**Objective: **Given two sorted arrays, Find intersection point between them.

**Examples:**

int[] a = { 1, 2, 3, 6, 8, 10 }; int[] b = { 4, 5, 6, 11, 15, 20 }; Output: Intersection point is : 6

**Approach:**

**Naive Approach: **Use 2 loops and compare each elements in both array and as soon as you find the intersection point, return it. Time Complexity – O(n^{2}).

**Better Approach: Time Complexity – O(n)**

- Say Arrays are arrA[] and arrB[] and indexes for navigation are x and y respectively.
- Since arrays are sorted, compare the first element of both the arrays.(x=0, y=0)
- If both elements are same, we have our intersection point, return it.
- Else if element of arrA[x] > element of arrB[y], increase the arrB[] index, y++.
- Else if element of arrA[x] < element of arrB[y], increase the arrA[] index, x++.
- If any of the array gets over that means you have not found the intersection point. return -1.

**Complete Code:**

public class IntersecionPoint2Arrays { | |

int intersectionPoint = –1; | |

int x; | |

int y; | |

public int intersection(int[] arrA, int[] arrB) { | |

while (x < arrA.length && y < arrB.length) { | |

if (arrA[x] > arrB[y]) | |

y++; | |

else if (arrA[x] < arrB[y]) | |

x++; | |

else { | |

intersectionPoint = arrA[x]; | |

return intersectionPoint; | |

} | |

} | |

return intersectionPoint; | |

} | |

public static void main(String[] args) throws java.lang.Exception { | |

int[] a = { 1, 2, 3, 6, 8, 10 }; | |

int[] b = { 4, 5, 6, 11, 15, 20 }; | |

IntersecionPoint2Arrays i = new IntersecionPoint2Arrays(); | |

System.out.println("Intersection point is : " + i.intersection(a, b)); | |

} | |

} |

**Output:**

Intersection point is : 6