1. Abridged problem statement
You are given a sequence of n grayscale pixel values c₁…cₙ (0 ≤ cᵢ ≤ 10⁶). You may delete any subset of pixels, preserving the original order of the remaining ones. After deletions you obtain a sequence r₁…rₘ.

Constraint: for every adjacent pair in the final sequence, |rᵢ − rᵢ₊₁| ≠ 1 (their colors must not differ by exactly 1).

Goal: Delete the minimum number of pixels (equivalently, keep a longest possible subsequence) such that no two adjacent remaining pixels differ by 1. Output:
- t = minimum number of deletions;
- one valid resulting sequence r of length m = n − t.

n ≤ 10⁵, colors ≤ 10⁶.

---

2. Detailed editorial

Reformulation
We need a longest subsequence (not necessarily contiguous) of the input array c[0…n−1], such that for all consecutive chosen indices i < j, |c[i] − c[j]| ≠ 1. Order must be preserved.

If we find the maximum length L of such a subsequence, the answer is:
- minimum deletions t = n − L,
- and any subsequence of length L satisfying the constraint.

This is a classic "longest subsequence with a constraint on adjacent chosen elements".

Dynamic programming idea
Let's process pixels from left to right and define:

- dp[pos] = length of the longest valid subsequence that ends exactly at position pos (0-based), i.e., we take pixel pos as the last pixel of the subsequence.

If the previous chosen pixel index is last < pos, we must have |c[pos] − c[last]| ≠ 1.

Thus:
dp[pos] = 1 + max over all last < pos, |c[pos] − c[last]| ≠ 1 of dp[last].

If we know the best "previous dp" we can transition. Directly this is O(n²), too slow (n up to 10⁵).

Transforming the transition
For each position pos, the recurrence doesn't depend on the distance in indices, only on colors:

dp[pos] = 1 + best_dp_for_previous_color, where "previous color" v is any color such that |v − c[pos]| ≠ 1.

We can maintain, for each color v, the best dp value achieved by any position with that color so far.

Let best[v] = max dp[last] over all last with c[last] = v.

At position pos with color x = c[pos], we want:
max over all v such that |v − x| ≠ 1 of best[v].

The forbidden previous colors are exactly x − 1 and x + 1 (and there is no restriction about same color x itself; colors can be equal). So allowed colors are:

- all colors v, except possibly v = x − 1 and v = x + 1.

So at pos:
best_over_allowed = maximum over all colors v of best[v]
                     but ignoring v = x−1 and v = x+1.

After we get best_over_allowed, we do:
dp[pos] = best_over_allowed + 1.

Then we must update best[x] = max(best[x], dp[pos]) since ending at pos gives a new candidate for color x.

The difficulty is computing best_over_allowed fast.

Segment tree approach
We need to query the maximum dp among all colors v except x−1 and x+1. That is "all colors" minus up to two points.

All colors v except x−1 and x+1 form these ranges (on color values):
- [0, x−2] and [x, x] and [x+2, MAXV]

Explanation:
- Remove x−1 and x+1.
- The color x itself IS allowed: |x−x| = 0, so we can place equal colors adjacent.
- For v < x−1: these all satisfy |v−x| ≥ 2.
- For v between x+2 and MAXV: also |v−x| ≥ 2.

So we can query at most three disjoint ranges on colors:
1) [0, x−2] if x ≥ 2
2) [x, x]
3) [x+2, MAXV] if x+2 ≤ MAXV

Take the maximum of those three range maxima.

Data we store in the segment tree
We need to later reconstruct which positions form the optimal subsequence, so we need to recall the previous index last that gave the best dp.

So for each color v we want to store:

- current best pair (best_dp_for_color_v, some_index_with_that_dp).

We'll keep in each tree node a pair<int,int>:

- first: dp value (the maximum for that node range),
- second: the index where this value is achieved.

The tree node value is simply the max (lexicographically) between its children: pair comparison first compares first, then second, which is fine.

Identity value (for an empty segment) is something like (-INF, -1) so that max(x, identity) = x.

Implementation details of DP:

- Let MAXV = 10⁶ + 42 (a bit larger than allowed colors).
- Build SegmentTree over indices 0..MAXV (color domain), initially all identity (very small).
- Arrays:
  - dp[n],
  - parent[n] = previous index in the optimal chain for each i (for path reconstruction).

Initialize first element:
- At position 0:
  - We always can start a subsequence of length 1 with c[0]:
    dp[0] = 1
    parent[0] = -1 (no previous)
    In segment tree, at position color = c[0], we set best for that color:
      update(c[0], max(current_value_at_color, (1, 0)))

For general i from 1..n−1:

Let x = c[i].

