# Minimum broadcast range required by M towers to reach N houses

Given an array a containing positions of N houses, and an array b containing positions of M radio towers, each placed along a horizontal line, the task is to find the minimum broadcast range such that each radio tower reaches every house.

Examples:

Input: a[] = {1, 5, 11, 20}, b[] = {4, 8, 15}
Output: 5
Explanation:
The minimum range required is 5, since that would be required for the tower at position 15 to reach the house at position 20

Input: a[] = {12, 13, 11, 80}, b[] = {4, 6, 15, 60}
Output: 20
Explanation:
The minimum range required is 20, since that would be required for the tower at position 60 to reach the house at position 80

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

Approach: Traverse both the arrays until the broadcast range for the last house is calculated. For every house, compare its distance from its left and right towers respectively and consider the minimum. Compare this minimum value with the maximum obtained so far and store the maximum.

Note: The distance of the left tower from the first house is considered Integer.MIN_VALUE. If we reach the end of towers, the distance of all remaining houses from the respective right tower is considered Integer.MAX_VALUE.

Below code implements the above approach:

## C++

 `// CPP program to implement the above approach ` `#include ` `using` `namespace` `std; ` ` `  `int` `minBroadcastRange(``int` `houses[], ``int` `towers[],``int` `n,``int` `m) ` `    ``{ ` `        ``// Initialize distance of left ` `        ``// tower from first house ` `        ``int` `leftTower = INT_MIN; ` ` `  `        ``// Initialize distance of right ` `        ``// tower from first house ` `        ``int` `rightTower = towers; ` ` `  `        ``// j: Index of houses[] ` `        ``// k: Index of towers[] ` `        ``int` `j = 0, k = 0; ` ` `  `        ``// Store the minimum required range ` `        ``int` `min_range = 0; ` ` `  `        ``while` `(j < n) { ` ` `  `            ``// If the house lies between ` `            ``// left and right towers ` `            ``if` `(houses[j] < rightTower) { ` ` `  `                ``int` `left = houses[j] - leftTower; ` `                ``int` `right = rightTower - houses[j]; ` ` `  `                ``// Compare the distance between the ` `                ``// left and right nearest towers ` `                ``int` `local_max = left < right ? left : right; ` ` `  `                ``if` `(local_max > min_range) ` ` `  `                    ``// updating the maximum value ` `                    ``min_range = local_max; ` `                ``j++; ` `            ``} ` `            ``else` `{ ` ` `  `                ``// updating the left tower ` `                ``leftTower = towers[k]; ` ` `  `                ``if` `(k < m - 1) { ` ` `  `                    ``k++; ` `                    ``// updating the right tower ` `                    ``rightTower = towers[k]; ` `                ``} ` `                ``else` `                    ``// updating right tower ` `                    ``// to maximum value after ` `                    ``// reaching the end of Tower array ` `                    ``rightTower = INT_MAX; ` `            ``} ` `        ``} ` `        ``return` `min_range; ` `    ``} ` ` `  `    ``// Driver code ` `    ``int` `main() ` `    ``{ ` `        ``int` `a[] = { 12, 13, 11, 80 }; ` `        ``int` `b[] = { 4, 6, 15, 60 }; ` `        ``int` `n = ``sizeof``(a)/``sizeof``(a); ` `        ``int` `m = ``sizeof``(b)/``sizeof``(b); ` `        ``int` `max = minBroadcastRange(a, b,n,m); ` `        ``cout<

## Java

 `// Java program to implement the above approach ` ` `  `import` `java.io.*; ` ` `  `class` `GFG { ` ` `  `    ``private` `static` `int` `minBroadcastRange( ` `        ``int``[] houses, ``int``[] towers) ` `    ``{ ` ` `  `        ``// Store no of houses ` `        ``int` `n = houses.length; ` ` `  `        ``// Store no of towers ` `        ``int` `m = towers.length; ` ` `  `        ``// Initialize distance of left ` `        ``// tower from first house ` `        ``int` `leftTower = Integer.MIN_VALUE; ` ` `  `        ``// Initialize distance of right ` `        ``// tower from first house ` `        ``int` `rightTower = towers[``0``]; ` ` `  `        ``// j: Index of houses[] ` `        ``// k: Index of towers[] ` `        ``int` `j = ``0``, k = ``0``; ` ` `  `        ``// Store the minimum required range ` `        ``int` `min_range = ``0``; ` ` `  `        ``while` `(j < n) { ` ` `  `            ``// If the house lies between ` `            ``// left and right towers ` `            ``if` `(houses[j] < rightTower) { ` ` `  `                ``int` `left = houses[j] - leftTower; ` `                ``int` `right = rightTower - houses[j]; ` ` `  `                ``// Compare the distance between the ` `                ``// left and right nearest towers ` `                ``int` `local_max = left < right ? left : right; ` ` `  `                ``if` `(local_max > min_range) ` ` `  `                    ``// updating the maximum value ` `                    ``min_range = local_max; ` `                ``j++; ` `            ``} ` `            ``else` `{ ` ` `  `                ``// updating the left tower ` `                ``leftTower = towers[k]; ` ` `  `                ``if` `(k < m - ``1``) { ` ` `  `                    ``k++; ` `                    ``// updating the right tower ` `                    ``rightTower = towers[k]; ` `                ``} ` `                ``else` `                    ``// updating right tower ` `                    ``// to maximum value after ` `                    ``// reaching the end of Tower array ` `                    ``rightTower = Integer.MAX_VALUE; ` `            ``} ` `        ``} ` `        ``return` `min_range; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int``[] a = { ``12``, ``13``, ``11``, ``80` `}; ` `        ``int``[] b = { ``4``, ``6``, ``15``, ``60` `}; ` `        ``int` `max = minBroadcastRange(a, b); ` `        ``System.out.println(max); ` `    ``} ` `} `

