Minimum and maximum count of elements at D distance from arr[i] in either direction

Given a sorted array arr[] and a positive integer D, the task is to find the minimum and the maximum number of array elements that lie over the distance of D from an array element arr[i] in either direction i.e., over the range [arr[i] – D, arr[i]] or [arr[i], arr[i] + D].

Examples:

Input arr[] = {2, 4, 7, 11, 13, 14}, D = 4
Output: 1 3
Explanation:
The minimum number of array elements is included is 1 from arr[0](= 2) as the there exists only 1 element that lies over the range [-2, 2].
The minimum number of array elements is included is 3 from arr[3](= 11) as the there exists only 3 elements that lies over the range [11, 15].
Therefore, print 1, 3.

Input: arr[] = {1, 3, 5, 9, 14}, D = 5
Output: 1 3

Approach: The given problem can be solved using the Greedy Technique by using the Binary Search to the left and right of every point to check how many points can be included in the range of distance D. Follow the steps below to solve the problem:

• Initialize two variables, say min and max to store minimum and maximum elements included in a range of distance D.
• Iterate the array arr[] and for each element perform the following:
  1. Initialize a variable dist to calculate the number of points included in a range of distance D.
  2. Perform the binary search on the left of arr[i] and find a number array elements over the range [arr[i] – D, arr[i]] using the following steps:
     • Initialize left = 0, right = i – 1 and at every iteration:
       • Find the value of mid = (left + right) / 2.
       • If arr[mid] < arr[i] – D, then update the value of left to mid + 1. Otherwise, update the value of dist to mid and update the value of right to mid – 1.
  3. Update the value of min and max according to the value of dist.
  4. Perform the binary search on the left of arr[i] and find the number of array elements over the range [arr[i], arr[i] + D] using the following steps:
     • Initialize left = i + 1, right = N – i and at every iteration:
       • Find the value of mid = (left + right) / 2.
       • If arr[mid] > arr[i] + D, then update the value of right to mid – 1. Otherwise, update the value of dist to mid and update the value of left to mid + 1.
  5. Update the value of min and max according to the value of dist.
• After completing the above steps, print the value of min and max as the resultant minimum and the maximum number of points covered.

Below is the implementation of the above approach:

C++

// c++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to perform the Binary
// Search to the left of arr[i]
// over the given range
int leftSearch(int arr[], int val, int i)
{

    // Base Case
    if (i == 0)
        return 1;

    int left = 0, right = i - 1;
    int ind = -1;

    // Binary Search for index to left
    while (left <= right) {

        int mid = (left + right) / 2;
        if (arr[mid] < val) {
            left = mid + 1;
        }
        else {
            right = mid - 1;

            // update index
            ind = mid;
        }
    }

    // Return the number of elements
    // by subtracting indices
    return ind != -1 ? i - ind + 1 : 1;
}

// Function to perform the Binary
// Search to the right of arr[i]
// over the given range
int rightSearch(int arr[], int val, int i, int N)
{

    // Base Case
    if (i == (N - 1))
        return 1;

    int left = i + 1;
    int right = N - 1;
    int ind = -1;

    // Binary Search for index to right
    while (left <= right) {

        int mid = (left + right) / 2;
        if (arr[mid] > val) {
            right = mid - 1;
        }
        else {
            left = mid + 1;

            // Update the index
            ind = mid;
        }
    }

    // Return the number of elements
    // by subtracting indices
    return ind != -1 ? ind - i + 1 : 1;
}
vector<int> minMaxRange(int arr[], int D, int N)
{

    // Stores the minimum and maximum
    // number of points that lies
    // over the distance of D
    int mx = 1, mn = N;

    // Iterate the array
    for (int i = 0; i < N; i++) {

        // Count of elements included
        // to left of point at index i
        int dist = leftSearch(arr, arr[i] - D, i);

        // Update the minimum number
        // of points
        mn = min(mn, dist);

        // Update the maximum number
        // of points
        mx = max(mx, dist);

        // Count of elements included
        // to right of point at index i
        dist = rightSearch(arr, arr[i] + D, i, N);

        // Update the minimum number
        // of points
        mn = min(mn, dist);

        // Update the maximum number
        // of points
        mx = max(mx, dist);
    }

    // Return the array
    vector<int> v;
    v.push_back(mn);
    v.push_back(mx);
    return v;
}

// Driver Code
int main()
{
    int arr[] = { 1, 3, 5, 9, 14 };
    int N = 5;
    int D = 4;
    vector<int> minMax = minMaxRange(arr, D, N);
    cout << minMax[0] << " " << minMax[1] << endl;
    return 0;
}

// This code is contributed by dwivediyash

Java

// Java program for the above approach

import java.io.*;
import java.lang.Math;
import java.util.*;

class GFG {

    // Function to find the minimum and
    // maximum number of points included
    // in a range of distance D
    public static int[] minMaxRange(
        int[] arr, int D, int N)
    {

