Hackerrank Almost Sorted Solution
.MathJax_SVG_Display {text-align: center; margin: 1em 0em; position: relative; display: block!important; text-indent: 0; max-width: none; max-height: none; min-width: 0; min-height: 0; width: 100%} .MathJax_SVG .MJX-monospace {font-family: monospace} .MathJax_SVG .MJX-sans-serif {font-family: sans-serif} .MathJax_SVG {display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 100%; font-size-adjust: none; text-indent: 0; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0; min-height: 0; border: 0; padding: 0; margin: 0} .MathJax_SVG * {transition: none; -webkit-transition: none; -moz-transition: none; -ms-transition: none; -o-transition: none} .mjx-svg-href {fill: blue; stroke: blue} .MathJax_SVG_LineBox {display: table!important} .MathJax_SVG_LineBox span {display: table-cell!important; width: 10000em!important; min-width: 0; max-width: none; padding: 0; border: 0; margin: 0}
Given an array of integers, determine whether the array can be sorted in ascending order using only one of the following operations one time.
- Swap two elements.
- Reverse one sub-segment.
Determine whether one, both or neither of the operations will complete the task. If both work, choose swap. For instance, given an array either swap the and , or reverse them to sort the array. Choose swap. The Output Format section below details requirements.
Function Description
Complete the almostSorted function in the editor below. It should print the results and return nothing.
almostSorted has the following parameter(s):
- arr: an array of integers
Input Format.MathJax_SVG_Display {text-align: center; margin: 1em 0em; position: relative; display: block!important; text-indent: 0; max-width: none; max-height: none; min-width: 0; min-height: 0; width: 100%} .MathJax_SVG .MJX-monospace {font-family: monospace} .MathJax_SVG .MJX-sans-serif {font-family: sans-serif} .MathJax_SVG {display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 100%; font-size-adjust: none; text-indent: 0; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0; min-height: 0; border: 0; padding: 0; margin: 0} .MathJax_SVG * {transition: none; -webkit-transition: none; -moz-transition: none; -ms-transition: none; -o-transition: none} .mjx-svg-href {fill: blue; stroke: blue} .MathJax_SVG_LineBox {display: table!important} .MathJax_SVG_LineBox span {display: table-cell!important; width: 10000em!important; min-width: 0; max-width: none; padding: 0; border: 0; margin: 0}
The first line contains a single integer , the size of .
The next line contains space-separated integers where .
Constraints.MathJax_SVG_Display {text-align: center; margin: 1em 0em; position: relative; display: block!important; text-indent: 0; max-width: none; max-height: none; min-width: 0; min-height: 0; width: 100%} .MathJax_SVG .MJX-monospace {font-family: monospace} .MathJax_SVG .MJX-sans-serif {font-family: sans-serif} .MathJax_SVG {display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 100%; font-size-adjust: none; text-indent: 0; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0; min-height: 0; border: 0; padding: 0; margin: 0} .MathJax_SVG * {transition: none; -webkit-transition: none; -moz-transition: none; -ms-transition: none; -o-transition: none} .mjx-svg-href {fill: blue; stroke: blue} .MathJax_SVG_LineBox {display: table!important} .MathJax_SVG_LineBox span {display: table-cell!important; width: 10000em!important; min-width: 0; max-width: none; padding: 0; border: 0; margin: 0}
All are distinct.
Output Format.MathJax_SVG_Display {text-align: center; margin: 1em 0em; position: relative; display: block!important; text-indent: 0; max-width: none; max-height: none; min-width: 0; min-height: 0; width: 100%} .MathJax_SVG .MJX-monospace {font-family: monospace} .MathJax_SVG .MJX-sans-serif {font-family: sans-serif} .MathJax_SVG {display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 100%; font-size-adjust: none; text-indent: 0; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0; min-height: 0; border: 0; padding: 0; margin: 0} .MathJax_SVG * {transition: none; -webkit-transition: none; -moz-transition: none; -ms-transition: none; -o-transition: none} .mjx-svg-href {fill: blue; stroke: blue} .MathJax_SVG_LineBox {display: table!important} .MathJax_SVG_LineBox span {display: table-cell!important; width: 10000em!important; min-width: 0; max-width: none; padding: 0; border: 0; margin: 0}
- If the array is already sorted, output yes on the first line. You do not need to output anything else.
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, and , output swap l r in the second line. and are the indices of the elements to be swapped, assuming that the array is indexed from to .
b. Otherwise, when reversing the segment , output reverse l r in the second line. and are the indices of the first and last elements of the subsequence to be reversed, assuming that the array is indexed from to .
represents the sub-sequence of the array, beginning at index and ending at index , both inclusive.
If an array can be sorted by either swapping or reversing, choose swap.
- If you cannot sort the array either way, output no on the first line.
