It is always optimal to add the smallest integer which is not already in the array.

The number of operations will never exceed $$$\sqrt{2 \cdot s}.$$$

Based on the first hint, the final array will be equal to $$${v_1,v_1,\ldots, v_n} \cup [1,x]$$$, for some $$$x$$$.

Hint 4: How can we find the value of $$$x$$$ faster than $$$O(\sqrt{s})$$$?

Lemma: The number of operations will never exceed $$$\sqrt{2 \cdot s}$$$.

Since we need to maximize the final length of the array, it is always optimal to add the smallest integer which is not already present in $$$a$$$.

The naive implementation of this idea has a runtime complexity of $$$O(n \cdot \sqrt s)$$$, although this can be easily optimized to $$$O((n+ \sqrt s)log(n))$$$, if binary search is used to check whether the new elements are already present in $$$a$$$.

The optimal solution has a runtime complexity of $$$O(N log S)$$$. Based on hints $$$3$$$ and $$$4$$$, we can find the value of $$$x$$$ with binary search. For some $$$x$$$, the sum of the elements in $$$a$$$ will be equal to $$$sum(x)=\frac{x\cdot(x+1)}{2}+\sum_1 v_i$$$, where $$$\sum_1 v_i$$$ is the sum of the elements greater than $$$x$$$.

Since the sum of elements $$$sum(x)$$$ is a monotonous function, we can use binary search on the interval $$$[1,\sqrt{2 \cdot s}]$$$ to find the exact value of $$$x$$$.

Final time complexity: $$$O(N log S)$$$

#include<bits/stdc++.h>

using namespace std;
typedef long long ll;
const ll NMAX=2e5+5;

ll v[NMAX];
int main()
{
    ios_base::sync_with_stdio(false); cin.tie(0);
    ll n,s;
    cin>>n>>s;

    for(ll i=0;i<n;i++)
        cin>>v[i];
    ll l=0,r=2e9,ans=0;

    while(l<=r){
        ll m=(l+r)/2;
        ll sum=m*(m+1)/2;

        for(ll i=0;i<n;i++)
            if(v[i]>m)
                sum+=v[i];

        if(sum>=s)
            ans=m,r=m-1;
        else
            l=m+1;
    }

    ll len=ans;

    for(ll i=0;i<n;i++)
        len+=(v[i]>ans);

    cout<<len;

    return 0;
}

The numbers are printed in ``layers''.

Try to represent the layers as a binary tree.

To better understand the solution, let's consider $$$n=12$$$. The final array is the BFS traversal starting from the root of the following binary tree:

This binary tree was generated by starting with the root $$$m=\lfloor \frac{n+1}{2} \rfloor=6$$$. Each node $$$m=\lfloor \frac{l+r}{2}\rfloor$$$ has up to two children: $$$m_1=\lfloor \frac{l+(m-1)}{2} \rfloor$$$ and $$$m_2=\lfloor \frac{r+(m+1)}{2} \rfloor$$$.

If our given integer $$$k$$$ is in a complete layer, then we can employ a binary search-type algorithm to find the position of $$$k$$$:

l = 1, r = n, position = 1, visited_nodes = 1
while(visited_nodes<=n){
    m = (l + r) / 2

    if(k==m){
        print position
        return
    }
    
    if(k<m) position = position * 2
    else position = position * 2 + 1

    visited_nodes = visited_nodes * 2 + 1
}

This algorithm has the added benefit of not printing anything if $$$k$$$ is not in a complete layer. In this case, the position of $$$k$$$ in the last layer can be determined from $$$position$$$, although this isn't very straightforward. If the binary tree were complete, then the position of $$$k$$$ in the last layer would be $$$position-\frac{\text{visited_nodes}-1}{2}$$$, since there are $$$\frac{\text{visited_nodes}-1}{2}$$$ nodes on the other layers, all of which will be printed before $$$k$$$.

By looking at the missing nodes from the binary tree for $$$n=12$$$, one may make the crucial observation that the general case position for $$$k$$$ in the last layer is equal to $$$k-(position-\frac{\text{visited_nodes}-1}{2})$$$.

In conclusion, the final position for $$$k$$$ will be $$$k-(position-\frac{\text{visited_nodes}-1}{2})+\frac{\text{visited_nodes}-1}{2}=k+\text{visited_nodes}-position-1$$$.

