Horje
Increasing Decreasing Update Queries (Akku and Arrays)

Akku has solved many problems, she is a genius. One day her friend gave her an Array of sizes n and asked her to perform some queries of the following type:

Each query consists of three integers

1 A B: Update the Array at index A by value B
2 A B: if the subarray from index A to B (both inclusive) is

  1. Both increasing(Non-decreasing) and decreasing(Non-increasing) print -1
  2. Only increasing(Non-decreasing) print 0
  3. Only decreasing(Non-increasing) print 1
  4. Neither increasing nor decreasing print -1

Akku needs your help, can you help her?

Example:

Input: nums = {1,5,7,4,3,5,9}, queries = {{2,1,3},{1,7,4},{2,6,7}}
Output: {0,1}
Explanation:
For the 1st query given : A = 1, B = 3. From 1 to 3(1,5,7) elements are in increasing order. So answer is 0.
For the 2nd query we have to update the 7th element of the array by 4. So new updated array will be {1,5,7,4,3,5,4}
For the 3rd query A = 6, B = 7. From 6 to 7 (5, 4) elements are in descending order. So answer is 1.

Input: nums = {3, 2, 5, 4, 7, 1, 6}, queries = {{2, 1, 3}, {1, 4, 6}, {2, 2, 5}, {1, 3, 8}}
Output: {-1, 0}

https://practice.horje.org/problems/akku-and-arrays4452/1

Approach:

The idea is to use a segment tree data structure to store information about the subarrays of the given array. Each node of the segment tree represents a subarray and has four attributes: dec, inc, lm, and rm. These attributes indicate whether the subarray is decreasing, increasing, the leftmost and rightmost elements of the subarray respectively.

We are going to use the below function to execute our logic:

We’ll use two functions: merge() and update(). The merge function takes two nodes of the segment tree and returns a new node that represents the merged subarray of the two nodes. The update function modifies the segment tree by changing the value of a given index in the array and updating the nodes affected by the change.

We’ll also use a function called query() that takes a range of indices and returns the node that represents the subarray in that range. The function uses the merge function to combine the nodes that cover the range in the segment tree.

We’ll also use a function called solveQueries() that takes the array and the queries as input and returns a vector of integers as output. Iterates over the queries and performs the appropriate operation based on the query type.

  • For query type 1, it calls the update function to change the value of the array at a given index.
  • For query type 2, it calls the query function to get the node for the given range and then checks the attributes of the node to determine the answer.

The answer is -1 if the subarray is neither increasing nor decreasing, 0 if it is only increasing, 1 if it is only decreasing, and -1 if it is both increasing and decreasing. Finally, pushes the answer to the output vector and returns it at the end.

Steps:

  • Initialize a segment tree data structure where each node stores information about a subarray of the input array. Each node should have four attributes: `dec`, `inc`, `lm`, and `rm`.
  • Create a function to merge two nodes in the segment tree.
    • It takes two nodes as input and returns a new node that represents the merged subarray. The new node’s attributes are determined based on the attributes of the input nodes.
  • Create a function to update the segment tree when a value in the array is changed.
    • It takes an index and a new value as input, changes the value at the given index in the array, and updates the corresponding nodes in the segment tree.
  • Create a function to perform a query on the segment tree.
    • It takes a range of indices as input and returns a node that represents the subarray in the given range. The returned node’s attributes are determined by merging the nodes that cover the range in the segment tree.
  • Iterate over the queries. For each query, check the query type.
    • If it’s an update query (type 1), call the update function with the given index and value.
    • If it’s a subarray query (type 2), call the query function with the given range of indices, then check the returned node’s attributes to determine the answer. Store the answers in an array.
  • After all queries have been processed, return the array of answers. Each answer corresponds to a subarray query in the input queries array, and is determined based on the attributes of the subarray in the segment tree.