## Python3

 `# Python 3 program to implement the above approach ` `import` `sys ` ` `  `def` `minBroadcastRange( houses, towers, n, m): ` ` `  `    ``# Initialize distance of left ` `    ``# tower from first house ` `    ``leftTower ``=` `-``sys.maxsize ``-` `1` ` `  `    ``# Initialize distance of right ` `    ``# tower from first house ` `    ``rightTower ``=` `towers[``0``] ` ` `  `    ``# j: Index of houses[] ` `    ``# k: Index of towers[] ` `    ``j , k ``=` `0` `, ``0` ` `  `    ``# Store the minimum required range ` `    ``min_range ``=` `0` ` `  `    ``while` `(j < n): ` ` `  `        ``# If the house lies between ` `        ``# left and right towers ` `        ``if` `(houses[j] < rightTower): ` ` `  `            ``left ``=` `houses[j] ``-` `leftTower ` `            ``right ``=` `rightTower ``-` `houses[j] ` ` `  `            ``# Compare the distance between the ` `            ``# left and right nearest towers ` `            ``if` `left < right : ` `                ``local_max ``=` `left  ` `            ``else``: ` `                ``local_max ``=` `right ` ` `  `            ``if` `(local_max > min_range): ` ` `  `                ``# updating the maximum value ` `                ``min_range ``=` `local_max ` `            ``j ``+``=` `1`  `         `  `        ``else``: ` ` `  `            ``# updating the left tower ` `            ``leftTower ``=` `towers[k] ` ` `  `            ``if` `(k < m ``-` `1``) : ` ` `  `                ``k ``+``=` `1`  ` `  `                ``# updating the right tower ` `                ``rightTower ``=` `towers[k] ` `             `  `            ``else``: ` `                ``# updating right tower ` `                ``# to maximum value after ` `                ``# reaching the end of Tower array ` `                ``rightTower ``=` `sys.maxsize ` `    ``return` `min_range ` ` `  `# Driver code ` `if` `__name__ ``=``=` `"__main__"``: ` `     `  `    ``a ``=` `[ ``12``, ``13``, ``11``, ``80` `] ` `    ``b ``=` `[ ``4``, ``6``, ``15``, ``60` `] ` `    ``n ``=` `len``(a) ` `    ``m ``=` `len``(b) ` `    ``max` `=` `minBroadcastRange(a, b,n,m) ` `    ``print``(``max``) ` ` `  `# This code is contributed by chitranayal `

## C#

 `// C# program to implement the above approach ` ` ``using` `System; ` ` `  `class` `GFG { ` `  `  `    ``private` `static` `int` `minBroadcastRange( ` `        ``int``[] houses, ``int``[] towers) ` `    ``{ ` `  `  `        ``// Store no of houses ` `        ``int` `n = houses.Length; ` `  `  `        ``// Store no of towers ` `        ``int` `m = towers.Length; ` `  `  `        ``// Initialize distance of left ` `        ``// tower from first house ` `        ``int` `leftTower = ``int``.MinValue; ` `  `  `        ``// Initialize distance of right ` `        ``// tower from first house ` `        ``int` `rightTower = towers; ` `  `  `        ``// j: Index of houses[] ` `        ``// k: Index of towers[] ` `        ``int` `j = 0, k = 0; ` `  `  `        ``// Store the minimum required range ` `        ``int` `min_range = 0; ` `  `  `        ``while` `(j < n) { ` `  `  `            ``// If the house lies between ` `            ``// left and right towers ` `            ``if` `(houses[j] < rightTower) { ` `  `  `                ``int` `left = houses[j] - leftTower; ` `                ``int` `right = rightTower - houses[j]; ` `  `  `                ``// Compare the distance between the ` `                ``// left and right nearest towers ` `                ``int` `local_max = left < right ? left : right; ` `  `  `                ``if` `(local_max > min_range) ` `  `  `                    ``// updating the maximum value ` `                    ``min_range = local_max; ` `                ``j++; ` `            ``} ` `            ``else` `{ ` `  `  `                ``// updating the left tower ` `                ``leftTower = towers[k]; ` `  `  `                ``if` `(k < m - 1) { ` `  `  `                    ``k++; ` `                    ``// updating the right tower ` `                    ``rightTower = towers[k]; ` `                ``} ` `                ``else` `                    ``// updating right tower ` `                    ``// to maximum value after ` `                    ``// reaching the end of Tower array ` `                    ``rightTower = ``int``.MaxValue; ` `            ``} ` `        ``} ` `        ``return` `min_range; ` `    ``} ` `  `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String[] args) ` `    ``{ ` `        ``int``[] a = { 12, 13, 11, 80 }; ` `        ``int``[] b = { 4, 6, 15, 60 }; ` `        ``int` `max = minBroadcastRange(a, b); ` `        ``Console.WriteLine(max); ` `    ``} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

Output:

```20
```

Time complexity: O(M + N)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.