Given an array arr[] of N integers representing the position of N points along a straight line and an integer K, the task is to find the minimum value of the maximum distance between adjacent points after adding K points anywhere in between, not necessarily on an integer position.
Examples:
Input: arr[] = {2, 4, 8, 10}, K = 1
Output: 2
Explanation: A point at position 6 can be added. So the new array of points become {2, 4, 6, 8, 10} and the maximum distance between two adjacent points is 2 which is minimum possible.Input: arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, K = 9
Output: 0.5
Approach: The given problem can be solved by using Binary Search. The idea is to perform a binary search on the value D in the range [0, 108] where D represents the value of the maximum distance between the adjacent points after adding K points. Follow the steps below to solve the given problem:
- Initialize variables, low = 1 and high = 108, where low represents the lower bound and high represents the upper bound of the binary search.
- Create a function isPossible(), which returns the boolean value of whether it is possible to add K points in the array such that the maximum distance between adjacent points is D. It is based on the observation that for two adjacent points (i, j), the number of points required to be placed in their middle such that the maximum distance between them is D = (j -i)/D.
- Therefore, traverse the range using the binary search algorithm discussed here, and if for the mid-value D in the range [X, Y], if isPossible(D) is false, then iterate in the upper half of the range i.e, [D, Y]. Otherwise, iterate on the lower half i.e, [X, D].
- Iterate a loop until (high – low) > 10-6.
- The value stored in low is the required answer.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <iostream>using namespace std;// Function to check if it is possible// to add K points such that the maximum// distance between adjacent points is Dbool isPossible(double D, int arr[], int N, int K){ // Stores the count of point used int used = 0; // Iterate over all given points for (int i = 0; i < N - 1; ++i) { // Add number of points required // to be placed between ith // and (i+1)th point used += (int)((arr[i + 1] - arr[i]) / D); } // Return answer return used <= K;}// Function to find the minimum value of// maximum distance between adjacent points// after adding K points any where betweendouble minMaxDist(int stations[], int N, int K){ // Stores the lower bound and upper // bound of the given range double low = 0, high = 1e8; // Perform binary search while (high - low > 1e-6) { // Find the middle value double mid = (low + high) / 2.0; if (isPossible(mid, stations, N, K)) { // Update the current range // to lower half high = mid; } // Update the current range // to upper half else { low = mid; } } return low;}// Driver Codeint main(){ int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }; int K = 9; int N = sizeof(arr) / sizeof(arr[0]); cout << minMaxDist(arr, N, K); return 0;} |
Java
// Java program for the above approachimport java.math.BigDecimal;class GFG { // Function to check if it is possible // to add K points such that the maximum // distance between adjacent points is D public static boolean isPossible(double D, int arr[], int N, int K) { // Stores the count of point used int used = 0; // Iterate over all given points for (int i = 0; i < N - 1; ++i) { // Add number of points required // to be placed between ith // and (i+1)th point used += (int) ((arr[i + 1] - arr[i]) / D); } // Return answer return used <= K; } // Function to find the minimum value of // maximum distance between adjacent points // after adding K points any where between public static double minMaxDist(int stations[], int N, int K) { // Stores the lower bound and upper // bound of the given range double low = 0, high = 1e8; // Perform binary search while (high - low > 1e-6) { // Find the middle value double mid = (low + high) / 2.0; if (isPossible(mid, stations, N, K)) { // Update the current range // to lower half high = mid; } // Update the current range // to upper half else { low = mid; } } // System.out.printf("Value: %.2f", low); return low; } // Driver Code public static void main(String args[]) { int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }; int K = 9; int N = arr.length; System.out.printf("%.1f", minMaxDist(arr, N, K)); }}// This code is contributed by _saurabh_jaiswal. |
Python3
# Python3 program for the above approach# Function to check if it is possible# to add K points such that the maximum# distance between adjacent points is Ddef isPossible(D, arr, N, K) : # Stores the count of point used used = 0; # Iterate over all given points for i in range(N - 1) : # Add number of points required # to be placed between ith # and (i+1)th point used += int((arr[i + 1] - arr[i]) / D); # Return answer return used <= K;# Function to find the minimum value of# maximum distance between adjacent points# after adding K points any where betweendef minMaxDist(stations, N, K) : # Stores the lower bound and upper # bound of the given range low = 0; high = 1e8; # Perform binary search while (high - low > 1e-6) : # Find the middle value mid = (low + high) / 2.0; if (isPossible(mid, stations, N, K)) : # Update the current range # to lower half high = mid; # Update the current range # to upper half else : low = mid; return round(low, 2);# Driver Codeif __name__ == "__main__" : arr = [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ]; K = 9; N = len(arr); print(minMaxDist(arr, N, K)); # This code is contributed by AnkThon |
C#
// C# program for the above approachusing System;public class GFG { // Function to check if it is possible // to add K points such that the maximum // distance between adjacent points is D public static bool isPossible(double D, int []arr, int N, int K) { // Stores the count of point used int used = 0; // Iterate over all given points for (int i = 0; i < N - 1; ++i) { // Add number of points required // to be placed between ith // and (i+1)th point used += (int) ((arr[i + 1] - arr[i]) / D); } // Return answer return used <= K; } // Function to find the minimum value of // maximum distance between adjacent points // after adding K points any where between public static double minMaxDist(int []stations, int N, int K) { // Stores the lower bound and upper // bound of the given range double low = 0, high = 1e8; // Perform binary search while (high - low > 1e-6) { // Find the middle value double mid = (low + high) / 2.0; if (isPossible(mid, stations, N, K)) { // Update the current range // to lower half high = mid; } // Update the current range // to upper half else { low = mid; } } // Console.Write("Value: %.2f", low); return low; } // Driver Code public static void Main(String []args) { int []arr = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }; int K = 9; int N = arr.Length; Console.Write("{0:F1}", minMaxDist(arr, N, K)); }}// This code is contributed by 29AjayKumar |
Javascript
<script> // JavaScript Program to implement // the above approach // Function to check if it is possible // to add K points such that the maximum // distance between adjacent points is D function isPossible(D, arr, N, K) { // Stores the count of point used let used = 0; // Iterate over all given points for (let i = 0; i < N - 1; ++i) { // Add number of points required // to be placed between ith // and (i+1)th point used += Math.floor((arr[i + 1] - arr[i]) / D); } // Return answer return used <= K; } // Function to find the minimum value of // maximum distance between adjacent points // after adding K points any where between function minMaxDist(stations, N, K) { // Stores the lower bound and upper // bound of the given range let low = 0, high = 1e8; // Perform binary search while (high - low > 1e-6) { // Find the middle value let mid = (low + high) / 2; if (isPossible(mid, stations, N, K)) { // Update the current range // to lower half high = mid; } // Update the current range // to upper half else { low = mid; } } return low.toFixed(1); } // Driver Code let arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]; let K = 9; let N = arr.length; document.write(minMaxDist(arr, N, K)); // This code is contributed by Potta Lokesh </script> |
Output
0.5
Time Complexity: O(N*log M), where the value of M is 1014.
Auxiliary Space: O(1)
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