        // Stores the minimum and maximum
        // number of points that lies
        // over the distance of D
        int max = 1, min = N;

        // Iterate the array
        for (int i = 0; i < N; i++) {

            // Count of elements included
            // to left of point at index i
            int dist = leftSearch(
                arr, arr[i] - D, i);

            // Update the minimum number
            // of points
            min = Math.min(min, dist);

            // Update the maximum number
            // of points
            max = Math.max(max, dist);

            // Count of elements included
            // to right of point at index i
            dist = rightSearch(
                arr, arr[i] + D, i);

            // Update the minimum number
            // of points
            min = Math.min(min, dist);

            // Update the maximum number
            // of points
            max = Math.max(max, dist);
        }

        // Return the array
        return new int[] { min, max };
    }

    // Function to perform the Binary
    // Search to the left of arr[i]
    // over the given range
    public static int leftSearch(
        int[] arr, int val, int i)
    {
        // Base Case
        if (i == 0)
            return 1;

        int left = 0, right = i - 1;
        int ind = -1;

        // Binary Search for index to left
        while (left <= right) {

            int mid = (left + right) / 2;
            if (arr[mid] < val) {
                left = mid + 1;
            }
            else {
                right = mid - 1;

                // update index
                ind = mid;
            }
        }

        // Return the number of elements
        // by subtracting indices
        return ind != -1 ? i - ind + 1 : 1;
    }

    // Function to perform the Binary
    // Search to the right of arr[i]
    // over the given range
    public static int rightSearch(
        int[] arr, int val, int i)
    {
        // Base Case
        if (i == arr.length - 1)
            return 1;

        int left = i + 1;
        int right = arr.length - 1;
        int ind = -1;

        // Binary Search for index to right
        while (left <= right) {

            int mid = (left + right) / 2;
            if (arr[mid] > val) {
                right = mid - 1;
            }
            else {
                left = mid + 1;

                // Update the index
                ind = mid;
            }
        }

        // Return the number of elements
        // by subtracting indices
        return ind != -1 ? ind - i + 1 : 1;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int[] arr = { 1, 3, 5, 9, 14 };
        int N = arr.length;
        int D = 4;
        int[] minMax = minMaxRange(arr, D, N);

        // Function Call
        System.out.print(
            minMax[0] + " " + minMax[1]);
    }
}

Python3

# Python Program to implement
# the above approach

# Function to find the minimum and
# maximum number of points included
# in a range of distance D
def minMaxRange(arr, D, N):

    # Stores the minimum and maximum
    # number of points that lies
    # over the distance of D
    Max = 1
    Min = N

    # Iterate the array
    for i in range(N):

        # Count of elements included
        # to left of point at index i
        dist = leftSearch(arr, arr[i] - D, i)

        # Update the minimum number
        # of points
        Min = min(Min, dist)

        # Update the maximum number
        # of points
        Max = max(Max, dist)

        # Count of elements included
        # to right of point at index i
        dist = rightSearch(arr, arr[i] + D, i)

        # Update the minimum number
        # of points
        Min = min(Min, dist)

        # Update the maximum number
        # of points
        Max = max(Max, dist)

    # Return the array
    return [Min, Max]


# Function to perform the Binary
# Search to the left of arr[i]
# over the given range
def leftSearch(arr, val, i):
    # Base Case
    if (i == 0):
        return 1

    left = 0
    right = i - 1
    ind = -1

    # Binary Search for index to left
    while (left <= right):

        mid = (left + right) // 2
        if (arr[mid] < val):
            left = mid + 1

        else:
            right = mid - 1

            # update index
            ind = mid

    # Return the number of elements
    # by subtracting indices
    return i - ind + 1 if ind != -1 else 1

# Function to perform the Binary
# Search to the right of arr[i]
# over the given range
def rightSearch(arr, val, i):

    # Base Case
    if (i == len(arr) - 1):
        return 1

    left = i + 1
    right = len(arr) - 1
    ind = -1

    # Binary Search for index to right
    while (left <= right):

        mid = (left + right) // 2

        if (arr[mid] > val):
            right = mid - 1
        else:
            left = mid + 1

            # Update the index
            ind = mid

    # Return the number of elements
    # by subtracting indices
    return ind - i + 1 if ind != -1 else 1

# Driver Code
arr = [1, 3, 5, 9, 14]
N = len(arr)
D = 4
minMax = minMaxRange(arr, D, N)

# Function Call
print(f"{minMax[0]}  {minMax[1]}")

# This code is contributed by gfgking

C#

// C# program for the above approach
using System;

class GFG
{

    // Function to find the minimum and
    // maximum number of points included
    // in a range of distance D
    public static int[] minMaxRange(int[] arr, int D, int N)
    {

        // Stores the minimum and maximum
        // number of points that lies
        // over the distance of D
        int max = 1, min = N;