Compute best candidate:

Initialize best = (0, -1)  // meaning subsequence length 0, no previous index

1) if x >= 2:
     best = max(best, segment_tree.query(0, x−2))

2) include same color:
     best = max(best, segment_tree.query(x, x))

3) if x + 2 ≤ MAXV:
     best = max(best, segment_tree.query(x+2, MAXV))

Now we have best = (best_dp_value, best_index).

Then:
dp[i] = best.first + 1
parent[i] = best.second

Now update the tree at color x with this new candidate:

current_color_val = segment_tree.get_pos(x)
segment_tree.update(x, max(current_color_val, (dp[i], i)))

At the end:

- The best global subsequence is the maximum dp[i] over i=0..n−1.
  Let:
    max_len = max(dp)
    end_pos = index achieving that max.

Reconstruct path:

Start from end_pos, and repeatedly go to parent[end_pos], until -1.

Store indices, then reverse to get them in increasing order.

Those indices describe the subsequence of pixels to keep. Its length is max_len.

Answer:

- Minimum deletions t = n − max_len
- Output the retained pixel values in order: c[p] for p in path.

Complexity analysis

- Segment tree size: O(MAXV), where MAXV ≈ 10⁶. Good under memory limits.
- For each of the n positions:
  - up to 3 range-maximum queries O(log MAXV),
  - 1 update O(log MAXV).
- Total: O(n log MAXV) ≈ 10⁵ * log(10⁶) ~ 10⁵ * 20 = 2⋅10⁶ operations, fine.

