# Number of pairs of lines having integer intersection points

Given two integer arrays **P[]** and **Q[]**, where * p_{i}* and

*for each*

**q**_{j}*0 <= i < size(P)*and

*0 <= j < size(Q)*represents the line equations

*and*

**x – y = -p**_{i}*respectively. The task is to find the number of pairs from*

**x + y = q**_{j}**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

**Approach:**

- The problem can be solved easily by solving the two equations and analyzing the condition for integer intersection points.
- The two equations are
and**x – y = -p**.**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
**p**and_{0}**p**be the number of even and odd_{1}**p**respectively._{i} - Similarly,
**q**and_{0}**q**for the number of even and odd_{1}**q**respectively._{i} - Therefore the required answers is
**p***_{0}**q**+_{0}**p***_{1}**q**._{1}

Below is the implementation of the above approach:

## C++

`// C++ program to Number of pairs of lines` `// having integer intersection points` `#include <bits/stdc++.h>` `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

`<script>` `// Javascript program to Number of` `// pairs of lines having integer` `// intersection points` `// Count number of pairs of lines` `// having integer intersection point` `function` `countPairs(P, Q, N, M)` `{` ` ` ` ` `// Initialize arrays to store counts` ` ` `var` `A = [0, 0], B = [0, 0];` ` ` `// Count number of odd and even Pi` ` ` `for` `(` `var` `i = 0; i < N; i++)` ` ` `A[P[i] % 2]++;` ` ` `// Count number of odd and even Qi` ` ` `for` `(` `var` `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` `var` `P = [ 1, 3, 2 ], Q = [ 3, 0 ];` `var` `N = P.length;` `var` `M = Q.length;` `document.write(countPairs(P, Q, N, M));` `// This code is contributed by rrrtnx` `</script>` |

**Output:**

3

**Time Complexity:** O(P + Q)

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