Time complexity per testcase: $$$O(log n)$$$

#include<bits/stdc++.h>
#define NMAX 1000005

using namespace std;
typedef long long ll;
void tc(){
    ll n,p;
    cin>>n>>p;
    ll cntLess=0,cnt=1,currentId=1;
    ll l=1,r=n;
    while(cnt<=n){
        ll m=(l+r)/2;
        cntLess=cntLess*2+(p>=m);
        if(m==p){
            cout<<currentId<<' ';
            return;
        }
        else if(p<m) r=m-1,currentId*=2;
        else l=m+1,currentId=currentId*2+1;
        cnt=cnt*2+1;
    }
    cout<<p+cnt/2-cntLess<<' ';
}
int main()
{
    ios_base::sync_with_stdio(false); cin.tie(0);
    int t;
    cin>>t;
    while(t--)
        tc();

    return 0;
}

Any final configuration can be uniquely determined by a permutation of $$${1,2,3,\ldots, k}$$$

$$$k \le 20$$$

Let dp[mask] be the minimum number of moves required to cover the first $$$\Sigma_{i=1}^k getBit(mask,i)*p_i$$$ slots with all plates of colours $$$c_1,c_2,\ldots$$$ where $$$getBit(mask,c_i)=1, \forall i$$$. Naturally, $$$dp[0]=0$$$.

Transitioning from $$$dp[mask]$$$ to $$$dp[mask+2^k]$$$, where $$$getBit(mask,k)=0$$$ can be done fairly easily. Let $$$pos=\Sigma_{i=1}^k getBit(mask,i)*p_i$$$ and $$$cntdiff(l,r,k)$$$ be the number of plates in the initial configuration $$$a_i$$$ which satisfy the following requirements:

• $$$l \le i \le r$$$;
• $$$a_i \neq 0$$$;
• $$$a_i \neq k$$$.

From these definitions, the value of $$$dp[mask+2^k]$$$ can be computed using the following formula:

$$$cntdiff(l,r,k)$$$ can be calculated in $$$O(1)$$$ per query using prefix sums. Calculating the prefix sums has a time complexity of $$$O(n \cdot k)$$$.

Intended time complexity: $$$O(n\cdot k + 2^k \cdot k)$$$

Reconstructing the final configuration will also require an auxiliary array $$$prev[mask]$$$ — the last colour added to $$$mask$$$.

#include<bits/stdc++.h>

using namespace std;
const int NMAX=1e5+1,KMAX=21;
int dp[1<<KMAX],lastAdded[1<<KMAX],cnt[KMAX],n,k;
bool visited[1<<KMAX];
int freq[NMAX][KMAX];
int cnteq(int l, int r, int num){
    if(r<l)
        return 0;
    return freq[r][num]-(l?freq[l-1][num]:0);
}
int cntdiff(int l, int r, int num){
    if(r<l)
        return 0;
    return r-l+1-cnteq(l,r,num)-cnteq(l,r,0);
}
void iterativeBt(){
    queue<pair<int,int>> q; /// (bitmask, pos)
    q.push({0,0});
    visited[0]=1;
    while(!q.empty()){
        int bm=q.front().first,pos=q.front().second;
        for(int i=0;i<k;i++){
            if(!(bm>>i&1)){
                if(dp[bm]+cntdiff(pos,pos+cnt[i]-1,i+1)<dp[bm|(1<<i)]){
                    lastAdded[bm|(1<<i)]=i;
                    dp[bm|(1<<i)]=dp[bm]+cntdiff(pos,pos+cnt[i]-1,i+1);
                }
                if(!visited[bm|(1<<i)]){
                    q.push({bm|(1<<i),pos+cnt[i]});
                }
                visited[bm|(1<<i)]=1;
            }
        }
        q.pop();
    }
}
int main()
{
    for(int i=1;i<(1<<KMAX);i++) dp[i]=INT_MAX; /// dp[0]=0
    scanf("%d%d",&n,&k);
    for(int i=0;i<n;i++){
        int x;
        scanf("%d",&x);
        if(i) for(int j=0;j<=k;j++)
            freq[i][j]=freq[i-1][j];
        freq[i][x]++;
    }
    for(int i=0;i<k;i++){
        scanf("%d",&cnt[i]);
    }
    iterativeBt();
    stack<int> ans; int pos=(1<<k)-1;
    printf("%d\n",dp[pos]);
    while(pos){
        ans.push(lastAdded[pos]);
        pos^=(1<<lastAdded[pos]);
    }
    while(!ans.empty()){
        for(int i=0;i<cnt[ans.top()];i++)
            printf("%d ",ans.top()+1);
        ans.pop();
    }
    return 0;
}