . . . r] represents the subarray that begins at index l and ends at index r, both inclusive.

If an array can be sorted both ways, by using either swap or reverse, choose swap.

  1. If the array cannot be sorted either way, output no on the first line.

Example

arr = [2, 3, 5, 4]

Either swap the 4 and 5 at indices 3 and 4, or reverse them to sort the array. As mentioned above, swap is preferred over reverse. Choose swap. On the first line, print yes. On the second line, print swap 3 4.

Function Description

Complete the almostSorted function in the editor below.

almostSorted has the following parameter(s):

  • int arr[n]: an array of integers

Prints

  • Print the results as described and return nothing.

Input Format

The first line contains a single integer n, the size of arr.
The next line contains n space-separated integers arr[i] where 1 <= i <= n.

Constraints

  • 2 <= n <= 100000
  • 0 <= arr[i] <= 1000000
  • All arr[i] are distinct.

Output Format

  1. If the array is already sorted, output yes on the first line. You do not need to output anything else.
  2. If you can sort this array using one single operation (from the two permitted operations) then output yes on the first line and then:
    a. If elements can be swapped, d[l] and d[r], output swap l r in the second line. l and r are the indices of the elements to be swapped, assuming that the array is indexed from 1 to n.
    b. Otherwise, when reversing the segment d[l . . . r], output reverse l r in the second line. l and r are the indices of the first and last elements of the subsequence to be reversed, assuming that the array is indexed from 1 to n. 
    d[l . . . r]represents the sub-sequence of the array, beginning at index l and ending at index r, both inclusive.

If an array can be sorted by either swapping or reversing, choose swap.

  1. If you cannot sort the array either way, output no on the first line.

Sample Input 1

STDIN   Function
-----   --------
2       arr[] size n = 2
4 2     arr = [4, 2]

Sample Output 1

yes  
swap 1 2

Explanation 1

You can either swap(1, 2) or reverse(1, 2). You prefer swap.

Sample Input 2

3
3 1 2

Sample Output 2

no

Explanation 2

It is impossible to sort by one single operation.

Sample Input 3

6
1 5 4 3 2 6

Sample Output 3

yes
reverse 2 5

Explanation 3

You can reverse the sub-array d[2…5] = “5 4 3 2”, then the array becomes sorted.

Solution – Almost Sorted – HackerRank Solution

C++

#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <cstdlib>
#include <cassert>
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
// Sort and compare.
// Do as you are told.
int main()
{
    int n;
    int arr[100000];
    scanf("%d",&n);
    for (int i=0;i<n;i++)
        scanf("%d",&arr[i]);
    
    int sorted[100000];
    for (int i=0;i<n;i++)
        sorted[i]=arr[i];
    sort(sorted,sorted+n);
    vector<int> diff;
    for (int i=0;i<n;i++)
        if (sorted[i]!=arr[i])
            diff.push_back(i);
    if (diff.size()==0)
        printf("yes\n");  
    else if (diff.size()==2)
        printf("yes\nswap %d %d\n",diff[0]+1,diff[1]+1); //Index starts at 1!!!
    else //Let's try reverse it!
    {
        int st = diff[0], ed = diff[diff.size()-1];
        while (st<ed)
        {
            swap(arr[st],arr[ed]);
            st++;
            ed--;
        }
        int flag=1;
        for (int i=0;i<n;i++)
            if (sorted[i]!=arr[i])
            {
                printf("no\n");  
                return 0;
            }
        printf("yes\nreverse %d %d\n",diff[0]+1,diff[diff.size()-1]+1); //Index starts at 1!!!
    }
}

Python

import sys

def is_sorted(arr):
    return all(arr[i] <= arr[i+1] for i in range(len(arr)-1))

def almostSorted(arr):
    swap_l = -1
    swap_r = -1
    for ind in range(1, len(arr)):
        if arr[ind - 1] > arr[ind]:
            swap_l = ind - 1
            break
            
    for ind in range(swap_l + 1, len(arr)):
        if ind == len(arr) - 1 or arr[ind + 1] > arr[swap_l]:
            swap_r = ind
            arr[swap_l], arr[swap_r] = arr[swap_r], arr[swap_l]
            break
            
    if is_sorted(arr):
        print("yes")
        print("swap {} {}".format(swap_l + 1, swap_r + 1))
        return True
    
    arr[swap_l], arr[swap_r] = arr[swap_r], arr[swap_l]
    
    rev_l = -1
    rev_r = -1
    for ind in range(len(arr) - 1):
        if rev_l == -1 and arr[ind] > arr[ind + 1]:
            rev_l = ind
        elif rev_l != -1 and arr[ind] < arr[ind + 1]:
            rev_r = ind
            break
    
    to_rev = arr[rev_l:rev_r+1]
    arr = arr[:rev_l] + to_rev[::-1] + arr[rev_r+1:]
    
    if is_sorted(arr):
        print("yes")
        print("reverse {} {}".format(rev_l + 1, rev_r + 1))
        return True
    
    print("no")
    return False

if __name__ == "__main__":
    n = int(input().strip())
    arr = list(map(int, input().strip().split(' ')))
    almostSorted(arr)

Java

import java.io.*;

public class Solution {
  public static void main(String[] args) throws IOException {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

    //Get input
    final int N = Integer.parseInt(br.readLine());
    final int[] arr = new int[N];
    String[] line = br.readLine().split(" ");
    for(int i = 0; i < N; ++i){
      arr[i] = Integer.parseInt(line[i]);
    }
    
    //Print output
    System.out.print(solve(arr, N));
  }
  
  private static String solve(final int[] A, final int N){
    int l = 0;
    int r = N - 1;
    
    //Check for out of place index from the left
    while(l < r && A[l] <= A[l+1]){
      ++l;
    }
    
    //Check if array already sorted
    if(l == r){
      return "yes";
    }
    
    //Check for out of place index from the right
    while(r > l && A[r] >= A[r-1]){
      --r;
    }
    
    //Check if swapping or reversing would NOT sort the array
    if((l > 0 && A[r] < A[l-1]) || (r < N-1 && A[l] > A[r+1])){
      return "no";
    }
        
    //Check if we're dealing with a reversal
    int m;
    for(m = l+1; m < r && A[m] >= A[m+1]; ++m){}
    if(m == r){
      return "yes\n" + ((r-l < 2) ? "swap " : "reverse ") + (l+1) + " " + (r+1);
    }
    
    //Check if we're NOT dealing with a swap
    if(m-l > 1 || A[l] < A[r-1] || A[r] > A[l+1]){
      return "no";
    }
    
    //Check if we're dealing with a swap
    for(int k = r-1; m < k && A[m] <= A[m+1]; ++m){}
    return (r-m > 1) ? "no" : "yes\nswap " + (l+1) + " " + (r+1);
  }
}

Note: This problem (Almost Sorted) is generated by HackerRank but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.

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