Hello Programmers, In this post, you will know how to solve the HackerRank Larrys Array Solution. This problem is a part of the HackerRank Algorithms Series.
One more thing to add, don’t directly look for the solutions, first try to solve the problems of Hackerrank by yourself. If you find any difficulty after trying several times, then you can look for solutions.
HackerRank Larrys Array Solution
Task
Larry has been given a permutation of a sequence of natural numbers incrementing from 1 as an array. He must determine whether the array can be sorted using the following operation any number of times:
- Choose any 3 consecutive indices and rotate their elements in such a way that ABC -> BCA -> CAB -> ABC.
For example, if A = {1, 6, 5, 2, 4, 3}:
A rotate [1,6,5,2,4,3] [6,5,2] [1,5,2,6,4,3] [5,2,6] [1,2,6,5,4,3] [5,4,3] [1,2,6,3,5,4] [6,3,5] [1,2,3,5,6,4] [5,6,4] [1,2,3,4,5,6] YES
On a new line for each test case, print YES if can be fully sorted. Otherwise, print NO.
Function Description
Complete the larrysArray function in the editor below. It must return a string, either YES or NO.
larrysArray has the following parameter(s):
- A: an array of integers
Input Format
The first line contains an integer t, the number of test cases.
The next t pairs of lines are as follows:
- The first line contains an integer n, the length of A.
- The next line contains n space-separated integers A[i].
Constraints
- 1 <= t <= 10
- 3 <= n <= 1000
- 1 <= A[i] <= n
- Asorted = integers that increment by 1 from 1 to n
Output Format
For each test case, print YES if A can be fully sorted. Otherwise, print NO.
Sample Input
3
3
3 1 2
4
1 3 4 2
5
1 2 3 5 4
Sample Output
YES
YES
NO
Explanation
In the explanation below, the subscript of A denotes the number of operations performed.
Test Case 0:
A0 = {3, 1, 2} -> rotate(3, 1, 2) -> A1 = {1, 2, 3}
A is now sorted, so we print YES on a new line.
Test Case 1:
A0 = {1, 3, 4, 2} -> rotate(3, 4, 2) -> A1 = {1, 4, 2, 3}.
A1 = {1, 4, 2, 3} -> rotate(4, 2, 3) -> A2 = {1, 2, 3, 4}.
A is now sorted, so we print YES on a new line.
Test Case 2:
No sequence of rotations will result in a sorted A. Thus, we print NO on a new line.
HackerRank Larrys Array Solution
Larrys Array Solution in C
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
unsigned t;
scanf ("%u", &t);
while (t--) {
unsigned n;
scanf ("%u", &n);
unsigned a[n];
for (unsigned i = 0; i < n; i++) {
scanf ("%u", a + i);
}
unsigned int num_inv = 0;
for (unsigned i = 0; i < n ; i++) {
for (unsigned j = i + 1; j < n; j++) {
if (a[i] > a[j]) {
num_inv++;
}
}
}
if (num_inv % 2) {
printf ("NO\n");
} else {
printf ("YES\n");
}
}
return 0;
}Larrys Array Solution in Cpp
#include <string>
#include <vector>
#include <algorithm>
#include <numeric>
#include <set>
#include <map>
#include <queue>
#include <iostream>
#include <sstream>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstring>
#include <cctype>
#include <cassert>
#include <limits>
#include <functional>
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define rer(i,l,u) for(int (i)=(int)(l);(i)<=(int)(u);++(i))
#define reu(i,l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i))
#if defined(_MSC_VER) || __cplusplus > 199711L
#define aut(r,v) auto r = (v)
#else
#define aut(r,v) __typeof(v) r = (v)
#endif
#define each(it,o) for(aut(it, (o).begin()); it != (o).end(); ++ it)
#define all(o) (o).begin(), (o).end()
#define pb(x) push_back(x)
#define mp(x,y) make_pair((x),(y))
#define mset(m,v) memset(m,v,sizeof(m))
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3fLL
using namespace std;
typedef vector<int> vi; typedef pair<int, int> pii; typedef vector<pair<int, int> > vpii; typedef long long ll;
template<typename T, typename U> inline void amin(T &x, U y) { if(y < x) x = y; }
template<typename T, typename U> inline void amax(T &x, U y) { if(x < y) x = y; }
int main() {
int T;
scanf("%d", &T);
