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

• Difficulty Level : Hard
• Last Updated : 29 Apr, 2021

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:
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

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[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 = 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[0]);``        ``int` `m = ``sizeof``(b)/``sizeof``(b[0]);``        ``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[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 = ``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`

## Javascript

 ``
Output:
`20`

Time complexity: O(M + N)

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