---

3. C++ Solution

```cpp
#include <bits/stdc++.h>
// #include <coding_library/data_structures/segment_tree.hpp>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

template<class T, T (*merge)(T, T), T (*e)()>
class SegmentTree {
  private:
    int n, size;
    vector<T> tr;

    void pull(int x) { tr[x] = merge(tr[2 * x], tr[2 * x + 1]); }

  public:
    SegmentTree() { init(vector<T>()); }
    SegmentTree(int _n) { init(vector<T>(_n, e())); }
    SegmentTree(const vector<T>& _a) { init(_a); }

    void init(const vector<T>& _a) {
        n = _a.size();
        size = 1;
        while(size < n) {
            size <<= 1;
        }

        tr.assign(2 * size, e());
        for(int i = 0; i < n; i++) {
            tr[size + i] = _a[i];
        }
        for(int i = size - 1; i > 0; i--) {
            pull(i);
        }
    }

    void update(int pos, T val) {
        pos += size;
        tr[pos] = val;
        for(pos >>= 1; pos > 0; pos >>= 1) {
            pull(pos);
        }
    }

    T get_pos(int pos) { return tr[pos + size]; }

    T query(int l, int r) {
        T ansl = e(), ansr = e();
        for(l += size, r += size + 1; l < r; l >>= 1, r >>= 1) {
            if(l & 1) {
                ansl = merge(ansl, tr[l++]);
            }
            if(r & 1) {
                ansr = merge(tr[--r], ansr);
            }
        }
        return merge(ansl, ansr);
    }

    T query_all() { return tr[1]; }

    template<bool (*f)(T)>
    int max_right(int l) const {
        return max_right(l, [](T x) { return f(x); });
    }
    template<class F>
    int max_right(int l, F f) const {
        if(l == n) {
            return n;
        }

        l += size;
        T sm = e();
        do {
            while(l % 2 == 0) {
                l >>= 1;
            }
            if(!f(merge(sm, tr[l]))) {
                while(l < size) {
                    l = (2 * l);
                    if(f(merge(sm, tr[l]))) {
                        sm = merge(sm, tr[l]);
                        l++;
                    }
                }
                return l - size;
            }
            sm = merge(sm, tr[l]);
            l++;
        } while((l & -l) != l);
        return n;
    }

    template<bool (*f)(T)>
    int min_left(int r) const {
        return min_left(r, [](T x) { return f(x); });
    }
    template<class F>
    int min_left(int r, F f) const {
        if(r == -1) {
            return 0;
        }

        r += size + 1;
        T sm = e();
        do {
            r--;
            while(r > 1 && (r % 2)) {
                r >>= 1;
            }
            if(!f(merge(tr[r], sm))) {
                while(r < size) {
                    r = (2 * r + 1);
                    if(f(merge(tr[r], sm))) {
                        sm = merge(tr[r], sm);
                        r--;
                    }
                }
                return r + 1 - size;
            }
            sm = merge(tr[r], sm);
        } while((r & -r) != r);
        return 0;
    }
};

pair<int, int> max_custom(pair<int, int> a, pair<int, int> b) {
    return max(a, b);
}
pair<int, int> max_e() { return {INT_MIN, -1}; }

const int MAXV = (int)1e6 + 42;

int n;
vector<int> c;

void read() {
    cin >> n;
    c.resize(n);
    for(int i = 0; i < n; i++) {
        cin >> c[i];
    }
}

void solve() {
    // We can solve this problem with DP - there is a direct N^2 solution that
    // keeps a dp[pos] as a state, and tries to select the largest dp[last],
    // such that last < pos and |c[pos] - c[last]| != 1. The cost of (pos - last
    // - 1), so overall:
    //
    //    dp[pos] = max dp[last] + pos - last - 1, over the eligible last
    //    positions
    //
    // We do the classic approach of separating what depends on last, and what
    // depends on pos, which in this case makes it easy as there are no terms
    // depending on both.
    //
    //    dp[pos] = pos - 1 + max (dp[last] - last)
    //
    // Essentially, for some pos, we are interested in the largest value of
    // dp[last] - last, excluding such positions that c[last] = c[pos]. We can
    // notice that this is two ranges: [0;c[pos]-2], [c[pos];c[pos]] and [c[pos]
    // + 2;MAXV], and so we can use a max segment tree to query this
    // efficiently. The problem does ask to recover the solution, so at index
    // c[last] in the segment tree, we will keep the largest value of {dp[last]
    // - last, last}.
    //
    // The overall complexity is O(N log MAXV).

    auto t = SegmentTree<pair<int, int>, max_custom, max_e>(MAXV + 1);

    vector<int> dp(n);
    vector<int> parent(n, -1);

    dp[0] = 1;
    t.update(c[0], {1, 0});

    for(int i = 1; i < n; i++) {
        pair<int, int> best = {0, -1};

        if(c[i] >= 2) {
            best = max(best, t.query(0, c[i] - 2));
        }
        best = max(best, t.query(c[i], c[i]));
        if(c[i] + 2 <= MAXV) {
            best = max(best, t.query(c[i] + 2, MAXV));
        }

        dp[i] = best.first + 1;
        parent[i] = best.second;

        auto current_val = make_pair(dp[i], i);
        t.update(c[i], max(t.get_pos(c[i]), current_val));
    }

    int max_len = *max_element(dp.begin(), dp.end());
    int end_pos = max_element(dp.begin(), dp.end()) - dp.begin();

    vector<int> path;
    int curr = end_pos;
    while(curr != -1) {
        path.push_back(curr);
        curr = parent[curr];
    }
    reverse(path.begin(), path.end());

    cout << n - max_len << '\n';
    for(int p: path) {
        cout << c[p] << ' ';
    }
    cout << '\n';
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

---

4. Python Solution

```python
import sys

# We will implement an iterative segment tree for range-max over pairs (dp, index).
# Python tuples compare lexicographically, so max() on tuples works directly.

INT_MIN = -10**18  # sufficiently small sentinel

MAXV = 10**6 + 42  # maximum color domain in segment tree


class SegmentTree:
    """
    Segment tree for range maximum over pairs (dp, index) on domain [0..n-1].
    merge: max of two tuples.
    identity: (INT_MIN, -1)
    """

    def __init__(self, n):
        # n is number of leaves (here n = MAXV+1).
        self.n = n
        # Find power-of-two base size >= n.
        size = 1
        while size < n:
            size <<= 1
        self.size = size
        # Allocate tree. Use list of tuples; initialize to identity.
        self.tr = [(INT_MIN, -1)] * (2 * size)

    def update(self, pos, val):
        """
        Point update at position pos with value val.
        We set tr[pos] = val (not taking max here; caller should do that).
        """
        i = pos + self.size
        self.tr[i] = val
        i >>= 1
        # Recompute all parents.
        while i > 0:
            left = self.tr[2 * i]
            right = self.tr[2 * i + 1]
            # Node value is maximum of children.
            self.tr[i] = left if left >= right else right
            i >>= 1

    def get_pos(self, pos):
        """Return the value stored at leaf pos (0-based)."""
        return self.tr[pos + self.size]

