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
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<bits/stdc++.h> 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<<max<<endl; } // This code is contributed by Surendra_Gangwar |
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
<script> // Javascript program to implement // the above approach function minBroadcastRange( houses, towers) { // Store no of houses let n = houses.length; // Store no of towers let m = towers.length; // Initialize distance of left // tower from first house let leftTower = Number.MIN_VALUE; // Initialize distance of right // tower from first house let rightTower = towers[0]; // j: Index of houses[] // k: Index of towers[] let j = 0, k = 0; // Store the minimum required range let min_range = 0; while (j < n) { // If the house lies between // left and right towers if (houses[j] < rightTower) { let left = houses[j] - leftTower; let right = rightTower - houses[j]; // Compare the distance between the // left and right nearest towers let 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 = Number.MAX_VALUE; } } return min_range; } // Driver Code let a = [12, 13, 11, 80 ]; let b = [ 4, 6, 15, 60 ]; let max = minBroadcastRange(a, b); document.write(max); </script> |
20
Time complexity: O(M + N)
Auxiliary Space: O(1)
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