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:
5
1
-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:
5
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++ 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 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 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# 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 |
<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> |
5 1 -1
Time Complexity: O(q)
Auxiliary Space: O(q)