    def query(self, l, r):
        """
        Range max on [l, r] inclusive, 0-based.
        """
        if l > r:
            return (INT_MIN, -1)
        n = self.size
        # Convert [l, r] to [l, r+1) on leaf indices.
        l += n
        r += n + 1
        resl = (INT_MIN, -1)
        resr = (INT_MIN, -1)
        while l < r:
            if l & 1:
                # Take current segment at l
                if self.tr[l] > resl:
                    resl = self.tr[l]
                l += 1
            if r & 1:
                r -= 1
                if self.tr[r] > resr:
                    resr = self.tr[r]
            l >>= 1
            r >>= 1
        # Combine left and right result.
        return resl if resl >= resr else resr


def solve():
    data = sys.stdin.read().strip().split()
    if not data:
        return
    it = iter(data)
    n = int(next(it))
    c = [int(next(it)) for _ in range(n)]

    # Build segment tree on color domain [0..MAXV].
    seg = SegmentTree(MAXV + 1)

    dp = [0] * n       # dp[i] = best subsequence length ending at i
    parent = [-1] * n  # parent[i] = previous index in subsequence

    # Initialize first element.
    dp[0] = 1
    parent[0] = -1
    # At color c[0], store (dp[0], index=0)
    old = seg.get_pos(c[0])
    # Keep maximum between old and new.
    best_pair = old if old >= (dp[0], 0) else (dp[0], 0)
    seg.update(c[0], best_pair)

    for i in range(1, n):
        col = c[i]
        # best = (best_dp, index)
        best_dp, best_idx = 0, -1

        # Query [0, col-2] if col >= 2.
        if col >= 2:
            cand = seg.query(0, col - 2)
            if cand[0] > best_dp:
                best_dp, best_idx = cand

        # Query [col, col] (same color).
        cand = seg.query(col, col)
        if cand[0] > best_dp:
            best_dp, best_idx = cand

        # Query [col+2, MAXV] if applicable.
        if col + 2 <= MAXV:
            cand = seg.query(col + 2, MAXV)
            if cand[0] > best_dp:
                best_dp, best_idx = cand

        # Compute dp[i] and parent[i].
        dp[i] = best_dp + 1
        parent[i] = best_idx

        # Update segment tree at this color with (dp[i], i).
        old = seg.get_pos(col)
        new_pair = (dp[i], i)
        best_pair = old if old >= new_pair else new_pair
        seg.update(col, best_pair)

    # Find global maximum dp and its index.
    max_len = max(dp)
    end_pos = max(range(n), key=lambda i: dp[i])

    # Reconstruct path of indices.
    path = []
    cur = end_pos
    while cur != -1:
        path.append(cur)
        cur = parent[cur]
    path.reverse()

    # Minimum deletions = n - max_len.
    out_lines = []
    out_lines.append(str(n - max_len))
    # Build the resulting sequence string.
    seq = ' '.join(str(c[i]) for i in path)
    out_lines.append(seq)
    sys.stdout.write('\n'.join(out_lines) + '\n')


if __name__ == "__main__":
    solve()
```

---

5. Compressed editorial

We need a longest subsequence of the given pixels such that any two consecutive kept pixels don't differ in color by exactly 1. Let c[i] be the color at position i.

Define dp[i] = length of the longest valid subsequence ending at position i. If the previous kept position is j < i, we must have |c[i] − c[j]| ≠ 1, so:

dp[i] = 1 + max over j<i, |c[i]−c[j]|≠1 of dp[j].

Since the restriction depends only on colors, not indices, maintain best[v] = best dp among indices with color v seen so far. At position i (color x = c[i]), we want:

max over all colors v with |v − x| ≠ 1 of best[v].

Forbidden colors are x−1 and x+1; x itself is allowed. Over the color axis [0..MAXV], allowed colors form up to three disjoint intervals: [0, x−2], [x, x], [x+2, MAXV]. We store best[v] in a segment tree over colors, where each node keeps (best_dp_for_color, index_that_achieves_it). Query these intervals to find the best predecessor:

- best = max(query(0, x−2), query(x, x), query(x+2, MAXV)), skipping ranges that are empty.

Then:

- dp[i] = best.first + 1,
- parent[i] = best.second (for reconstruction),
- update the tree at color x with max(current_stored_value, (dp[i], i)).

Initialize dp[0] = 1 and store (1,0) at color c[0]. After processing all positions, take max dp[i] and its index end_pos. Reconstruct the optimal subsequence by following parent pointers from end_pos back to −1. Minimum deletions are n − max_dp.

Complexity: O(n log MAXV) with MAXV ≈ 10⁶, which fits easily in time and memory limits.
