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K-th term from given N merged Arithmetic Progressions
  • Last Updated : 09 Dec, 2020

Given an integer K and an array arr[] of N integers, each of which is the first term and common difference of an Arithemetic Progression, the task is to find the Kth element of the set S formed by merging the N arithmetic progressions.

Examples:  

Input: arr[] = {2, 3}, K = 10 
Output: 15 
Explanation: 
There are 2 arithmetic progressions. 
First-term and the common difference of the first A.P is 2. Therefore AP1 = {2, 4, 6, …} 
First term and the common difference of the second A.P is 3. Therefore AP2 = {3, 6, 9, …} 
Set S contains AP1 and AP2, i.e. { 2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20…… }. 
Hence 10th term is: 15

Input: arr[] = {2, 3, 5, 7, 11}, K = 8 
Output:
Explanation: 
There are 5 arithmetic progressions. 
First-term and the common difference of the first A.P is 2. Therefore AP1 = {2, 4, 6, …} 
First term and the common difference of the second A.P is 3. Therefore AP2 = {3, 6, 9, …} 
First-term and the common difference of the third A.P is 5. Therefore AP3 = {5, 10, 15, …} 
First term and the common difference of the fourth A.P is 7. Therefore AP4 = {7, 14, 21, …} 
First-term and the common difference of the fifth A.P is 11. Therefore AP5 = {11, 22, 33, …} 
Thus set S contains { 2, 3, 4, 5, 6, 7, 8, 9, 10, …. }. 
Hence 8th term is: 9 

Approach: 
Follow the steps below to solve the problem:  



  • Consider a range of [1, maxm] and compute mid of that range.
  • Check if K elements can be obtained by merging the N series’. If the number of terms exceeds K, reduce R to mid. Otherwise, update L to mid. Now, proceed further until we find a mid for which exactly K terms in the series can be obtained.
  • To find how many elements will occur before any particular value, we need to apply Inclusion and Exclusion principle to obtain union of all A.P.s.

Below is the implementation of the above approach: 

C++




// C++ program to find k-th term of
// N merged Arithmetic Progressions
#include <bits/stdc++.h>
using namespace std;
#define maxm 1000000000
 
// Function to count and return the
// number of values less than equal
// to N present in the set
int count(vector<int> v, int n)
{
    int i, odd = 0, even = 0;
    int j, d, count;
 
    int t = (int)1 << v.size();
    int size = v.size();
 
    for (i = 1; i < t; i++) {
        d = 1, count = 0;
        for (j = 0; j < size; j++) {
 
            // Check whether j-th bit
            // is set bit or not
            if (i & (1 << j)) {
                d *= v[j];
                count++;
            }
        }
        if (count & 1)
            odd += n / d;
        else
            even += n / d;
    }
 
    return (odd - even);
}
 
// Function to implement Binary
// Search to find K-th element
int BinarySearch(int l, int r,
                 vector<int> v,
                 int key)
{
    int mid;
 
    while (r - l > 1) {
 
        // Find middle index of
        // the array
        mid = (l + r) / 2;
 
        // Search in the left half
        if (key <= count(v, mid)) {
            r = mid;
        }
 
        // Search in the right half
        else {
            l = mid;
        }
    }
 
    // If exactly K elements
    // are present
    if (key == count(v, l))
        return l;
    else
        return r;
}
 
// Driver Code
int main()
{
    int N = 2, K = 10;
 
    vector<int> v = { 2, 3 };
 
    cout << BinarySearch(1, maxm, v, K)
         << endl;
 
    return 0;
}

Java




// Java program to find k-th term of
// N merged Arithmetic Progressions
import java.util.*;
class GFG{
static final int maxm = 1000000000;
 
// Function to count and return the
// number of values less than equal
// to N present in the set
static int count(int []v, int n)
{
    int i, odd = 0, even = 0;
    int j, d, count;
 
    int t = (int)1 << v.length;
    int size = v.length;
 
    for (i = 1; i < t; i++)
    {
        d = 1;
        count = 0;
        for (j = 0; j < size; j++)
        {
 