Below is the implementation of the above approach:

C++

#include <bits/stdc++.h>
using namespace std;
 
// Define a class for the nodes in the segment tree
class Node {
public:
    bool dec
        = false; // Indicates if the subarray is decreasing
    bool inc
        = false; // Indicates if the subarray is increasing
    int lm = -1; // The leftmost element of the subarray
    int rm = -1; // The rightmost element of the subarray
};
 
class Solution {
public:
    vector<Node> seg; // The segment tree
 
    // Function to merge two nodes in the segment tree
    Node merge(Node& n1, Node& n2)
    {
        // If either of the nodes represents an empty
        // subarray, return the other node
        if (n1.lm == -1 || n2.lm == -1)
            return n1.lm == -1 ? n2 : n1;
 
        // Merge information from two nodes to update the
        // parent node
        Node ans;
        ans.lm = n1.lm;
        ans.rm = n2.rm;
        if (n1.inc && n2.inc && n1.rm <= n2.lm)
            ans.inc = true;
        if (n1.dec && n2.dec && n1.rm >= n2.lm)
            ans.dec = true;
        return ans;
    }
 
    // Function to update the segment tree
    void update(int idx, int l, int r, int& id, int& val)
    {
        // If the leaf node is reached, update it with the
        // new value
        if (l == r) {
            seg[idx] = { true, true, val, val };
            return;
        }
 
        // Otherwise, recursively update the appropriate
        // child node
        int mid = (l + r) / 2;
        if (id <= mid)
            update(2 * idx + 1, l, mid, id, val);
        else
            update(2 * idx + 2, mid + 1, r, id, val);
 
        // After updating the child nodes, merge their
        // information to update the current node
        seg[idx]
            = merge(seg[2 * idx + 1], seg[2 * idx + 2]);
    }
 
    // Function to perform a query on the segment tree
    Node query(int idx, int l, int r, int& ql, int& qr)
    {
        // If the current segment is completely outside the
        // query range, return an empty node
        if (ql > r || qr < l)
            return { false, false, -1, -1 };
 
        // If the current segment is completely inside the
        // query range, return the node representing it
        if (ql <= l && qr >= r)
            return seg[idx];
 
        // Otherwise, recursively query the child nodes and
        // merge their information
        int mid = (l + r) / 2;
        Node lq = query(2 * idx + 1, l, mid, ql, qr);
        Node rq = query(2 * idx + 2, mid + 1, r, ql, qr);
        return merge(lq, rq);
    }
 
    // Function to solve the queries
    vector<int> solveQueries(vector<int> nums,
                            vector<vector<int> > queries)
    {
        vector<int> ans;
        int n = nums.size();
        seg.resize(4 * n
                + 1); // Resize the segment tree to
                        // accommodate the given array
 
        // Initializing the segment tree with the initial
        // array
        for (int i = 0; i < n; i++) {
            update(0, 0, n - 1, i, nums[i]);
        }
 
        // Processing each query
        for (auto it : queries) {
            int q = it[0];
            int a = it[1] - 1;
            int b = it[2] - 1;
            if (q == 1) {
                // Update query: Increment the value at
                // index 'a' by 'b'
                update(0, 0, n - 1, a, ++b);
            }
            else {
                // Subarray query: Check properties of the
                // subarray from index 'a' to 'b'
                Node res = query(0, 0, n - 1, a, b);
                if (res.inc == res.dec)
                    ans.push_back(-1); // Both increasing
                                    // and decreasing
                else if (res.dec)
                    ans.push_back(1); // Only decreasing
                else
                    ans.push_back(0); // Only increasing
            }
        }
        return ans;
    }
};
 
int main()
{
    Solution solution;
    vector<int> nums = { 1, 5, 7, 4, 3, 5, 9 };
    vector<vector<int> > queries
        = { { 2, 1, 3 }, { 1, 7, 4 }, { 2, 6, 7 } };
    vector<int> result
        = solution.solveQueries(nums, queries);
 
    cout << "Result: ";
    for (int res : result) {
        cout << res << " ";
    }
    cout << endl;
 
    return 0;
}

Java

import java.util.*;
 
class Node {
    boolean dec
        = false; // Indicates if the subarray is decreasing
    boolean inc
        = false; // Indicates if the subarray is increasing
    int lm = -1; // The leftmost element of the subarray
    int rm = -1; // The rightmost element of the subarray
}
 
public class Solution {
    ArrayList<Node> seg
        = new ArrayList<>(); // The segment tree
 
