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Find a point whose sum of distances from all given points on a line is K

  • Last Updated : 17 Jun, 2021

Given a sorted array arr[] consisting of N integers, representing points on a line and an integer K, the task is to find any point P between the first and last point such that the sum of distances of all given points from P is equal to K. If no such point exists, then print “-1”.

Examples: 

Input: arr[] = {1, 3, 6, 7, 11}, K = 18
Output: 8
Explanation:
Consider the value of P as 8. Therefore, the sum of distance of P(= 8) from all points is (8 – 1) + (8 – 3) + (8 – 6) + (8 – 7) + (11 – 8) = 18( =K) which is equal to the given value K and the point 8 lies between the first(= 1) and the last(= 11) point.

Input: arr[] = {-10, -2, 1, 2}, K= 29
Output: -9

Approach: The given problem can be solved based on the observation that the sum of distances will be minimum at the median of the array and the distance increases when moved from the median towards any of the extremities. So, the idea is to perform a binary search on the both the halves of the array and check if any point has a distance equal to K. Follow the steps below to solve the problem:



  • Declare a function that calculates the sum of distances of all points from a given point.
  • Perform a binary search on the right half of the array as:
    • If the value of N is odd, then update the value of left as arr[N / 2]. Otherwise, update the value of left as arr[N / 2 – 1] + 1.
    • If the value of N is even, then update the value of right as arr[N – 1].
    • Find the sum of distances say temp from mid = (left + right) / 2 and check if the value of temp is equal to K or not. IF found to be true, then print the value of mid as the result.
    • If the value of K < temp, then update the value of right as mid – 1. Otherwise, update the value of left as mid + 1.
  • Perform a binary search on the left half of the array:
    • Set the value of left = arr[0] and right = arr[N / 2] – 1.
    • Find the sum of distances say temp from mid = (left + right) / 2 and check if temp is equal to K or not. If found to be true, then print the value of mid as the result.
    • If the value of K > temp, then update the value of right = mid – 1. Otherwise, update the value of left = mid + 1.
  • If there is no value found in the left and right half then print “-1” as the result.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the sum of distances
// of all points from a given point
int findSum(int* arr, int N, int pt)
{
    // Stores sum of distances
    int sum = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
        sum += abs(arr[i] - pt);
    }
 
    // Return the sum
    return sum;
}
 
// Function to find such a point having
// sum of distances of all other points
// from this point equal to K
void findPoint(int* arr, int N, int K)
{
    // If N is odd keep left as arr[n / 2]
    // else keep left as arr[n / 2 - 1] + 1;
    int left;
 
    if (N % 2) {
        left = arr[N / 2];
    }
    else {
        left = arr[N / 2 - 1] + 1;
    }
 
    // Keep right as arr[N - 1]
    int right = arr[N - 1];
 
    // Perform binary search in the
    // right half
    while (left <= right) {
 
        // Calculate the mid index
        // of the range
        int mid = (left + right) / 2;
 
        int temp = findSum(arr, N, mid);
 
        // If temp is equal to K
        if (temp == K) {
 
            // Print the value of mid
            cout << mid << endl;
            return;
        }
 
        // If the value of K < temp
        else if (K < temp) {
 
            // Update right to mid - 1
            right = mid - 1;
        }
 
        // If the value of K > temp
        else {
 
            // Update left to mid + 1
            left = mid + 1;
        }
    }
 
    // Update the value of left
    left = arr[0];
 
    // Update the value of right
    right = arr[N / 2] - 1;
 
    // Perform binary search on the
    // left half
    while (left <= right) {
 
        // Calculate the mid index
        // of the range
        int mid = (left + right) / 2;
 
        int temp = findSum(arr, N, mid);
 
