Block Decomposition: Guide To The First 2 Problems

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Are you diving into the world of block decomposition and feeling a bit overwhelmed? You're not alone! Block decomposition is a powerful technique used in computer science to optimize solutions for various problems, especially in data structures and algorithm design. This article will walk you through the first two problems in a series of nine introductory block decomposition challenges, providing clear explanations, code snippets, and practical insights to help you grasp the fundamentals. Let's get started!

Understanding Block Decomposition

Before we jump into the problems, let's briefly discuss what block decomposition is and why it's so useful. Block decomposition, also known as square root decomposition, is a technique that involves dividing a data structure (like an array) into smaller, more manageable blocks. The size of these blocks is typically the square root of the total number of elements. This division allows us to perform operations more efficiently by combining operations on individual blocks with operations on the entire structure.

The primary advantage of block decomposition is that it strikes a balance between the time complexity of brute-force approaches and more complex data structures like segment trees or binary indexed trees. It's particularly effective for problems involving range queries and updates, where we need to perform operations on a specific range of elements.

Now, let's dive into the first problem and see how block decomposition works in practice.

Problem 1: Sum of Range Queries

The first problem focuses on efficiently calculating the sum of elements within a given range in an array. We'll also need to handle updates, where we add a value to all elements within a specified range. This is a classic scenario where block decomposition shines.

Problem Statement

Given an array A of N integers, implement two types of operations:

  1. Range Update: Add a value d to all elements in the range [l, r].
  2. Range Query: Calculate the sum of all elements in the range [l, r].

Approach

To solve this problem using block decomposition, we'll divide the array into blocks of size √N. We'll maintain two additional arrays:

  • sum: Stores the sum of elements in each block.
  • add: Stores the lazy update value for each block.

When we perform a range update, we need to consider three cases:

  1. The range [l, r] falls entirely within a single block. In this case, we directly update the elements in the array and update the sum for that block.
  2. The range [l, r] spans multiple blocks. We update the partial blocks at the beginning and end of the range as in case 1. For the complete blocks in between, we update the add value for those blocks.
  3. The range [l, r] is the same as range, we add to range.

When we perform a range query, we again consider three cases:

  1. The range [l, r] falls entirely within a single block. We calculate the sum by iterating through the elements in the array and adding the add value for that block.
  2. The range [l, r] spans multiple blocks. We calculate the sum for the partial blocks at the beginning and end of the range as in case 1. For the complete blocks in between, we add the sum value for those blocks.
  3. The range [l, r] is the same as range, we query to range.

Code Explanation

Let's break down the C++ code provided for this problem:

  • Includes and Type Definitions:

    #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<int> vi;
#define IOS                  \n        ios::sync_with_stdio(0); \
    cin.tie(0);              \
    cout.tie(0);
#define rep(i, x, n) for (ll i = x; i <= n; i++)
#define endl '\n'
const int N = 1e5 + 10, M = 350;

    This section includes necessary headers, defines aliases for data types (ll for long long), and sets up input/output optimizations (IOS). It also defines macros for loops (rep) and endlines (endl), and declares constants for array sizes (N, M).

  • Global Variables:

    ll n, m, len;
ll a[N], add[M], sum[M];

    Here, we declare global variables:

    • n: The size of the array.
    • m: The number of operations.
    • len: The block size (√N).
    • a[N]: The input array.
    • add[M]: The array to store lazy update values for each block.
    • sum[M]: The array to store the sum of each block.

  • get(ll i) Function:

    ll get(ll i)
{
    return i / len;
}

    This function returns the block index for a given array index i.

  • change(ll l, ll r, ll d) Function:

    void change(ll l, ll r, ll d)
{
    if (get(l) == get(r))
    {
        rep(i, l, r) a[i] += d, sum[get(i)] += d;
    }
    else
    {
        ll i = l, j = r;
        while (get(i) == get(l))
            a[i] += d, sum[get(i)] += d, i++;
        while (get(j) == get(r))
            a[j] += d, sum[get(j)] += d, j--;
        rep(k, get(i), get(j))
        {
            sum[k] += len * d;
            add[k] += d;
        }
    }
}

    This function performs the range update operation. It handles the three cases mentioned earlier:

    1. If l and r are in the same block, it directly updates the elements in a and the block sum in sum.
    2. If l and r span multiple blocks, it updates the partial blocks and uses lazy propagation for the complete blocks by updating add.