    // Function to merge two nodes in the segment tree
    Node merge(Node n1, Node n2)
    {
        // If either of the nodes represents an empty
        // subarray, return the other node
        if (n1.lm == -1 || n2.lm == -1)
            return n1.lm == -1 ? n2 : n1;
 
        // Merge information from two nodes to update the
        // parent node
        Node ans = new Node();
        ans.lm = n1.lm;
        ans.rm = n2.rm;
        if (n1.inc && n2.inc && n1.rm <= n2.lm)
            ans.inc = true;
        if (n1.dec && n2.dec && n1.rm >= n2.lm)
            ans.dec = true;
        return ans;
    }
 
    // Function to update the segment tree
    void update(int idx, int l, int r, int id, int val)
    {
        // If the leaf node is reached, update it with the
        // new value
        if (l == r) {
            seg.set(idx, new Node());
            seg.get(idx).inc = true;
            seg.get(idx).dec = true;
            seg.get(idx).lm = val;
            seg.get(idx).rm = val;
            return;
        }
 
        // Otherwise, recursively update the appropriate
        // child node
        int mid = (l + r) / 2;
        if (id <= mid)
            update(2 * idx + 1, l, mid, id, val);
        else
            update(2 * idx + 2, mid + 1, r, id, val);
 
        // After updating the child nodes, merge their
        // information to update the current node
        seg.set(idx, merge(seg.get(2 * idx + 1),
                           seg.get(2 * idx + 2)));
    }
 
    // Function to perform a query on the segment tree
    Node query(int idx, int l, int r, int ql, int qr)
    {
        // If the current segment is completely outside the
        // query range, return an empty node
        if (ql > r || qr < l)
            return new Node();
 
        // If the current segment is completely inside the
        // query range, return the node representing it
        if (ql <= l && qr >= r)
            return seg.get(idx);
 
        // Otherwise, recursively query the child nodes and
        // merge their information
        int mid = (l + r) / 2;
        Node lq = query(2 * idx + 1, l, mid, ql, qr);
        Node rq = query(2 * idx + 2, mid + 1, r, ql, qr);
        return merge(lq, rq);
    }
 
    // Function to solve the queries
    ArrayList<Integer>
    solveQueries(ArrayList<Integer> nums,
                 ArrayList<ArrayList<Integer> > queries)
    {
        ArrayList<Integer> ans = new ArrayList<>();
        int n = nums.size();
        seg = new ArrayList<>(Collections.nCopies(
            4 * n + 1,
            new Node())); // Resize the segment tree to
                          // accommodate the given array
 
        // Initializing the segment tree with the initial
        // array
        for (int i = 0; i < n; i++) {
            update(0, 0, n - 1, i, nums.get(i));
        }
 
        // Processing each query
        for (ArrayList<Integer> it : queries) {
            int q = it.get(0);
            int a = it.get(1) - 1;
            int b = it.get(2) - 1;
            if (q == 1) {
                // Update query: Increment the value at
                // index 'a' by 'b'
                update(0, 0, n - 1, a, ++b);
            }
            else {
                // Subarray query: Check properties of the
                // subarray from index 'a' to 'b'
                Node res = query(0, 0, n - 1, a, b);
                if (res.inc == res.dec)
                    ans.add(-1); // Both increasing and
                                 // decreasing
                else if (res.dec)
                    ans.add(1); // Only decreasing
                else
                    ans.add(0); // Only increasing
            }
        }
        return ans;
    }
 
    public static void main(String[] args)
    {
        Solution solution = new Solution();
        ArrayList<Integer> nums = new ArrayList<>(
            Arrays.asList(1, 5, 7, 4, 3, 5, 9));
        ArrayList<ArrayList<Integer> > queries
            = new ArrayList<>();
        queries.add(
            new ArrayList<>(Arrays.asList(2, 1, 3)));
        queries.add(
            new ArrayList<>(Arrays.asList(1, 7, 4)));
        queries.add(
            new ArrayList<>(Arrays.asList(2, 6, 7)));
        ArrayList<Integer> result
            = solution.solveQueries(nums, queries);
 
        System.out.print("Result: ");
        for (int res : result) {
            System.out.print(res + " ");
        }
        System.out.println();
    }
}

