# K-th term from given N merged Arithmetic Progressions

Given an integer K and an array arr[] of N integers, each of which is the first term and common difference of an Arithmetic 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 ` `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`

## Javascript

 ``

Output:

`15`

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

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