  • query(ll l, ll r) Function:

    ll query(ll l, ll r)
{
    ll ret = 0;
    if (get(l) == get(r))
    {
        rep(i, l, r) ret += a[i] + add[get(i)];
    }
    else
    {
        ll i = l, j = r;
        while (get(i) == get(l))
            ret += a[i] + add[get(i)], i++;
        while (get(j) == get(r))
            ret += a[j] + add[get(j)], j--;
        rep(k, get(i), get(j))
        {
            ret += sum[k];
        }
    }
    return ret;
}

    This function performs the range query operation. It also handles the three cases:

    1. If l and r are in the same block, it calculates the sum by iterating through the elements in a and adding the lazy update value add.
    2. If l and r span multiple blocks, it calculates the sum for the partial blocks and adds the block sums sum for the complete blocks.

  • solve() Function:

    void solve()
{
    cin >> n;
    m = n;
    rep(i, 1, n) cin >> a[i];
    len = sqrt(n);
    rep(i, 1, n)
    {
        sum[get(i)] += a[i];
    }
    while (m--)
    {
        string op;
        cin >> op;
        if (op == "0")
        {
            ll l, r, d;
            cin >> l >> r >> d;
            change(l, r, d);
        }
        else
        {
            ll l, r, d;
            cin >> l >> r >> d;
            ll ans = query(r, r);
            cout << ans << endl;
        }
    }
}

    This function reads the input, initializes the len and sum arrays, and processes the queries. It reads the operation type (op) and the corresponding parameters, then calls change for updates and query for queries.

  • main() Function:

    int main()
{
    IOS;
    int _ = 1;
    while (_--)
    {
        solve();
    }
    return 0;
}

    The main function sets up input/output optimizations and calls the solve function to handle the problem.

Time Complexity

The time complexity for both the change and query operations is O(√N), where N is the size of the array. This is because we iterate through at most √N elements in the partial blocks and perform constant-time operations on the complete blocks. The overall time complexity for M operations is O(M√N).

Problem 2: Range Updates and Less Than Queries

The second problem builds on the first by introducing a new type of query: counting the number of elements in a range that are less than a certain value. This requires a slightly more sophisticated approach within the block decomposition framework.

Problem Statement

Given an array A of N integers, implement two types of operations:

  1. Range Update: Add a value d to all elements in the range [l, r].
  2. Less Than Query: Count the number of elements in the range [l, r] that are less than c². Note that c is the input, and we need to compare with c².

Approach

For this problem, we'll use block decomposition similar to the first problem. However, we need an efficient way to count the number of elements less than c² within a block. To do this, we'll maintain a sorted array for each block.

We'll have the following data structures:

  • a[N]: The input array.
  • add[M]: The lazy update value for each block.
  • sorted[M]: A vector of sorted elements for each block.

When we perform a range update, we need to:

  1. Update the elements in the partial blocks at the beginning and end of the range. After updating, we need to rebuild the sorted array for these blocks.
  2. Update the add value for the complete blocks in between.

When we perform a less than query, we need to:

  1. Count the elements less than c² in the partial blocks by iterating through the elements and adding the add value for the block.
  2. For the complete blocks, we use binary search on the sorted array to find the number of elements less than c² - add[i], where add[i] is the lazy update value for the i-th block.

Code Explanation

Let's examine the C++ code for this problem:

  • Includes and Type Definitions: (Same as Problem 1)

  • Global Variables:

    ll n, len;
ll a[N], add[M];
vector<ll> sorted[M];

    Here, we have the same variables as before, with the addition of sorted[M], which is a vector of vectors to store the sorted elements for each block.

  • get(ll i) Function:

    ll get(ll i)
{
    return (i - 1) / len + 1;
}

    This function calculates the block index for a given array index i.

