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.

Naive approach  
If we kept best[v] in an array of size up to MAXV = 10⁶ and scanned all colors every time, each step is O(MAXV) → O(n·MAXV) impossible.

We need something like a segment tree or Fenwick tree to get range maxima quickly.

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.

We can’t directly subtract a point from a segment tree max query. But we can express:

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.

Correctness arguments

- DP correctness: Each dp[i] is defined as 1 + best possible dp[last] among all last < i with |c[i] − c[last]| ≠ 1. We never consider invalid predecessors because the tree query explicitly excludes colors x−1 and x+1; we include all others (three ranges cover all allowed colors).
- Using color-based segmentation is valid since the only condition depends on color difference, not on position difference.
- Segment tree always holds correct best dp-per-color up to current i−1, and since we update immediately after computing dp[i], by induction it is correct for all steps.

- Reconstruction: parent[i] always points to some previous index last that achieved the dp value used to compute dp[i]. Starting from any index achieving global maximum and following parents yields a chain of indices with strictly increasing order and valid color constraints. The chain length equals dp[end_pos], which is max_len, so it’s an optimal subsequence.

Edge cases

- n = 1: dp[0] = 1, answer is 0 deletions and the single pixel.
- All pixels equal: we can keep all, since |x−x|=0. DP will keep extending using same color range.
- Colors are such that every adjacent pair has difference 1: we can still pick a subsequence skipping some pixels; DP finds the longest valid subsequence automatically.

---

3. C++ solution with detailed comments

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

// Overload stream operators for convenience (pairs and vectors).

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

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

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

// Print all elements of a vector separated by spaces
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for (auto x: a) {
        out << x << ' ';
    }
    return out;
};

// Generic iterative segment tree template for type T.
// merge: function to combine two children nodes.
// e: function returning the identity element for merge.
template<class T, T (*merge)(T, T), T (*e)()>
class SegmentTree {
  private:
    int n;          // number of "real" elements (original array size)
    int size;       // size of the internal power-of-two segment tree base
    vector<T> tr;   // tree array: indices [1 .. 2*size-1]

    // Recompute node x from its children.
    void pull(int x) { tr[x] = merge(tr[2 * x], tr[2 * x + 1]); }

  public:
    // Default constructor: builds empty tree.
    SegmentTree() { init(vector<T>()); }

    // Build tree from size n, initializing all values with e().
    SegmentTree(int _n) { init(vector<T>(_n, e())); }

    // Build tree from an existing array _a.
    SegmentTree(const vector<T>& _a) { init(_a); }

    // Initialize / rebuild tree from array _a.
    void init(const vector<T>& _a) {
        n = (int)_a.size();
        // Find smallest power of two >= n.
        size = 1;
        while (size < n) {
            size <<= 1;
        }

        // Allocate tree with identity values.
        tr.assign(2 * size, e());

        // Copy array into leaves (starting at index size).
        for (int i = 0; i < n; i++) {
            tr[size + i] = _a[i];
        }

        // Build internal nodes bottom-up.
        for (int i = size - 1; i > 0; i--) {
            pull(i);
        }
    }

    // Point update: set position pos to val (0-based on original array).
    void update(int pos, T val) {
        pos += size;        // move to leaf
        tr[pos] = val;      // assign
        // Move upward and recompute parents.
        for (pos >>= 1; pos > 0; pos >>= 1) {
            pull(pos);
        }
    }

    // Get the value stored at a single position (0-based).
    T get_pos(int pos) { return tr[pos + size]; }

    // Range query on [l, r] inclusive (0-based).
    T query(int l, int r) {
        // ansl accumulates from the left, ansr from the right.
        T ansl = e(), ansr = e();

        // Convert to leaf indices on [l, r+1)
        for (l += size, r += size + 1; l < r; l >>= 1, r >>= 1) {
            // If l is a right child, its segment is disjoint from the left side accumulated so far.
            if (l & 1) {
                ansl = merge(ansl, tr[l++]);
            }
            // If r is about to become a right bound, take the left sibling's segment.
            if (r & 1) {
                ansr = merge(tr[--r], ansr);
            }
        }
        // Merge left and right accumulations.
        return merge(ansl, ansr);
    }

    // Query the entire range [0, n-1].
    T query_all() { return tr[1]; }

    // Additional helpers (max_right/min_left) are not used in this problem.
    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;
    }
};

// Merge function for segment tree: take maximum of two (dp, index) pairs.
pair<int, int> max_custom(pair<int, int> a, pair<int, int> b) {
    return max(a, b);  // uses std::pair lexicographic comparison
}

// Identity element for max over pairs: very small dp value, invalid index.
pair<int, int> max_e() {
    return {INT_MIN, -1};
}

// Maximum color value allowed in input is 1e6,
// we add some slack to safely allow x+2 queries.
const int MAXV = (int)1e6 + 42;

int n;              // number of pixels
vector<int> c;      // colors

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

void solve() {
    // Segment tree over color domain [0 .. MAXV].
    // Each node stores (best_dp_for_color, index_of_pixel).
    auto t = SegmentTree<pair<int, int>, max_custom, max_e>(MAXV + 1);

    vector<int> dp(n);      // dp[i] = best subsequence length ending at i
    vector<int> parent(n, -1);  // parent pointer for reconstruction

    // Initialize first position: we can always start with it.
    dp[0] = 1;
    parent[0] = -1;
    // At color c[0], store (dp[0], index=0)
    t.update(c[0], {1, 0});

    // Process subsequent positions.
    for (int i = 1; i < n; i++) {
        pair<int, int> best = {0, -1};  // (best_dp_so_far, index_with_that_dp)

        int col = c[i];

        // Range 1: [0, col-2] if col >= 2.
        if (col >= 2) {
            best = max(best, t.query(0, col - 2));
        }

        // Range 2: [col, col] -- same color is allowed.
        best = max(best, t.query(col, col));

        // Range 3: [col+2, MAXV] if applicable.
        if (col + 2 <= MAXV) {
            best = max(best, t.query(col + 2, MAXV));
        }

        // dp[i] = best_dp + 1 (or 1 if best_dp = 0 and best.index = -1).
        dp[i] = best.first + 1;
        parent[i] = best.second;    // previous index that led to this dp value

        // Update color col in segment tree with new candidate (dp[i], i).
        auto current_val = t.get_pos(col);
        // Keep the best pair at this color.
        t.update(col, max(current_val, make_pair(dp[i], i)));
    }

    // Find global best subsequence length and ending position.
    int max_len = *max_element(dp.begin(), dp.end());
    int end_pos = int(max_element(dp.begin(), dp.end()) - dp.begin());

    // Reconstruct the path of indices by following parent pointers.
    vector<int> path;
    int curr = end_pos;
    while (curr != -1) {
        path.push_back(curr);
        curr = parent[curr];
    }
    reverse(path.begin(), path.end());  // now in increasing index order

    // Output number of deletions.
    cout << n - max_len << '\n';
    // Output the resulting sequence values.
    for (int p: path) {
        cout << c[p] << ' ';
    }
    cout << '\n';
}

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

    int T = 1;
    // The problem has a single test case.
    // If there were multiple, we'd read T here.
    for (int test = 1; test <= T; test++) {
        read();
        solve();
    }

    return 0;
}
```

---

4. Python solution with detailed comments

```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.