We can easily think of Brute Force Solution of getting mex for each index with a Complexity of $$$O(n^2)$$$ resulting in TLE. So For each index $$$i$$$, the $$$mex_i$$$ has to be determined in $$$O(1)$$$ or $$$O(\log{}{n})$$$. For this, we have to determine the mex of the whole array & for each index we will compare with this mex to get $$$mex_i$$$.

Let, $$$mex = mex(a_1,a_2,\ldots,a_n)$$$. That means all numbers from $$$0$$$ to $$$mex-1$$$ remains in the array!

Now for index $$$i$$$, if occurrence of $$$a_i$$$, $$$occ[a[i]]\gt 1$$$, then it does not affect the mex. So here $$$mex_i = mex$$$. But if $$$occ[a[i]]==1$$$, then its absence will change the mex if $$$mex\gt a[i]$$$. Then $$$mex_i = a[i]$$$. But if $$$mex\le a[i], mex_i=mex$$$. Ultimately, we get $$$mex_i=min(mex,a[i])$$$.

Complexity:

$$$O(n)$$$ [If you use array for occurrence of each element of the array]

$$$O(n\cdot \log{}{n})$$$ [If you use map for occurrence of each element of the array]

//﷽ [Bismillaahir Rahmaanir Raheem]

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

#define ll long long int
#define ul unsigned long long int
#define ld long double
#define TT int TC;cin>>TC;for(int ti=1;ti<=TC;ti++)
#define TT1 int TC=1;for(int ti=1;ti<=TC;ti++)
#define fastio ios_base::sync_with_stdio(0); cin.tie(0);
#define YES cout<<"YES"<<endl
#define NO cout<<"NO"<<endl
#define ans_YN if(ans)YES;else NO;
#define pb push_back
#define pr_q priority_queue
#define F first
#define S second
#define MP make_pair
#define UB upper_bound
#define LB lower_bound
#define B begin()
#define E end()
//#define endl "\n"
ll ceill(ll a,ll b){if(a>=0){if(a%b==0)return a/b;else return a/b+1;}else return a/b;}//Here b>0 for being divisor
ll flr(ll a, ll b){if(a>=0){return a/b;} else {if(a%b==0) return a/b; else return a/b-1;}}//Here b>0 for being divisor
ll GCD(ll a, ll b){ ll ans=__gcd(a,b);if(ans<0)ans=-ans;return ans;}
ll LCM(ll a, ll b){ return (a*b)/(GCD(a,b));}
ll rem(ll a, ll b){ if(a>=0) return a%b; else{a=-a; return (b-a%b)%b;}}
int const mod=1e9+7;
int const AND=1073741823;

int main()
{
    fastio
    TT
    {
        ll n;
        cin>>n;
        ll a[n];
        ll mex;
        map<ll,ll>occ;
        for(int i=0;i<n;i++){
            cin>>a[i];
            occ[a[i]]++;
        }

        for(ll i=0;i<=n;i++){
            if(occ[i]==0){
                mex=i;
                break;
            }
        }

        for(int i=0;i<n;i++){
            ll mexx=mex;
            if(occ[a[i]]==1){
                mexx=min(mex,a[i]);
            }
            cout<<mexx<<" ";
        }
        cout<<endl;

        
    }
    return 0;
}