        // If temp is equal to K
        if (temp == K) {
 
            // Print mid
            cout << mid << endl;
            return;
        }
 
        // if K > temp
        else if (K > temp) {
 
            // Update right to mid - 1
            right = mid - 1;
        }
 
        // If K < temp
        else {
 
            // Update left to mid + 1
            left = mid + 1;
        }
    }
 
    // If no such point found
    cout << "-1" << endl;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 3, 6, 7, 11 };
    int K = 18;
    int N = sizeof(arr) / sizeof(arr[0]);
    findPoint(arr, N, K);
 
    return 0;
}

Java




// Java program for the above approach
import java.lang.*;
 
class GFG{
 
// Function to find the sum of distances
// of all points from a given point
public static int findSum(int arr[], int N,
                          int pt)
{
     
    // Stores sum of distances
    int sum = 0;
 
    // Traverse the array
    for(int i = 0; i < N; i++)
    {
        sum += Math.abs(arr[i] - pt);
    }
 
    // Return the sum
    return sum;
}
 
// Function to find such a point having
// sum of distances of all other points
// from this point equal to K
public static void findPoint(int arr[], int N, int K)
{
     
    // If N is odd keep left as arr[n / 2]
    // else keep left as arr[n / 2 - 1] + 1;
    int left;
 
    if (N % 2 != 0)
    {
        left = arr[N / 2];
    }
    else
    {
        left = arr[N / 2 - 1] + 1;
    }
 
    // Keep right as arr[N - 1]
    int right = arr[N - 1];
 
    // Perform binary search in the
    // right half
    while (left <= right)
    {
         
        // Calculate the mid index
        // of the range
        int mid = (left + right) / 2;
 
        int temp = findSum(arr, N, mid);
 
        // If temp is equal to K
        if (temp == K)
        {
             
            // Print the value of mid
            System.out.println(mid);
            return;
        }
 
        // If the value of K < temp
        else if (K < temp)
        {
             
            // Update right to mid - 1
            right = mid - 1;
        }
 
        // If the value of K > temp
        else
        {
             
            // Update left to mid + 1
            left = mid + 1;
        }
    }
 
    // Update the value of left
    left = arr[0];
 
    // Update the value of right
    right = arr[N / 2] - 1;
 
    // Perform binary search on the
    // left half
    while (left <= right)
    {
         
        // Calculate the mid index
        // of the range
        int mid = (left + right) / 2;
 
        int temp = findSum(arr, N, mid);
 
        // If temp is equal to K
        if (temp == K)
        {
             
            // Print mid
            System.out.println(mid);
            return;
        }
 
        // if K > temp
        else if (K > temp)
        {
             
            // Update right to mid - 1
            right = mid - 1;
        }
 
        // If K < temp
        else
        {
 
            // Update left to mid + 1
            left = mid + 1;
        }
    }
 
    // If no such point found
    System.out.println( "-1" );
}
 
// Driver Code
public static void main(String args[])
{
    int arr[] = { 1, 3, 6, 7, 11 };
    int K = 18;
    int N = arr.length;
     
    findPoint(arr, N, K);
}
}
 
// This code is contributed by SoumikMondal

Python3




# python 3 program for the above approach
 
# Function to find the sum of distances
# of all points from a given point
def findSum(arr, N, pt):
   
    # Stores sum of distances
    sum = 0
 
    # Traverse the array
    for i in range(N):
        sum += abs(arr[i] - pt)
 
    # Return the sum
    return sum
 
# Function to find such a point having
# sum of distances of all other points
# from this point equal to K
def findPoint(arr, N, K):
    # If N is odd keep left as arr[n / 2]
    # else keep left as arr[n / 2 - 1] + 1;
    left = 0
 
    if (N % 2):
        left = arr[N // 2]
    else:
        left = arr[N // 2 - 1] + 1
 
    # Keep right as arr[N - 1]
    right = arr[N - 1]
 
    # Perform binary search in the
    # right half
    while (left <= right):
 
        # Calculate the mid index
        # of the range
        mid = (left + right) // 2
 
        temp = findSum(arr, N, mid)
 
        # If temp is equal to K
        if (temp == K):
            # Print the value of mid
            print(mid)
            return
 
