# 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: 3
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: 8

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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:

## CPP

 `// 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 = { 0 }, B = { 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 * B + A * B); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `P[] = { 1, 3, 2 }, Q[] = { 3, 0 }; ` `    ``int` `N = ``sizeof``(P) / ``sizeof``(P); ` `    ``int` `M = ``sizeof``(Q) / ``sizeof``(Q); ` ` `  `    ``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 `

## Python

 `# 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``; ` `        ``int` `[]B = ``new` `int``; ` `     `  `        ``// 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 * B + A * B); ` `    ``} ` `     `  `    ``// 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 `

Output:

```3
```

Time Complexity: O(P + Q)

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