            // Check whether j-th bit
            // is set bit or not
            if ((i & (1 << j)) > 0)
            {
                d *= v[j];
                count++;
            }
        }
        if (count % 2 == 1)
            odd += n / d;
        else
            even += n / d;
    }
 
    return (odd - even);
}
 
// Function to implement Binary
// Search to find K-th element
static int BinarySearch(int l, int r,
                        int []v,
                        int key)
{
    int mid;
 
    while (r - l > 1)
    {
 
        // Find middle index of
        // the array
        mid = (l + r) / 2;
 
        // Search in the left half
        if (key <= count(v, mid))
        {
            r = mid;
        }
 
        // Search in the right half
        else
        {
            l = mid;
        }
    }
 
    // If exactly K elements
    // are present
    if (key == count(v, l))
        return l;
    else
        return r;
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 2, K = 10;
 
    int []v = { 2, 3 };
 
    System.out.print(BinarySearch(1, maxm, v, K) + "\n");
}
}
 
// This code is contributed by Rajput-Ji

Python3




# Python3 program to find k-th term of
# N merged Arithmetic Progressions
maxm = 1000000000
 
# Function to count and return the
# number of values less than equal
# to N present in the set
def count(v, n):
     
    odd, even = 0, 0
 
    t = 1 << len(v)
    size = len(v)
 
    for i in range(1, t):
        d, count = 1, 0
         
        for j in range(0, size):
 
            # Check whether j-th bit
            # is set bit or not
            if (i & (1 << j)):
                d *= v[j]
                count += 1
 
        if (count & 1):
            odd += n // d
        else:
            even += n // d
 
    return (odd - even)
 
# Function to implement Binary
# Search to find K-th element
def BinarySearch(l, r, v, key):
 
    while (r - l > 1):
 
        # Find middle index of
        # the array
        mid = (l + r) // 2
 
        # Search in the left half
        if (key <= count(v, mid)):
            r = mid
 
        # Search in the right half
        else:
            l = mid
 
    # If exactly K elements
    # are present
    if (key == count(v, l)):
        return l
    else:
        return r
         
# Driver Code
N, K = 2, 10
 
v = [ 2, 3 ]
 
print(BinarySearch(1, maxm, v, K))
 
# This code is contributed by divyeshrabadiya07

C#




// C# program to find k-th term of
// N merged Arithmetic Progressions
using System;
 
class GFG{
     
static readonly int maxm = 1000000000;
 
// Function to count and return the
// number of values less than equal
// to N present in the set
static int count(int []v, int n)
{
    int i, odd = 0, even = 0;
    int j, d, count;
 
    int t = (int)1 << v.Length;
    int size = v.Length;
 
    for(i = 1; i < t; i++)
    {
       d = 1;
       count = 0;
       for(j = 0; j < size; j++)
       {
            
          // Check whether j-th bit
          // is set bit or not
          if ((i & (1 << j)) > 0)
          {
              d *= v[j];
              count++;
          }
       }
        
       if (count % 2 == 1)
           odd += n / d;
       else
           even += n / d;
    }
    return (odd - even);
}
 
// Function to implement Binary
// Search to find K-th element
static int BinarySearch(int l, int r,
                        int []v, int key)
{
    int mid;
 
    while (r - l > 1)
    {
         
        // Find middle index of
        // the array
        mid = (l + r) / 2;
 
        // Search in the left half
        if (key <= count(v, mid))
        {
            r = mid;
        }
 
        // Search in the right half
        else
        {
            l = mid;
        }
    }
 
    // If exactly K elements
    // are present
    if (key == count(v, l))
        return l;
    else
        return r;
}
 
// Driver Code
public static void Main(String[] args)
{
    //int N = 2;
    int K = 10;
 
    int []v = { 2, 3 };
 
    Console.Write(BinarySearch(
                  1, maxm, v, K) + "\n");
}
}
 
// This code is contributed by gauravrajput1
Output: 
15

 

Time Complexity: O(N * 2N) 
Auxiliary Space: O(1)
 

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