Hello Programmers, In this post, you will know how to solve the HackerRank Sequence Equation 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 Sequence Equation Solution
Task
Given a sequence of n integers, p(1), p(2), . . . , p(n) where each element is distinct and satisfies 1 <= p(x) <= n. For eachx where 1 <= x <= n, that is x increments from 1 to n, find any integer y such that p(p(y)) = x and keep a history of the values of y in a return array.
Example
p = [5, 2, 1, 3, 4]
Each value of x between 1 and 5, the length of the sequence, is analyzed as follows:
- x = 1 = p[3],p[4] = 3, so p[p[4]] = 1
- x = 2 = p[2],p[2] = 2, so p[p[2]] = 2
- x = 3 = p[4],p[5] = 4, so p[p[5]] = 3
- x = 4 = p[5],p[1] = 5, so p[p[1]] = 4
- x = 5 = p[1],p[3] = 1, so p[p[3]] = 5
The values for y are [4, 2, 5, 1, 3].
Function Description
Complete the permutationEquation function in the editor below.
permutationEquation has the following parameter(s):
- int p[n]: an array of integers
Returns
- int[n]: the values of y for all x in the arithmetic sequence 1 to n
Input Format
The first line contains an integer n, the number of elements in the sequence.
The second line contains n space–separated integers p[i] where 1 <= i <= n.
Constraints
- 1 <= n <= 50
- 1 <= p[i] <= 50, where 1 <= i <= n.
- Each element in the sequence is distinct.
Sample Input 0
3 2 3 1
Sample Output 0
2 3 1
Explanation 0
Given the values of p(1) = 2, p(2) = 3, and p(3) = 1, we calculate and print the following values for each x from 1 to n:
- x = 1 = p(3) = p(p(2)) = p(p(y)), so we print the value of y = 2 on a new line.
- x = 2 = p(1) = p(p(3)) = p(p(y)), so we print the value of y = 3 on a new line.
- x = 3 = p(2) = p(p(1)) = p(p(y)), so we print the value of y =1 on a new line.
Sample Input 1
5 4 3 5 1 2
Sample Output 1
1 3 5 4 2
HackerRank Sequence Equation Solution
Sequence Equation Solution in C
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>
int main() {
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
int i,n,a[100],j;
scanf("%d",&n);
for(i=1;i<=n;i++)
scanf("%d",&a[i]);
for(j=1;j<=n;j++)
{
for(i=1;i<=n;i++)
if(a[a[i]]==j)
break;
printf("%d\n",i);
}
return 0;
}Sequence Equation Solution in Cpp
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
#ifdef ONPC
freopen("a.in", "r", stdin);
//freopen("a.out", "w", stdout);
#else
//freopen("a.in", "r", stdin);
//freopen("a.out", "w", stdout);
#endif
ios::sync_with_stdio(0);
int n;
cin >> n;
vector <int> p(n + 1);
for (int i = 1; i <= n; i++)
{
cin >> p[i];
}
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= n; j++)
{
if (p[p[j]] == i)
{
cout << j << endl;
break;
}
}
}
}Sequence Equation Solution in Java
import java.io.*;
import java.util.*;
class Main {
InputReader in;
PrintWriter out;
void main() {
int n = in.nextInt();
int nxt[] = new int[n];
for (int i = 0; i < n; ++i) {
int y = in.nextInt();
nxt[--y] = i;
}
for (int i = 0; i < n; ++i)
out.println(nxt[nxt[i]] + 1);
}
public static void main(String[] args) {
new Main();
}
public Main() {
in = new InputReader(System.getProperty("ONLINE_JUDGE") != null ? null : "main.inp");
out = new PrintWriter(System.out);
main();
out.close();
}
class InputReader {
BufferedReader bf;
StringTokenizer st = null;
InputReader(String filename) {
try {
bf = new BufferedReader(filename == null
? new InputStreamReader(System.in)
: new FileReader(filename)
);
} catch (IOException e) {
e.printStackTrace();
System.err.println("use stdin instead");
bf = new BufferedReader(new InputStreamReader(System.in));
}
}
String nextString() {
try {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(bf.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
return st.nextToken();
}
long nextLong() {return Long.parseLong(nextString());}
int nextInt() {return Integer.parseInt(nextString());}
double nextDouble() {return Double.parseDouble(nextString());}
}
}
Sequence Equation Solution in Python
rr = raw_input
rrM = lambda: map(int, rr().split())
N = int(rr())
A = rrM()
A = [x-1 for x in A]
B = [0] * N
for i,u in enumerate(A):
B[i] = A[u]
C = [0]*N
for i,u in enumerate(B):
C[u] = i+1
for x in C:
print xSequence Equation Solution using JavaScript
function processData(input) {
var lines = input.split("\n");
var data = lines[1].split(' ').map(Number);
var all = {};
for(var i=0;i<data.length;i++){
all[data[i]] = i+1;
}
for(var i=1;i<=data.length;i++){
console.log(all[all[i]]);
}
}
process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
_input += input;
});
process.stdin.on("end", function () {
processData(_input);
});Sequence Equation Solution in Scala
// main method in "Solution" will be run as your answer
object Solution {
def main(args: Array[String]) {
//Enter your code here. Read input from STDIN. Print output to STDOUT
val n = io.StdIn.readLine().toInt
val px = io.StdIn.readLine().split(" ").toList.map(_.toInt)
val revP = px.zip(1 to n).toMap
(1 to n) map(revP) map(revP) foreach(println)
}
}Sequence Equation Solution in Pascal
VAR i,n,j,l:integer; a:array[1..50] of integer; Begin Readln(n); for i:=1 to n do Begin Read(a[i]); end; for i:=1 to n do Begin for j:=1 to n do Begin if i = a[j] then Begin for l:=1 to n do Begin if a[l] = j then Writeln(l); end; end; end; end; Readln; END.
Disclaimer: This problem (Sequence Equation) is generated by HackerRank but the Solution is Provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.