Python3

class Node:
    def __init__(self):
        self.dec = False  # Indicates if the subarray is decreasing
        self.inc = False  # Indicates if the subarray is increasing
        self.lm = -1      # The leftmost element of the subarray
        self.rm = -1      # The rightmost element of the subarray
 
 
class Solution:
    def __init__(self):
        self.seg = []  # The segment tree
 
    # Function to merge two nodes in the segment tree
    def merge(self, n1, n2):
        # If either of the nodes represents an empty
        # subarray, return the other node
        if n1.lm == -1 or n2.lm == -1:
            return n2 if n1.lm == -1 else n1
 
        # Merge information from two nodes to update the
        # parent node
        ans = Node()
        ans.lm = n1.lm
        ans.rm = n2.rm
        if n1.inc and n2.inc and n1.rm <= n2.lm:
            ans.inc = True
        if n1.dec and n2.dec and n1.rm >= n2.lm:
            ans.dec = True
        return ans
 
    # Function to update the segment tree
    def update(self, idx, l, r, id, val):
        # If the leaf node is reached, update it with the
        # new value
        if l == r:
            self.seg[idx] = Node()
            self.seg[idx].dec = True
            self.seg[idx].inc = True
            self.seg[idx].lm = val
            self.seg[idx].rm = val
            return
 
        # Otherwise, recursively update the appropriate
        # child node
        mid = (l + r) // 2
        if id <= mid:
            self.update(2 * idx + 1, l, mid, id, val)
        else:
            self.update(2 * idx + 2, mid + 1, r, id, val)
 
        # After updating the child nodes, merge their
        # information to update the current node
        self.seg[idx] = self.merge(self.seg[2 * idx + 1], self.seg[2 * idx + 2])
 
    # Function to perform a query on the segment tree
    def query(self, idx, l, r, ql, qr):
        # If the current segment is completely outside the
        # query range, return an empty node
        if ql > r or qr < l:
            return Node()
 
        # If the current segment is completely inside the
        # query range, return the node representing it
        if ql <= l and qr >= r:
            return self.seg[idx]
 
        # Otherwise, recursively query the child nodes and
        # merge their information
        mid = (l + r) // 2
        lq = self.query(2 * idx + 1, l, mid, ql, qr)
        rq = self.query(2 * idx + 2, mid + 1, r, ql, qr)
        return self.merge(lq, rq)
 
    # Function to solve the queries
    def solveQueries(self, nums, queries):
        ans = []
        n = len(nums)
        self.seg = [Node()] * (4 * n + 1)  # Resize the segment tree to accommodate the given array
 
        # Initializing the segment tree with the initial
        # array
        for i in range(n):
            self.update(0, 0, n - 1, i, nums[i])
 
        # Processing each query
        for it in queries:
            q, a, b = it
            a -= 1
            b -= 1
            if q == 1:
                # Update query: Increment the value at
                # index 'a' by 'b'
                self.update(0, 0, n - 1, a, b + 1)
            else:
                # Subarray query: Check properties of the
                # subarray from index 'a' to 'b'
                res = self.query(0, 0, n - 1, a, b)
                if res.inc == res.dec:
                    ans.append(-1)  # Both increasing and decreasing
                elif res.dec:
                    ans.append(1)  # Only decreasing
                else:
                    ans.append(0)  # Only increasing
        return ans
 
 
solution = Solution()
nums = [1, 5, 7, 4, 3, 5, 9]
queries = [[2, 1, 3], [1, 7, 4], [2, 6, 7]]
result = solution.solveQueries(nums, queries)
 
print("Result:", " ".join(map(str, result)))

C#

using System;
using System.Collections.Generic;
 
public class Node
{
    public bool Dec = false;  // Indicates if the subarray is decreasing
    public bool Inc = false;  // Indicates if the subarray is increasing
    public int Lm = -1;       // The leftmost element of the subarray
    public int Rm = -1;       // The rightmost element of the subarray
}
 
public class Solution
{
    private List<Node> seg;  // The segment tree
 
    // Function to merge two nodes in the segment tree
    private Node Merge(Node n1, Node n2)
    {
        // If either of the nodes represents an empty subarray, return the other node
        if (n1.Lm == -1 || n2.Lm == -1)
            return n1.Lm == -1 ? n2 : n1;
 