Sample Input 1
2
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 in java8
Approach 1.
import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.regex.*;
public class Solution {
// Complete the almostSorted function below.
static void almostSorted(int[] arr) {
int count=0;
int[] a=new int[arr.length];
for(int i=0;i<arr.length;i++)
{
a[i]=arr[i];
}
Arrays.sort(a);
int[] c=new int[arr.length];int j=0;
for(int i=0;i<arr.length;i++)
{
if(a[i]!=arr[i]){
c[j]=i;j++;
count++;}
}
int k=j,count1=0;
if(count==2)
{
System.out.println("yes");
System.out.println("swap "+(c[0]+1)+" "+(c[1]+1));
}else
{for(int i=0,l=k-1;(i<=k-1)&&(l>=0);i++,l--){
if(arr[c[i]]==a[c[l]]){
count1++;
}}
if(count1==j){
System.out.println("yes");
System.out.println("reverse "+(c[0]+1)+" "+(c[j-1]+1));}
else
System.out.println("no");
}
}
private static final Scanner scanner = new Scanner(System.in);
public static void main(String[] args) {
int n = scanner.nextInt();
scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
int[] arr = new int[n];
String[] arrItems = scanner.nextLine().split(" ");
scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
for (int i = 0; i < n; i++) {
int arrItem = Integer.parseInt(arrItems[i]);
arr[i] = arrItem;
}
almostSorted(arr);
scanner.close();
}
}
Approach 2.
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);
}
}Approach 3.
import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.regex.*;
import java.util.stream.Collectors;
public class Solution {
// Complete the almostSorted function below.
static void almostSorted(int[] arr) {
int[] sortedArr = Arrays.stream(arr).sorted().toArray();
if(arr.equals(sortedArr)){
System.out.println("yes");
return;
}
int count = 0;
List<Integer> ins = new LinkedList<Integer>();
for(int i=0; i<arr.length; i++){
if(arr[i]!=sortedArr[i]){
count++;
ins.add(i);
}
}
if(count == 2){
System.out.println("yes");
System.out.println("swap " + Integer.valueOf(ins.get(0)+1) + " " + Integer.valueOf(ins.get(1)+1));
return;
}
ins = ins.stream().sorted().collect(Collectors.toList());
int a = ins.get(0);
int b = ins.get(ins.size()-1);
boolean sortable = true;
for(int i=a; i<b; i++){
if(arr[i+1]>arr[i]){
sortable = false;
break;
}
}
if(sortable){
System.out.println("yes");
System.out.println("reverse " + Integer.valueOf(a+1) + " " + Integer.valueOf(b+1));
}
else{
System.out.println("no");
}
}
private static final Scanner scanner = new Scanner(System.in);
public static void main(String[] args) {
int n = scanner.nextInt();
scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
int[] arr = new int[n];
String[] arrItems = scanner.nextLine().split(" ");
scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
for (int i = 0; i < n; i++) {
int arrItem = Integer.parseInt(arrItems[i]);
arr[i] = arrItem;
}
almostSorted(arr);
scanner.close();
}
}
Solution in python3
Approach 1.
n=int(input())
l=list(map(int,input().split()))
sl=sorted(l)
diffcount = 0
diff1 = -1
diff2 = -1
for i in range(len(l)):
if sl[i] != l[i]:
diffcount += 1
if diff1 == -1:
diff1 = i
elif diffcount > 1:
diff2 = i
lastdiff = i
if diffcount == 2:
l[diff1], l[diff2] = l[diff2], l[diff1]
if l == sl:
print("yes")
print("swap {} {}".format(diff1 + 1, diff2 + 1))
else:
print("no")
elif diffcount > 2:
l = l[:diff1] + l[diff1:diff2 + 1][::-1] + l[diff2 + 1:]
if l == sl:
print("yes")
print("reverse {} {}".format(diff1 + 1, diff2 + 1))
else:
print("no")
elif l == sl:
print("yes")
Approach 2.
#!/bin/python3
import math
import os
import random
import re
import sys
# Complete the almostSorted function below.
def almostSorted(arr):
sArr = sorted(arr)
misplaced = []
for i in range(len(arr)):
if sArr[i] != arr[i]:
misplaced.append(i)
if(len(misplaced)==0):
print('yes')
elif len(misplaced)==2:
print('yes')
print('swap',misplaced[0]+1, misplaced[1]+1)
else:
if sArr[misplaced[0]:misplaced[len(misplaced)-1]+1] == list(reversed(arr[misplaced[0]:misplaced[len(misplaced)-1]+1])):
print('yes')
print('reverse', misplaced[0]+1,misplaced[len(misplaced)-1]+1)
else:
print('no')
if __name__ == '__main__':
n = int(input())
arr = list(map(int, input().rstrip().split()))
almostSorted(arr)
Approach 3.
