Incomplete Color (Editorial)

Incomplete Color#13095

Idea

We need to count subarrays containing exactly one incomplete color.

Recall that a color is incomplete if the subarray contains at least one, but not all of its occurrences in the original array.

A straightforward solution would enumerate every subarray and maintain the number of incomplete colors. This takes $O (n^{2})$ O(n^2) time, which is too slow.

Instead, let's fix the right endpoint $r$ and count all valid left endpoints $l$ .

Key Observation

For a fixed right endpoint $r$ , let $f (l)$ be the number of incomplete colors in the subarray $[l, r]$ [l,r].

Our goal is to count positions $l$ for which

f (l) = 1.

f(l)=1.

We will maintain these values while moving $r$ from left to right.

For each color $c$ , calculate:

$f i r s t_{c}$ first_c — the position of its first occurrence.
$l a s t_{c}$ last_c — the position of its last occurrence.
$p r e v_{c}$ prev_c — the position of its latest occurrence before the current position, initially $0$ .

Suppose we extend the current subarray by adding $a_{r} = c$ a_r=c.

Consider all left endpoints satisfying

p r e v_{c} < l \leq r .

prev_c<l\leq r.

For these subarrays, color $c$ was previously absent and has now appeared.

Therefore, we can temporarily consider it incomplete and increase $f (l)$ by one on the interval

[p r e v_{c} + 1, r] .

[prev_c+1,r].

However, if $r = l a s t_{c}$ r=last_c, we have just reached the final occurrence of this color.

For every $l \leq f i r s t_{c}$ l\leq first_c, the subarray now contains all occurrences of $c$ . Thus, this color is no longer incomplete.

We must decrease $f (l)$ by one on

[1, f i r s t_{c}] .

[1,first_c].

Notice that these two updates also handle colors appearing only once: the updates simply cancel each other.

So, for every position $r$ , we perform at most two range additions.

Solution

We need a data structure supporting:

Adding $1$ or $- 1$ to all values on an interval.
Counting how many values are equal to $1$ .

We can use a segment tree with lazy propagation.

For every node, maintain:

$mn$ — the minimum value in the segment.
$c n t$ — the number of occurrences of this minimum.
$c n t 1$ — the number of values equal to $mn + 1$ mn+1.
$l a z y$ — the pending addition.

When we add a constant to the entire segment, only $mn$ and $l a z y$ change. The values of $c n t$ and $c n t 1$ remain unchanged.

We can also merge two child nodes in constant time.

The minimum of the parent is the smaller minimum of its children.

To calculate $c n t$ , we sum the minimum counts of children whose minimum equals the parent's minimum.

To calculate $c n t 1$ , a child contributes:

Its $c n t 1$ , if its minimum equals the parent's minimum.
Its $c n t$ , if its minimum is exactly one greater than the parent's minimum.
Nothing otherwise.

Thus, every range addition takes $O (lo g n)$ O(\log n) time.

Initially, all values $f (l)$ are zero.

After processing position $r$ , we only modify positions $l \leq r$ l\leq r. The remaining positions still have value zero and do not contribute to the answer.

Moreover, after processing the entire array, $f (1) = 0$ f(1)=0, because the whole array contains all occurrences of every color.

Therefore, the global minimum always equals zero.

This means that $c n t 1$ at the root of the segment tree is exactly the number of subarrays ending at $r$ that contain one incomplete color.

We add this value to the answer after every position.

Complexity Analysis

We process all $n$ positions, performing at most two segment tree updates for each position.

Each update takes $O (lo g n)$ O(\log n) time.

Therefore, the total time complexity is $O (n lo g n)$ O(n\log n).

The segment tree and auxiliary arrays require $O (n)$ additional memory.

The answer can be as large as $O (n^{2})$ O(n^2), so we use a 64-bit integer.

Implementation

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

typedef long long ll;

const int N = 2e5 + 7;

int a[N], fst[N], lst[N], prv[N];
int mn[4 * N], cnt[4 * N], cnt1[4 * N], lazy[4 * N];

void build(int v, int l, int r) {
    cnt[v] = r - l + 1;

    if (l == r) return;

    int mid = (l + r) / 2;
    build(v * 2, l, mid);
    build(v * 2 + 1, mid + 1, r);
}

void pull(int v) {
    int x = v * 2, y = v * 2 + 1;

    mn[v] = min(mn[x], mn[y]);
    cnt[v] = cnt1[v] = 0;

    if (mn[x] == mn[v]) {
        cnt[v] += cnt[x];
        cnt1[v] += cnt1[x];
    } else if (mn[x] == mn[v] + 1) {
        cnt1[v] += cnt[x];
    }

    if (mn[y] == mn[v]) {
        cnt[v] += cnt[y];
        cnt1[v] += cnt1[y];
    } else if (mn[y] == mn[v] + 1) {
        cnt1[v] += cnt[y];
    }
}

void push(int v) {
    if (!lazy[v]) return;

    int x = v * 2, y = v * 2 + 1;

    mn[x] += lazy[v];
    mn[y] += lazy[v];

    lazy[x] += lazy[v];
    lazy[y] += lazy[v];

    lazy[v] = 0;
}

void upd(int v, int l, int r, int ql, int qr, int val) {
    if (ql > qr) return;

    if (ql <= l && r <= qr) {
        mn[v] += val;
        lazy[v] += val;
        return;
    }

    push(v);

    int mid = (l + r) / 2;

    if (ql <= mid)
        upd(v * 2, l, mid, ql, qr, val);

    if (qr > mid)
        upd(v * 2 + 1, mid + 1, r, ql, qr, val);

    pull(v);
}

void solve() {
    int n;
    cin >> n;

    for (int i = 1; i <= n; i++) {
        cin >> a[i];

        if (!fst[a[i]]) fst[a[i]] = i;
        lst[a[i]] = i;
    }

    build(1, 1, n);

    ll ans = 0;

    for (int r = 1; r <= n; r++) {
        int c = a[r];

        upd(1, 1, n, prv[c] + 1, r, 1);

        if (r == lst[c]) {
            upd(1, 1, n, 1, fst[c], -1);
        }

        prv[c] = r;

        ans += cnt1[1];
    }

    cout << ans << '\n';
}

signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);

    solve();

    return 0;
}

C++