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Minimize the maximum distance between adjacent points after adding K points anywhere in between

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  • Last Updated : 26 May, 2022

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 D
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
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;
        }
    }
 
    return low;
}
 
// Driver Code
int 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 approach
import 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 D
def 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 between
def 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 Code
if __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 approach
using 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)

Alternate Approach: The problem boils down to minimizing the maximum distance K times. Every time, to get the maximum distance we can use max heap.

Step -1 : Iterate through the array elements and store the adjacent array element’s distances in a max heap.

Step-2 : For K times, poll the max element from the heap and add to the heap max/2, max/2 i.e., we’re each time reducing the maximum distance to two equal halves.

Step-3 : Return the max element after iterating for K times.

Below is the implementation of the above approach:

C++




#include<bits/stdc++.h>
using namespace std;
 
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
    int K = 9;
    int N = sizeof(arr)/sizeof(arr[0]);
  
    // Max heap initialisation
    priority_queue<float>pq;
   
    // Add adjacent distances to max heap
    for (int i = 1; i < N; i++) {
        pq.push((float)arr[i] - (float)arr[i - 1]);
    }
  
    // For K times, half the maximum distance
    for (int i = 0; i < K; i++) {
        float temp = pq.top();
        pq.pop();
        pq.push(temp / 2);
        pq.push(temp / 2);
    }
    cout<<pq.top()<<endl;
    return 0;
}
 
// This code is contributed by shinjanpatra.


Java




import java.util.Collections;
import java.util.PriorityQueue;
 
class GFG {
    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;
 
        // Max heap initialisation
        PriorityQueue<Float> pq = new PriorityQueue<>(
            N + K, Collections.reverseOrder());
 
        // Add adjacent distances to max heap
        for (int i = 1; i < N; i++) {
            pq.add((float)arr[i] - (float)arr[i - 1]);
        }
 
        // For K times, half the maximum distance
        for (int i = 0; i < K; i++) {
            float temp = pq.poll();
            pq.add(temp / 2);
            pq.add(temp / 2);
        }
        System.out.println(pq.peek());
    }
}
// This code is contributed by _govardhani


Python3




# Python3 program for the above approach
 
# importing "bisect" for bisection operations
import bisect
 
arr = [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ]
K = 9
N = len(arr)
 
# Max heap initialisation
pq = []
 
# Add adjacent distances to max heap
for i in range(1, N):
    bisect.insort(pq,float(arr[i])-float(arr[i-1]))
 
# For K times, half the maximum distance
for i in range(K):
    temp=pq[0]
    pq.pop(0)
    pq.append(temp/2)
    pq.append(temp/2)
 
print(pq[0])
 
# This code is contributed by Pushpesh Raj


Time Complexity: O(NlogN)
Auxiliary Space: O(N+K)


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