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Queries On Array with disappearing and reappearing elements

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  • Difficulty Level : Expert
  • Last Updated : 24 Feb, 2022
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Given an array, arr[] of size N. Each second, an integer disappears for N seconds and after N seconds, it reappears at its original position. Integer disappear in the order from left to right arr[0], arr[1], …, arr[N – 1]. After all the integers disappear, they start reappearing until all integers reappear. Once N elements appear again, the process starts again. 
Now given Q queries, each consists of two integers t and M. The task is to determine the Mth element from the left at tth second. If the array does not exist till M, then print -1.

Examples: 

Input: arr[] = {1, 2, 3, 4, 5}, Q = {{1, 4}, {6, 1}, {3, 5}} 
Output: 


-1 
At time, 
t1 -> {2, 3, 4, 5} 
t2 -> {3, 4, 5} 
t3 -> {4, 5} 
t4 -> {5} 
t5 -> {} 
t6 -> {1}

Input: arr[] = {5, 4, 3, 4, 5}, Q = {{2, 3}, {100000000, 2}} 
Output: 

4

Approach: The main approach is that it is required to check whether the array is empty or full and that can be seen by dividing the number of turns by the size of the array. If the remainder is 0, then it can be either one of the cases ( empty or fill ). 
By observation, it is seen that in the odd turn the array is reducing and in even turns the array is expanding and using this observation it will be checked that M is outside of the index or inside the array.
Below is the implementation of the above approach: 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to perform the queries
void PerformQueries(vector<int>& a,
                    vector<pair<long long, int> >& vec)
{
 
    vector<int> ans;
 
    // Size of the array with
    // 1-based indexing
    int n = (int)a.size() - 1;
 
    // Number of queries
    int q = (int)vec.size();
 
    // Iterating through the queries
    for (int i = 0; i < q; ++i) {
 
        long long t = vec[i].first;
        int m = vec[i].second;
 
        // If m is more than the
        // size of the array
        if (m > n) {
            ans.push_back(-1);
            continue;
        }
 
        // Count of turns
        int turn = t / n;
 
        // Find the remainder
        int rem = t % n;
 
        // If the remainder is 0 and turn is
        // odd then the array is empty
        if (rem == 0 and turn % 2 == 1) {
            ans.push_back(-1);
            continue;
        }
 
        // If the remainder is 0 and turn is
        // even then array is full and
        // is in its initial state
        if (rem == 0 and turn % 2 == 0) {
            ans.push_back(a[m]);
            continue;
        }
 
        // If the remainder is not 0
        // and the turn is even
        if (turn % 2 == 0) {
 
            // Current size of the array
            int cursize = n - rem;
 
            if (cursize < m) {
                ans.push_back(-1);
                continue;
            }
            ans.push_back(a[m + rem]);
        }
        else {
 
            // Current size of the array
            int cursize = rem;
            if (cursize < m) {
                ans.push_back(-1);
                continue;
            }
            ans.push_back(a[m]);
        }
    }
 
    // Print the result
    for (int i : ans)
        cout << i << "\n";
}
 
// Driver code
int main()
{
 
    // The initial array, -1 is for
    // 1 base indexing
    vector<int> a = { -1, 1, 2, 3, 4, 5 };
 
    // Queries in the form of the pairs of (t, M)
    vector<pair<long long, int> > vec = {
        { 1, 4 },
        { 6, 1 },
        { 3, 5 }
    };
 
    PerformQueries(a, vec);
 
    return 0;
}

Java




// Java implementation of the approach
 
import java.util.*;
 
class GFG
{
 
    // Function to perform the queries
    static void PerformQueries(int[] a, int[][] vec)
    {
 
        Vector<Integer> ans = new Vector<>();
 
        // Size of the array with
        // 1-based indexing
        int n = (int) a.length - 1;
 
        // Number of queries
        int q = (int) vec.length;
 
        // Iterating through the queries
        for (int i = 0; i < q; ++i)
        {
 
            long t = vec[i][0];
            int m = vec[i][1];
 
            // If m is more than the
            // size of the array
            if (m > n)
            {
                ans.add(-1);
                continue;
            }
 
            // Count of turns
            int turn = (int) (t / n);
 
            // Find the remainder
            int rem = (int) (t % n);
 
            // If the remainder is 0 and turn is
            // odd then the array is empty
            if (rem == 0 && turn % 2 == 1)
            {
                ans.add(-1);
                continue;
            }
 
            // If the remainder is 0 and turn is
            // even then array is full and
            // is in its initial state
            if (rem == 0 && turn % 2 == 0)
            {
                ans.add(a[m]);
                continue;
            }
 
            // If the remainder is not 0
            // and the turn is even
            if (turn % 2 == 0)
            {
 
                // Current size of the array
                int cursize = n - rem;
 
                if (cursize < m)
                {
                    ans.add(-1);
                    continue;
                }
                ans.add(a[m + rem]);
            }
            else
            {
 
                // Current size of the array
                int cursize = rem;
                if (cursize < m)
                {
                    ans.add(-1);
                    continue;
                }
                ans.add(a[m]);
            }
        }
 
        // Print the result
        for (int i : ans)
            System.out.print(i + "\n");
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        // The initial array, -1 is for
        // 1 base indexing
        int[] a = { -1, 1, 2, 3, 4, 5 };
 
        // Queries in the form of the pairs of (t, M)
        int[][] vec = { { 1, 4 }, { 6, 1 }, { 3, 5 } };
 
        PerformQueries(a, vec);
 
    }
}
 
// This code is contributed by 29AjayKumar

Python3




# Python3 implementation of the approach
 
# Function to perform the queries
def PerformQueries(a, vec) :
 
    ans = [];
 
