# Number of pairs of lines having integer intersection points

Given two integer arrays P[] and Q[], where pi and qj for each 0 <= i < size(P) and 0 <= j < size(Q) represents the line equations x – y = -pi and x + y = qj respectively. The task is to find the number of pairs from P[] and Q[] having integer intersection points.

Examples:

Input: P[] = {1, 3, 2}, Q[] = {3, 0}
Output:
The pairs of lines (p, q) having integer intersection points are (1, 3), (2, 0) and (3, 3). Here p is the line parameter of P[] and q is the that of Q[].

Input: P[] = {1, 4, 3, 2}, Q[] = {3, 6, 10, 11}
Output:

Approach:

• The problem can be solved easily by solving the two equations and analyzing the condition for integer intersection points.
• The two equations are x – y = -p and x + y = q.
• Solving for x and y we get, x = (q-p)/2 and y = (p+q)/2.
• It is clear that integer intersection point is possible if and only if p and q have same parity.
• Let p0 and p1 be the number of even and odd pi respectively.
• Similarly, q0 and q1 for the number of even and odd qi respectively.
• Therefore the required answers is p0 * q0 + p1 * q1.

Below is the implementation of the above approach:

## C++

 `// C++ program to Number of pairs of lines` `// having integer intersection points`   `#include ` `using` `namespace` `std;`   `// Count number of pairs of lines` `// having integer intersection point` `int` `countPairs(``int``* P, ``int``* Q, ``int` `N, ``int` `M)` `{` `    ``// Initialize arrays to store counts` `    ``int` `A[2] = { 0 }, B[2] = { 0 };`   `    ``// Count number of odd and even Pi` `    ``for` `(``int` `i = 0; i < N; i++)` `        ``A[P[i] % 2]++;`   `    ``// Count number of odd and even Qi` `    ``for` `(``int` `i = 0; i < M; i++)` `        ``B[Q[i] % 2]++;`   `    ``// Return the count of pairs` `    ``return` `(A[0] * B[0] + A[1] * B[1]);` `}`   `// Driver code` `int` `main()` `{` `    ``int` `P[] = { 1, 3, 2 }, Q[] = { 3, 0 };` `    ``int` `N = ``sizeof``(P) / ``sizeof``(P[0]);` `    ``int` `M = ``sizeof``(Q) / ``sizeof``(Q[0]);`   `    ``cout << countPairs(P, Q, N, M);`   `    ``return` `0;` `}`

## Java

 `// Java program to Number of pairs of lines` `// having integer intersection points` `class` `GFG` `{`   `// Count number of pairs of lines` `// having integer intersection point` `static` `int` `countPairs(``int` `[]P, ``int` `[]Q, ` `                      ``int` `N, ``int` `M)` `{` `    ``// Initialize arrays to store counts` `    ``int` `[]A = ``new` `int``[``2``], B = ``new` `int``[``2``];`   `    ``// Count number of odd and even Pi` `    ``for` `(``int` `i = ``0``; i < N; i++)` `        ``A[P[i] % ``2``]++;`   `    ``// Count number of odd and even Qi` `    ``for` `(``int` `i = ``0``; i < M; i++)` `        ``B[Q[i] % ``2``]++;`   `    ``// Return the count of pairs` `    ``return` `(A[``0``] * B[``0``] + A[``1``] * B[``1``]);` `}`   `// Driver code` `public` `static` `void` `main(String[] args)` `{` `    ``int` `[]P = { ``1``, ``3``, ``2` `};` `    ``int` `[]Q = { ``3``, ``0` `};` `    ``int` `N = P.length;` `    ``int` `M = Q.length;`   `    ``System.out.print(countPairs(P, Q, N, M));` `}` `}`   `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 program to Number of pairs of lines` `# having eger ersection pos`   `# Count number of pairs of lines` `# having eger ersection po` `def` `countPairs(P, Q, N, M):` `    `  `    ``# Initialize arrays to store counts` `    ``A ``=` `[``0``] ``*` `2` `    ``B ``=` `[``0``] ``*` `2`   `    ``# Count number of odd and even Pi` `    ``for` `i ``in` `range``(N):` `        ``A[P[i] ``%` `2``] ``+``=` `1`   `    ``# Count number of odd and even Qi` `    ``for` `i ``in` `range``(M):` `        ``B[Q[i] ``%` `2``] ``+``=` `1`   `    ``# Return the count of pairs` `    ``return` `(A[``0``] ``*` `B[``0``] ``+` `A[``1``] ``*` `B[``1``])`   `# Driver code`   `P ``=` `[``1``, ``3``, ``2``]` `Q ``=` `[``3``, ``0``]` `N ``=` `len``(P)` `M ``=` `len``(Q)`   `print``(countPairs(P, Q, N, M))`   `# This code is contributed by mohit kumar 29`

## C#

 `// C# program to Number of pairs of lines` `// having integer intersection points` `using` `System;`   `class` `GFG` `{` `    `  `    ``// Count number of pairs of lines` `    ``// having integer intersection point` `    ``static` `int` `countPairs(``int` `[]P, ``int` `[]Q, ` `                        ``int` `N, ``int` `M)` `    ``{` `        ``// Initialize arrays to store counts` `        ``int` `[]A = ``new` `int``[2];` `        ``int` `[]B = ``new` `int``[2];` `    `  `        ``// Count number of odd and even Pi` `        ``for` `(``int` `i = 0; i < N; i++)` `            ``A[P[i] % 2]++;` `    `  `        ``// Count number of odd and even Qi` `        ``for` `(``int` `i = 0; i < M; i++)` `            ``B[Q[i] % 2]++;` `    `  `        ``// Return the count of pairs` `        ``return` `(A[0] * B[0] + A[1] * B[1]);` `    ``}` `    `  `    ``// Driver code` `    ``public` `static` `void` `Main()` `    ``{` `        ``int` `[]P = { 1, 3, 2 };` `        ``int` `[]Q = { 3, 0 };` `        ``int` `N = P.Length;` `        ``int` `M = Q.Length;` `    `  `        ``Console.Write(countPairs(P, Q, N, M));` `    ``}` `}`   `// This code is contributed by AnkitRai01`

## Javascript

 ``

Output:

`3`

Time Complexity: O(P + Q)

Auxiliary Space: O(1)

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