        // Merge information from two nodes to update the parent node
        Node ans = new Node();
        ans.Lm = n1.Lm;
        ans.Rm = n2.Rm;
        if (n1.Inc && n2.Inc && n1.Rm <= n2.Lm)
            ans.Inc = true;
        if (n1.Dec && n2.Dec && n1.Rm >= n2.Lm)
            ans.Dec = true;
        return ans;
    }
 
    // Function to update the segment tree
    private void Update(int idx, int l, int r, int id, int val)
    {
        // If the leaf node is reached, update it with the new value
        if (l == r)
        {
            seg[idx] = new Node { Inc = true, Dec = true, Lm = val, Rm = val };
            return;
        }
 
        // Otherwise, recursively update the appropriate child node
        int mid = (l + r) / 2;
        if (id <= mid)
            Update(2 * idx + 1, l, mid, id, val);
        else
            Update(2 * idx + 2, mid + 1, r, id, val);
 
        // After updating the child nodes, merge their information to update the current node
        seg[idx] = Merge(seg[2 * idx + 1], seg[2 * idx + 2]);
    }
 
    // Function to perform a query on the segment tree
    private Node Query(int idx, int l, int r, int ql, int qr)
    {
        // If the current segment is completely outside the query range, return an empty node
        if (ql > r || qr < l)
            return new Node();
 
        // If the current segment is completely inside the query range, return the node representing it
        if (ql <= l && qr >= r)
            return seg[idx];
 
        // Otherwise, recursively query the child nodes and merge their information
        int mid = (l + r) / 2;
        Node lq = Query(2 * idx + 1, l, mid, ql, qr);
        Node rq = Query(2 * idx + 2, mid + 1, r, ql, qr);
        return Merge(lq, rq);
    }
 
    // Function to solve the queries
    private List<int> SolveQueries(List<int> nums, List<List<int>> queries)
    {
        List<int> ans = new List<int>();
        int n = nums.Count;
        seg = new List<Node>();
 
        // Initializing the segment tree with the initial array
        for (int i = 0; i < 4 * n + 1; i++)
        {
            seg.Add(new Node());
        }
 
        for (int i = 0; i < n; i++)
        {
            Update(0, 0, n - 1, i, nums[i]);
        }
 
        // Processing each query
        foreach (List<int> it in queries)
        {
            int q = it[0];
            int a = it[1] - 1;
            int b = it[2] - 1;
            if (q == 1)
            {
                // Update query: Increment the value at index 'a' by 'b'
                Update(0, 0, n - 1, a, ++b);
            }
            else
            {
                // Subarray query: Check properties of the subarray from index 'a' to 'b'
                Node res = Query(0, 0, n - 1, a, b);
                if (res.Inc == res.Dec)
                    ans.Add(-1);  // Both increasing and decreasing
                else if (res.Dec)
                    ans.Add(1);   // Only decreasing
                else
                    ans.Add(0);   // Only increasing
            }
        }
        return ans;
    }
 
    public static void Main(string[] args)
    {
        Solution solution = new Solution();
        List<int> nums = new List<int> { 1, 5, 7, 4, 3, 5, 9 };
        List<List<int>> queries = new List<List<int>>
        {
            new List<int> { 2, 1, 3 },
            new List<int> { 1, 7, 4 },
            new List<int> { 2, 6, 7 }
        };
        List<int> result = solution.SolveQueries(nums, queries);
 
        Console.Write("Result: ");
        foreach (int res in result)
        {
            Console.Write(res + " ");
        }
        Console.WriteLine();
    }
}

Javascript

class Node {
    constructor() {
        this.Dec = false; // Indicates if the subarray is decreasing
        this.Inc = false; // Indicates if the subarray is increasing
        this.Lm = -1;     // The leftmost element of the subarray
        this.Rm = -1;     // The rightmost element of the subarray
    }
}
 
class Solution {
    constructor() {
        this.seg = []; // The segment tree
    }
 
