Sequence Equation


Problem Statement :


Given a sequence of n integers, p(1), p(2),......, P(n)  where each element is distinct and satisfies 1 <= p(x) <= n. For each x 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 u 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:

1.  x = 1 = p[3], p[4] = 3, so p[p[4]] = 1
2.  x = 2 = p[2], p[2] = 2, so p[p[2]] = 2
3.  x = 3 = p[4], p[5] = 4, so p[p[5]] = 3
4.  x = 4 = p[5], p[1] = 5, so p[p[1]] = 4
5.  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.


Solution :



title-img 


                            Solution in C :

python 3  :

n = int(input())
ps = [int(x) for x in input().split()]

for x in range(1, n+1):
    for y in range(n):
        if ps[ps[y]-1] == x:
            print(y+1)
            break











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[] ) throws Exception {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int arr[]=new int[n];
        for(int i = 0 ;i<n;i++)
            arr[i]=sc.nextInt();
        for(int i = 0 ;i<n;i++)
            {
            int pos = 0;
            for(int j = 0 ; j<n;j++)
                {
                if(arr[j]==i+1)
                {
                pos = j+1 ;
                    break;
                }
              }
            int pos1=0;
            for(int j = 0 ; j<n;j++)
                {
                if(arr[j]==pos)
                {
                pos1 = j ;break;
            }
            }
            System.out.println(pos1+1);
        }
    }
}












C ++  :

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
typedef pair<int, int> II;

int main() {
    #ifdef LOCAL
        freopen("Data.inp", "r", stdin);
        freopen("Data.out", "w", stdout);
    #endif

    int n, a[100];
    cin >> n;
    for (int i = 1; i <= n; ++i) cin >> a[i];

    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) if (a[a[j]] == i) {
            cout << j << endl;
            break;
        }
    }

    return 0;
}










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;
}








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