    # Size of the array with
    # 1-based indexing
    n = len(a) - 1;
 
    # Number of queries
    q = len(vec);
 
    # Iterating through the queries
    for i in range(q) :
 
        t = vec[i][0];
        m = vec[i][1];
 
        # If m is more than the
        # size of the array
        if (m > n) :
            ans.append(-1);
            continue;
 
        # Count of turns
        turn = t // n;
 
        # Find the remainder
        rem = t % n;
 
        # If the remainder is 0 and turn is
        # odd then the array is empty
        if (rem == 0 and turn % 2 == 1) :
            ans.append(-1);
            continue;
 
        # If the remainder is 0 and turn is
        # even then array is full and
        # is in its initial state
        if (rem == 0 and turn % 2 == 0) :
            ans.append(a[m]);
            continue;
 
        # If the remainder is not 0
        # and the turn is even
        if (turn % 2 == 0) :
 
            # Current size of the array
            cursize = n - rem;
 
            if (cursize < m) :
                ans.append(-1);
                continue;
     
            ans.append(a[m + rem]);
             
        else :
 
            # Current size of the array
            cursize = rem;
             
            if (cursize < m) :
                ans.append(-1);
                continue;
         
            ans.append(a[m]);
 
    # Print the result
    for i in ans :
        print(i) ;
 
# Driver code
if __name__ == "__main__" :
 
    # The initial array, -1 is for
    # 1 base indexing
    a = [ -1, 1, 2, 3, 4, 5 ];
 
    # Queries in the form of the pairs of (t, M)
    vec = [
        [ 1, 4 ],
        [ 6, 1 ],
        [ 3, 5 ]
    ];
 
    PerformQueries(a, vec);
 
# This code is contributed by AnkitRai01

C#




// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
    // Function to perform the queries
    static void PerformQueries(int[] a, int[,] vec)
    {
 
        List<int> ans = new List<int>();
 
        // Size of the array with
        // 1-based indexing
        int n = (int) a.Length - 1;
 
        // Number of queries
        int q = (int) vec.GetLength(0);
 
        // Iterating through the queries
        for (int i = 0; i < q; ++i)
        {
 
            long t = vec[i, 0];
            int m = vec[i, 1];
 
            // If m is more than the
            // size of the array
            if (m > n)
            {
                ans.Add(-1);
                continue;
            }
 
            // Count of turns
            int turn = (int) (t / n);
 
            // Find the remainder
            int rem = (int) (t % n);
 
            // If the remainder is 0 and turn is
            // odd then the array is empty
            if (rem == 0 && turn % 2 == 1)
            {
                ans.Add(-1);
                continue;
            }
 
            // If the remainder is 0 and turn is
            // even then array is full and
            // is in its initial state
            if (rem == 0 && turn % 2 == 0)
            {
                ans.Add(a[m]);
                continue;
            }
 
            // If the remainder is not 0
            // and the turn is even
            if (turn % 2 == 0)
            {
 
                // Current size of the array
                int cursize = n - rem;
 
                if (cursize < m)
                {
                    ans.Add(-1);
                    continue;
                }
                ans.Add(a[m + rem]);
            }
            else
            {
 
                // Current size of the array
                int cursize = rem;
                if (cursize < m)
                {
                    ans.Add(-1);
                    continue;
                }
                ans.Add(a[m]);
            }
        }
 
        // Print the result
        foreach (int i in ans)
            Console.Write(i + "\n");
    }
 
    // Driver code
    public static void Main(String[] args)
    {
 
        // The initial array, -1 is for
        // 1 base indexing
        int[] a = { -1, 1, 2, 3, 4, 5 };
 
        // Queries in the form of the pairs of (t, M)
        int[,] vec = { { 1, 4 }, { 6, 1 }, { 3, 5 } };
 
        PerformQueries(a, vec);
 
    }
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
// Javascript implementation of the approach
 
 
// Function to perform the queries
function PerformQueries(a, vec) {
 
    let ans = new Array();
 
    // Size of the array with
    // 1-based indexing
    let n = a.length - 1;
 
    // Number of queries
    let q = vec.length;
 
    // Iterating through the queries
    for (let i = 0; i < q; ++i) {
 
        let t = vec[i][0];
        let m = vec[i][1];
 
        // If m is more than the
        // size of the array
        if (m > n) {
            ans.push(-1);
            continue;
        }
 
        // Count of turns
        let turn = Math.floor(t / n);
 
        // Find the remainder
        let rem = t % n;
 
        // If the remainder is 0 and turn is
        // odd then the array is empty
        if (rem == 0 && turn % 2 == 1) {
            ans.push(-1);
            continue;
        }
 
        // If the remainder is 0 and turn is
        // even then array is full and
        // is in its initial state
        if (rem == 0 && turn % 2 == 0) {
            ans.push(a[m]);
            continue;
        }
 
        // If the remainder is not 0
        // and the turn is even
        if (turn % 2 == 0) {
 
            // Current size of the array
            let cursize = n - rem;
 
            if (cursize < m) {
                ans.push(-1);
                continue;
            }
            ans.push(a[m + rem]);
        }
        else {
 
            // Current size of the array
            let cursize = rem;
            if (cursize < m) {
                ans.push(-1);
                continue;
            }
            ans.push(a[m]);
        }
    }
 
    // Print the result
    for (let i of ans)
        document.write(i + "<br>");
}
 
// Driver code
 
 
// The initial array, -1 is for
// 1 base indexing
let a = [-1, 1, 2, 3, 4, 5];
 
// Queries in the form of the pairs of (t, M)
let vec = [
    [1, 4],
    [6, 1],
    [3, 5]];
 
PerformQueries(a, vec);
</script>

Output: 

5
1
-1

 

Time Complexity: O(q)

Auxiliary Space: O(q)


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