  • build(ll block) Function:

    void build(ll block)
{
    ll st = (block - 1) * len + 1;
    ll ed = min((block)*len, n);
    sorted[block].clear();
    rep(i, st, ed)
    {
        sorted[block].push_back(a[i]);
    }
    sort(all(sorted[block]));
}

    This function builds the sorted array for a given block. It iterates through the elements in the block, adds them to the sorted vector, and then sorts the vector.

  • change(ll l, ll r, ll d) Function:

    void change(ll l, ll r, ll d)
{
    ll bl = get(l), br = get(r);
    if (bl == br)
    {
        rep(i, l, r) a[i] += d;
        build(bl);
    }
    else
    {
        rep(i, l, (bl)*len) a[i] += d;
        build(bl);
        rep(i, (br - 1) * len + 1, r) a[i] += d;
        build(br);
        rep(i, bl + 1, br - 1) add[i] += d;
    }
}

    This function performs the range update operation. If the range falls within a single block, it updates the elements and rebuilds the sorted array for that block. If the range spans multiple blocks, it updates the partial blocks and rebuilds their sorted arrays, and updates the add values for the complete blocks.

  • query_less(ll l, ll r, ll c) Function:

    ll query_less(ll l, ll r, ll c)
{
    ll x = c * c;
    ll bl = get(l), br = get(r);
    ll ans = 0;
    if (bl == br)
    {
        rep(i, l, r)
        {
            if (a[i] + add[bl] < x)
                ans++;
        }
    }
    else
    {
        rep(i, l, bl * len)
        {
            if (a[i] + add[bl] < x)
                ans++;
        }
        rep(i, (br - 1) * len + 1, r)
        {
            if (a[i] + add[br] < x)
                ans++;
        }
        rep(i, bl + 1, br - 1)
        {
            ll d = x - add[i];
            ll pos = lower_bound(all(sorted[i]), d) - sorted[i].begin();
            ans += pos;
        }
    }
    return ans;
}

    This function performs the less than query. It calculates c² and then counts the elements less than c² in the given range. For partial blocks, it iterates through the elements and checks the condition. For complete blocks, it uses lower_bound to perform a binary search on the sorted array, finding the number of elements less than c² - add[i].

  • solve() Function:

    void solve()
{
    cin >> n;
    len = sqrt(n);
    ll t = get(n);
    rep(i, 1, n) cin >> a[i];
    rep(i, 1, t)
    {
        add[i] = 0;
        build(i);
    }
    rep(i, 1, n)
    {
        ll op, l, r, d;
        cin >> op >> l >> r >> d;
        if (op == 0)
        {
            change(l, r, d);
        }
        else
        {
            ll ans = query_less(l, r, d);
            cout << ans << endl;
        }
    }
}

    This function reads the input, initializes the len, add, and sorted arrays, and processes the queries. It reads the operation type (op) and the corresponding parameters, then calls change for updates and query_less for queries.

  • main() Function: (Same as Problem 1)

Time Complexity

  • Range Update: O(√N) for updating partial blocks and rebuilding sorted arrays, and O(1) for updating add values in complete blocks. Overall, O(√N).
  • Less Than Query: O(√N) for iterating through partial blocks and O(log √N) for binary search on each complete block. Overall, O(√N log N).

The overall time complexity for M operations is O(M√N log N).

Key Takeaways

  • Block decomposition is a versatile technique for handling range queries and updates efficiently.
  • Dividing the data into blocks of size √N provides a good balance between update and query performance.
  • Lazy propagation helps optimize range updates by deferring updates to individual elements.
  • Maintaining sorted arrays within blocks allows for efficient counting of elements within a certain range.

Conclusion

We've explored the first two problems in a series of introductory block decomposition challenges. By understanding the core concepts and techniques, you're well-equipped to tackle more complex problems. Remember, block decomposition is a powerful tool in your algorithm design arsenal, and practice is key to mastering it. Keep exploring, experimenting, and pushing your boundaries!

For further reading and advanced topics on block decomposition, check out CP-Algorithms, a trusted resource for competitive programming algorithms and data structures.