Inverse Permutation (GFG)

 Given an array A of size n of integers in the range from 1 to n, we need to find the inverse permutation of that array.

Inverse Permutation is a permutation which you will get by inserting position of an element at the position specified by the element value in the array. For better understanding, consider the following example:
Suppose we found element 4 at position 3 in an array, then in reverse permutation, we insert 3 (position of element 4 in the array) in position 4 (element value).

Input:
The first line of the input contains an integer T denoting the number of test cases. For each test case, the first line contains an integer n, denoting the size of the array A  followed by n-space separated integers i.e elements of array A.

Output:
For each test case, the output is the array after performing inverse permutation on A.

Constraints:
1<=T<=100
1<=n<=50
1<=A[i]<=50
Note: Array should contain unique elements and should have elements from 1 to n. 

Example:
Input:
3
4
1 4 3 2
5
2 3 4 5 1
5
2 3 1 5 4

Output:
1 4 3 2
5 1 2 3 4
3 1 2 5 4

Explanation:
1. For element 1 we insert position of 1 from arr1 i.e 1 at position 1 in arr2. For element 4 in arr1, we insert 2 from arr1 at position 4 in arr2. Similarly, for element 2 in arr1, we insert position of 2 i.e 4 in arr2.
2. As index 1 has value 2 so 1 will b placed at index 2, similarly 2 has 3 so 2 will be placed at index 3 similarly 3 has 4 so placed at 4, 4 has 5 so 4 placed at 5 and 5 has 1 so placed at index 1.   

Solution:

t=int(input())

for i in range (t):

    n=int(input())

    a=list(map(int,input().split()))

    b=sorted(a)

    for i in range(n):

        for j in range(n):

            if b[i]==a[j]:

                print(j+1,end=" ")

    print()

    


or

t=int(input())

for i in range(t):

    n=int(input())

    a=list(map(int,input().split()))

    l=[0]*n

    for i in range(n):

        x=a.index(a[i])

        l[a[i]-1]=x+1

    print(*l,end="")

    print()

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