        // Iterate the array
        for (int i = 0; i < N; i++)
        {

            // Count of elements included
            // to left of point at index i
            int dist = leftSearch(
                arr, arr[i] - D, i);

            // Update the minimum number
            // of points
            min = Math.Min(min, dist);

            // Update the maximum number
            // of points
            max = Math.Max(max, dist);

            // Count of elements included
            // to right of point at index i
            dist = rightSearch(
                arr, arr[i] + D, i);

            // Update the minimum number
            // of points
            min = Math.Min(min, dist);

            // Update the maximum number
            // of points
            max = Math.Max(max, dist);
        }

        // Return the array
        return new int[] { min, max };
    }

    // Function to perform the Binary
    // Search to the left of arr[i]
    // over the given range
    public static int leftSearch(
        int[] arr, int val, int i)
    {
        // Base Case
        if (i == 0)
            return 1;

        int left = 0, right = i - 1;
        int ind = -1;

        // Binary Search for index to left
        while (left <= right)
        {

            int mid = (left + right) / 2;
            if (arr[mid] < val)
            {
                left = mid + 1;
            }
            else
            {
                right = mid - 1;

                // update index
                ind = mid;
            }
        }

        // Return the number of elements
        // by subtracting indices
        return ind != -1 ? i - ind + 1 : 1;
    }

    // Function to perform the Binary
    // Search to the right of arr[i]
    // over the given range
    public static int rightSearch(
        int[] arr, int val, int i)
    {
        // Base Case
        if (i == arr.Length - 1)
            return 1;

        int left = i + 1;
        int right = arr.Length - 1;
        int ind = -1;

        // Binary Search for index to right
        while (left <= right)
        {

            int mid = (left + right) / 2;
            if (arr[mid] > val)
            {
                right = mid - 1;
            }
            else
            {
                left = mid + 1;

                // Update the index
                ind = mid;
            }
        }

        // Return the number of elements
        // by subtracting indices
        return ind != -1 ? ind - i + 1 : 1;
    }

    // Driver Code
    public static void Main()
    {
        int[] arr = { 1, 3, 5, 9, 14 };
        int N = arr.Length;
        int D = 4;
        int[] minMax = minMaxRange(arr, D, N);

        // Function Call
        Console.Write(minMax[0] + " " + minMax[1]);
    }
}

// This code is contributed by gfgking.

Javascript

<script>
        // JavaScript Program to implement
        // the above approach

        // Function to find the minimum and
        // maximum number of points included
        // in a range of distance D
        function minMaxRange(
            arr, D, N) {

            // Stores the minimum and maximum
            // number of points that lies
            // over the distance of D
            let max = 1, min = N;

            // Iterate the array
            for (let i = 0; i < N; i++) {

                // Count of elements included
                // to left of point at index i
                let dist = leftSearch(
                    arr, arr[i] - D, i);

                // Update the minimum number
                // of points
                min = Math.min(min, dist);

                // Update the maximum number
                // of points
                max = Math.max(max, dist);

                // Count of elements included
                // to right of point at index i
                dist = rightSearch(
                    arr, arr[i] + D, i);

                // Update the minimum number
                // of points
                min = Math.min(min, dist);

                // Update the maximum number
                // of points
                max = Math.max(max, dist);
            }

            // Return the array
            return [min, max];
        }

        // Function to perform the Binary
        // Search to the left of arr[i]
        // over the given range
        function leftSearch(
            arr, val, i) {
            // Base Case
            if (i == 0)
                return 1;

            let left = 0, right = i - 1;
            let ind = -1;

            // Binary Search for index to left
            while (left <= right) {

                let mid = Math.floor((left + right) / 2);
                if (arr[mid] < val) {
                    left = mid + 1;
                }
                else {
                    right = mid - 1;

                    // update index
                    ind = mid;
                }
            }

            // Return the number of elements
            // by subtracting indices
            return ind != -1 ? i - ind + 1 : 1;
        }

        // Function to perform the Binary
        // Search to the right of arr[i]
        // over the given range
        function rightSearch(
            arr, val, i) {
            // Base Case
            if (i == arr.length - 1)
                return 1;

            let left = i + 1;
            let right = arr.length - 1;
            let ind = -1;

            // Binary Search for index to right
            while (left <= right) {

                let mid = Math.floor((left + right) / 2);
                if (arr[mid] > val) {
                    right = mid - 1;
                }
                else {
                    left = mid + 1;

                    // Update the index
                    ind = mid;
                }
            }

            // Return the number of elements
            // by subtracting indices
            return ind != -1 ? ind - i + 1 : 1;
        }

        // Driver Code

        let arr = [1, 3, 5, 9, 14];
        let N = arr.length;
        let D = 4;
        let minMax = minMaxRange(arr, D, N);

        // Function Call
        document.write(
            minMax[0] + " " + minMax[1]);


// This code is contributed by Potta Lokesh
    </script>

Output:

1 3

Time Complexity: O(N*log N)
Auxiliary Space: O(1)