    // Function to merge two nodes in the segment tree
    Merge(n1, n2) {
        // If either of the nodes represents an empty subarray, return the other node
        if (!n1 || !n2 || n1.Lm == -1 || n2.Lm == -1)
            return n1 && n1.Lm == -1 ? n2 : n1;
 
        // Merge information from two nodes to update the parent node
        const ans = new Node();
        ans.Lm = n1.Lm;
        ans.Rm = n2.Rm;
        if (n1.Inc && n2.Inc && n1.Rm <= n2.Lm)
            ans.Inc = true;
        if (n1.Dec && n2.Dec && n1.Rm >= n2.Lm)
            ans.Dec = true;
        return ans;
    }
 
    // Function to update the segment tree
    Update(idx, l, r, id, val) {
        // If the leaf node is reached, update it with the new value
        if (l === r) {
            this.seg[idx] = new Node();
            this.seg[idx].Inc = true;
            this.seg[idx].Dec = true;
            this.seg[idx].Lm = val;
            this.seg[idx].Rm = val;
            return;
        }
 
        // Otherwise, recursively update the appropriate child node
        const mid = Math.floor((l + r) / 2);
        if (id <= mid)
            this.Update(2 * idx + 1, l, mid, id, val);
        else
            this.Update(2 * idx + 2, mid + 1, r, id, val);
 
        // After updating the child nodes, merge their information to update the current node
        this.seg[idx] = this.Merge(this.seg[2 * idx + 1], this.seg[2 * idx + 2]);
    }
 
    // Function to perform a query on the segment tree
    Query(idx, l, r, ql, qr) {
        // If the current segment is completely outside the query range, return an empty node
        if (ql > r || qr < l)
            return new Node();
 
        // If the current segment is completely inside the query range, return the node representing it
        if (ql <= l && qr >= r)
            return this.seg[idx];
 
        // Otherwise, recursively query the child nodes and merge their information
        const mid = Math.floor((l + r) / 2);
        const lq = this.Query(2 * idx + 1, l, mid, ql, qr);
        const rq = this.Query(2 * idx + 2, mid + 1, r, ql, qr);
        return this.Merge(lq, rq);
    }
 
    // Function to solve the queries
    SolveQueries(nums, queries) {
        const ans = [];
        const n = nums.length;
        this.seg = new Array(4 * n + 1).fill(null);
 
        // Initializing the segment tree with the initial array
        for (let i = 0; i < n; i++) {
            this.Update(0, 0, n - 1, i, nums[i]);
        }
 
        // Processing each query
        for (const it of queries) {
            const q = it[0];
            let a = it[1] - 1;
            let b = it[2] - 1;
            if (q === 1) {
                // Update query: Increment the value at index 'a' by 'b'
                this.Update(0, 0, n - 1, a, ++b);
            } else {
                // Subarray query: Check properties of the subarray from index 'a' to 'b'
                const res = this.Query(0, 0, n - 1, a, b);
                if (res.Inc === res.Dec)
                    ans.push(-1); // Both increasing and decreasing
                else if (res.Dec)
                    ans.push(1); // Only decreasing
                else
                    ans.push(0); // Only increasing
            }
        }
        return ans;
    }
}
 
const solution = new Solution();
const nums = [1, 5, 7, 4, 3, 5, 9];
const queries = [
    [2, 1, 3],
    [1, 7, 4],
    [2, 6, 7]
];
const result = solution.SolveQueries(nums, queries);
 
console.log("Result:", result.join(" "));
//This code is contributed by Adarsh

Output

Result: 0 1

Time Complexity: O(N+QlogN), where N is the size of the input array, and Q is the number of queries.
Auxiliary Space: O(N)


 


Reffered: https://www.geeksforgeeks.org

Arrays

Related
Nitika and her queries Nitika and her queries
Maximum Sum of Array with given MEX Maximum Sum of Array with given MEX
Minimum cost to buy all items at least once Minimum cost to buy all items at least once
Optimal Placement of People in Circular Locations Optimal Placement of People in Circular Locations
Minimum Operations to Sort Array with Custom Shuffling Minimum Operations to Sort Array with Custom Shuffling

Type:
 Geek
Category:
 Coding
Sub Category:
 Tutorial
Uploaded by:
 Admin
Views:
 13



 

Horje