#!/bin/python3
import math
import os
import random
import re
import sys
def swapped(arr, i, j):
arr=arr[:]
arr[i], arr[j]=arr[j], arr[i]
return arr
def reverse(arr, i, j):
aa=arr[:i]+list(reversed(arr[i:j+1]))
if j!=len(arr)-1 :
aa+=arr[j+1:]
return aa
# Complete the almostSorted function below.
def almostSorted(arr):
sar=sorted(arr)
i, j = 0, len(arr)-1
while arr[i]<=arr[i+1]:i+=1
while arr[j-1]<=arr[j]:j-=1
if j<i:
print ("yes")
elif swapped(arr, i, j)==sar:
print ("""yes
swap %d %d
""" % (i+1, j+1))
elif reverse(arr, i, j)==sar:
print ("""yes
reverse %d %d
""" % (i+1, j+1))
else:
print("no")
if __name__ == '__main__':
n = int(input())
arr = list(map(int, input().rstrip().split()))
almostSorted(arr)
Solution in cpp
Approach 1.
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
int n;
cin >> n;
vector<int> v(n + 1);
for (int z = 1; z < n + 1; z++) {
cin >> v[z];
}
int i = 0, s = 0;
for (i = 1; i <= n - 1; i++) {
if (v[i] > v[i + 1]) {
s = i;
break;
}
}
int j, e;
for (j = i + 1; j <= n; j++) {
if (v[s] < v[j]) {
j -= 1;
break;
}
}
e = j;
if (j > n) {
e -= 1;
}
int k;
for (k = e + 1; k < n - 1; k++) {
if (v[k] > v[k + 1]) {
cout << "no" << endl;
return 0;
}
}
if (s - 1 >= 0 && v[s - 1] > v[e] ) {
cout << "no" << endl;
return 0;
}
else if (e - s == 1 && v[s] > v[e]) {
cout << "yes" << endl;
cout << "swap" << " " << s << " " << e << endl;
}
else if (v[e - 1] > v[e]) {
cout << "yes" << endl;
for (k = e - 1; k >= s + 1; k--) {
if (v[k - 1] < v[k]) {
cout << "swap" << " " << s << " " << e << endl;
return 0;
}
}
cout << "reverse" << " " << s << " " << e << endl;
}
else {
cout << "no" << endl;
return 0;
}
return 0;
}
Approach 2.
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
bool checksorted(vector <long> a){
long last = -1;
for(int i=0;i<a.size();i++){
//cout << i<<"/"<<a.size() <<" : " << a[i]<<endl;
if(a[i] < last){
return false;
}
last = a[i];
}
return true;
}
int main() {
int n,last(-1),l(-1),r(0);
vector <int> falls;
long temp;
cin>>n;
vector <long> a(n);
//bool sorted = true;
for(int i=0;i<n;i++){
cin>>a[i];
if(a[i]<last && l==-1){
l = i-1;
}
if(a[i]<last && l!=-1 && i>r ){
r = i;
}
last = a[i];
}
if(l==-1){
cout << "yes"<<endl;
return 0;
}
vector <long> a2(a);
a2[l] = a[r];
a2[r] = a[l];
if(checksorted(a2)){
cout << "yes"<<endl<<"swap "<<l+1<<" "<<r+1<<endl;
return 0;
}
for(int i=0;i< (r-l) ;i++){
a2[l+i]=a[r-i];
}
if(checksorted(a2)){
cout << "yes"<<endl<<"reverse "<<l+1<<" "<<r+1<<endl;
return 0;
}
cout << "no"<<endl;
return 0;
}
Approach 3.
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
long n,a[100000],flag[100000],i,j=0,min;
cin>>n;
for(i=0;i<n;i++){
cin>>a[i];
if(i==0)
min=a[0];
else{
if(min>=a[i]){
flag[j]=i;
j++;
}
min=a[i];
}
}
//cout<<"j="<<j<<endl;
if(j>2){
long first=flag[0],count=1;
for(i=1;i<j;i++){
//cout<<flag[i]<<" ";
if(flag[i]==first+1){
first=flag[i];
count++;
}
else{
cout<<"no";
return 0;
}
}
cout<<"yes\n"<<"reverse "<<flag[0]<<" "<<flag[j-1]+1;
}
else if(j==1){
if(n>2){
j=flag[0];
if(a[j]>a[j-2])
cout<<"yes\n"<<"swap "<<j<<" "<<j+1;
else
cout<<"no";
}
else
cout<<"yes\n"<<"swap "<<1<<" "<<2;
}
else{
cout<<"yes\n"<<"swap "<<flag[0]<<" "<<flag[1]+1;
}
return 0;
}