## Replace for X  codechef October long challenge solution -

This Problem is taken from CodeChef October long challenge.

## Problem Statement-

You are given an array of $N$ integers ${A}_{1},{A}_{2},\dots ,{A}_{N}$ and three more integers $X,p,$ and $k$.

An operation on the array is defined to be: replace the $k$-th smallest integer in the array with any integer you want.

Output the minimum number of operations you must perform on the array (possibly $0$ operations) to make the $p$-th smallest element of the array equal to $X$. If it is impossible to do so output $-1$.

Note: the $k$-th smallest number in an array is the $k$-th number from the left when the array is sorted in non-decreasing order.

### Input

• The first line of the input contains a single integer $T$ denoting the number of test cases. The description of $T$ test cases follows.
• The first line of each test case contains four integers $N,X,p,k$ respectively.
• The second line contains $N$ space-separated integers ${A}_{1},{A}_{2},\dots ,{A}_{N}$.

### Output

For each test case, print a single line containing one integer ― the minimum number of operations you must perform to make $X$ the $p$-th smallest element or $-1$ if its impossible to do so.

### Constraints

• $1\le T\le 100$
• $1\le p,k\le N$
• $0\le X\le {10}^{9}$
• The sum of $N$ over all test cases does not exceed $4\ast {10}^{5}$
• $0\le {A}_{i}\le {10}^{9}$ for each valid $i$

Subtask #1 (10 points): $N\le 5$

Subtask #2 (40 points): The sum of $N$ over all test cases does not exceed $3\ast {10}^{3}$

Subtask #3 (50 points): Original constraints

### Example Input

2
5 4 3 4
4 9 7 0 8
2 25 1 2
100 20


### Example Output

1
-1


### Explanation

Example case 1:

• We can perform one operation in the array. Take the $k$-th smallest integer of the current array (which is $8$ in this case) and replace it with $0$. The array then becomes $\left[4,9,7,0,0\right]$ which now makes $4$ as the 3rd smallest number of the array.

Example case 2:

• It is impossible to make $25$ as the smallest number of the array.

### Solution-

SOlution-
Since a note is mentioned in question that
Note: the k-th smallest number in an array is the k-th number from the left when the array is sorted in non-decreasing order.
So first of all we will sort our array/vector in non-decreasing order so that we can find kth smalles number.

then  p=p-1;
k=k-1;
because index is starting from 0 in my code so i did previous step.
I have written a solution function which return minumum number of operation reequired.
Now Let's jump into logic.
1-first of all i will check is(v[p]==X) if it is true i will simply return 0 because in this case i already have my answer.
2-now i will maintain a count variable which count total operation required.
3-Main logic start from here since out array is sorted till now so  we have three choices from here if(k>p),if(k<p) and if(k==p) and
in every 3 case these 3 case also contain two-two cases if(v[p]<x) and if(v[p]>x).

4- if(k>p) and if(v[p]<x) in this case we never reach to final solution so i will return -1 in this case.
if (k>p) && v[p]> x in this case i can reach to my solution and i will try to add some smaller value at k so that i can shift
value at p toward k ans every time i increase my value by one.
5- same if (k<p) this case is totally opposite of case 4.
6- if (k==p)
in this case i can reach to my solution
if(v[p]>x) i have to shift pth value to left
and if(v[p]<x) i will shift pth value  to right
and count ans.
finllay i return ans which tells minimum way to get final solution.
code of this problem

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