for(int ii = 0; ii < T; ++ ii) {
int N;
scanf("%d", &N);
vector<int> A(N);
for(int i = 0; i < N; ++ i)
scanf("%d", &A[i]);
int p = 0;
rep(i, N) rep(j, i)
p += A[i] < A[j];
bool ans = p % 2 == 0;
puts(ans ? "YES" : "NO");
}
return 0;
}
Larrys Array Solution in Java
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();
int n;
List<Integer> a;
for (int i = 0; i < t; i++) {
n = in.nextInt();
a = new ArrayList<Integer>(n);
for (int j = 0; j < n; j++) {
a.add(in.nextInt());
}
for (int curElem = 1; curElem <= n - 2; curElem++) {
int curIdx = a.indexOf(curElem);
System.err.println(a);
while (curIdx != curElem - 1) {
if (curIdx == a.size() - 1) {
int tmp = a.get(curIdx - 2);
a.set(curIdx - 2, a.get(curIdx - 1));
a.set(curIdx - 1, curElem);
a.set(curIdx, tmp);
} else {
int tmp = a.get(curIdx - 1);
a.set(curIdx - 1, curElem);
a.set(curIdx, a.get(curIdx + 1));
a.set(curIdx + 1, tmp);
}
curIdx--;
}
}
System.err.println(a);
boolean printed = false;
for (int j = 1; j < a.size(); j++) {
if (a.get(j) < a.get(j-1)) {
System.out.println("NO");
printed = true;
}
}
if (!printed) System.out.println("YES");
}
}
}Larrys Array Solution in Python
# Enter your code here. Read input from STDIN. Print output to STDOUT
x=int(raw_input())
for _ in range(x):
raw_input()
arr=[int(i) for i in raw_input().split()]
count=0
for ind,val in enumerate(arr):
count+=sum(i>val for i in arr[:ind])
if count%2==0:
print "YES"
else:
print "NO"Larrys Array Solution using JavaScript
process.stdin.resume();
process.stdin.setEncoding('ascii');
var input_stdin = "";
var input_stdin_array = "";
var input_currentline = 0;
process.stdin.on('data', function (data) {
input_stdin += data;
});
process.stdin.on('end', function () {
input_stdin_array = input_stdin.split("\n");
main();
});
function readLine() {
return input_stdin_array[input_currentline++];
}
/////////////// ignore above this line ////////////////////
function main() {
var T = +readLine();
while(T--) {
var N = +readLine();
var P = readLine().split(' ');
var C = 0,i,j,t;
//console.log('>>',P);
for(i=0;i<P.length;i++) {
if (P[i]!=i+1) {
for(j=i+1;j<P.length;j++) if (P[j]==i+1) {
break;
};
while(j>i) {
t = P[j-1];
P[j-1] = P[j];
P[j] = t;
//console.log(P);
j--;
C++;
};
};
};
//console.log('--',C);
if (C%2==0) {
console.log('YES');
} else {
console.log('NO');
}
};
}
Larrys Array Solution in Scala
object Solution {
def main(args: Array[String]) {
val sc = new java.util.Scanner(System.in)
val t = sc.nextInt()
for (_ <- 1 to t) {
val n = sc.nextInt
val ar = (1 to n).map(_ => sc.nextInt())
if (isSolvable(ar.toList)) {
println("YES")
} else {
println("NO")
}
}
}
def isSolvable(ar: List[Int]):Boolean = {
ar.tails.filter(_.length > 1).map({
case h::t => t.count(_ < h)
case _ => 0
}).sum % 2 == 0
}
}
Larrys Array Solution in Pascal
type
TArr = array of integer;
var
Tests, i: integer;
a: TArr;
n, j, k, temp, Start, Min: integer;
function MinN: integer;
var
i: integer;
begin
MinN := start+1;
for i := Start +1 to n-1 do
if a[i] < a[MinN] then MinN := i;
end;
function Sort: boolean;
var
i: integer;
begin
Sort := true;
for i := 0 to n-2 do
if a[i] > a[i+1] then Sort := false;
end;
begin
readln(Tests);
for i := 1 to Tests do
begin
readln(n);
SetLength(a, n);
for j := 0 to n-1 do
read(a[j]);
// 2 1 0 3
// Находим минимальный элемент
k := 0;
Start := -1;
Min := MinN;
while (start < n-3)do
begin
{
k := k + 1;
if k > n-start then break;
}
if Min = Start +1 then
begin
start := start + 1;
Min := MinN;
k := 0;
end
else
begin
if min < n-1 then
begin
temp := a[min-1];
a[min-1] := a[min];
a[min] := a[min+1];
a[min+1] := temp;
Min := Min - 1;
end
else
begin
temp := a[min-2];
a[min-2] := a[min];
a[min] := a[min-1];
a[min-1] := temp;
Min := Min-2;
end;
end;
end;
if Sort then writeln('YES') else writeln('NO');
end;
end.Disclaimer: This problem (Larrys Array) is generated by HackerRank but the Solution is Provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.