        # If the value of K < temp
        elif (K < temp):
            # Update right to mid - 1
            right = mid - 1
 
        # If the value of K > temp
        else:
            # Update left to mid + 1
            left = mid + 1
 
    # Update the value of left
    left = arr[0]
 
    # Update the value of right
    right = arr[N // 2] - 1
 
    # Perform binary search on the
    # left half
    while (left <= right):
        # Calculate the mid index
        # of the range
        mid = (left + right) // 2
 
        temp = findSum(arr, N, mid)
 
        # If temp is equal to K
        if (temp == K):
 
            # Print mid
            print(mid)
            return
 
        # if K > temp
        elif(K > temp):
            # Update right to mid - 1
            right = mid - 1
 
        # If K < temp
        else:
            # Update left to mid + 1
            left = mid + 1
 
    # If no such point found
    print("-1")
 
# Driver Code
if __name__ == '__main__':
    arr = [1, 3, 6, 7, 11]
    K = 18
    N = len(arr)
    findPoint(arr, N, K)
 
    # This code is contributed by SURENDRA_GANGWAR.

C#




// C# program for the above approach
using System;
 
class GFG{
 
// Function to find the sum of distances
// of all points from a given point
public static int findSum(int[] arr, int N, int pt)
{
 
    // Stores sum of distances
    int sum = 0;
 
    // Traverse the array
    for(int i = 0; i < N; i++)
    {
        sum += Math.Abs(arr[i] - pt);
    }
 
    // Return the sum
    return sum;
}
 
// Function to find such a point having
// sum of distances of all other points
// from this point equal to K
public static void findPoint(int[] arr, int N, int K)
{
     
    // If N is odd keep left as arr[n / 2]
    // else keep left as arr[n / 2 - 1] + 1;
    int left;
 
    if (N % 2 != 0)
    {
        left = arr[N / 2];
    }
    else
    {
        left = arr[N / 2 - 1] + 1;
    }
 
    // Keep right as arr[N - 1]
    int right = arr[N - 1];
 
    // Perform binary search in the
    // right half
    while (left <= right)
    {
         
        // Calculate the mid index
        // of the range
        int mid = (left + right) / 2;
 
        int temp = findSum(arr, N, mid);
 
        // If temp is equal to K
        if (temp == K)
        {
 
            // Print the value of mid
            Console.WriteLine(mid);
            return;
        }
 
        // If the value of K < temp
        else if (K < temp)
        {
             
            // Update right to mid - 1
            right = mid - 1;
        }
 
        // If the value of K > temp
        else
        {
             
            // Update left to mid + 1
            left = mid + 1;
        }
    }
 
    // Update the value of left
    left = arr[0];
 
    // Update the value of right
    right = arr[N / 2] - 1;
 
    // Perform binary search on the
    // left half
    while (left <= right)
    {
         
        // Calculate the mid index
        // of the range
        int mid = (left + right) / 2;
 
        int temp = findSum(arr, N, mid);
 
        // If temp is equal to K
        if (temp == K)
        {
 
            // Print mid
            Console.WriteLine(mid);
            return;
        }
 
        // if K > temp
        else if (K > temp)
        {
 
            // Update right to mid - 1
            right = mid - 1;
        }
 
        // If K < temp
        else
        {
 
            // Update left to mid + 1
            left = mid + 1;
        }
    }
 
    // If no such point found
    Console.WriteLine("-1");
}
 
// Driver Code
public static void Main(string[] args)
{
    int[] arr = { 1, 3, 6, 7, 11 };
    int K = 18;
    int N = arr.Length;
 
    findPoint(arr, N, K);
}
}
 
// This code is contributed by ukasp

Javascript




<script>
 
// Jaascript program for the above approach
 
// Function to find the sum of distances
// of all points from a given point
function findSum(arr, N, pt)
{
     
    // Stores sum of distances
    var sum = 0;
    var i;
 
    // Traverse the array
    for(i = 0; i < N; i++)
    {
        sum += Math.abs(arr[i] - pt);
    }
 
    // Return the sum
    return sum;
}
 
// Function to find such a point having
// sum of distances of all other points
// from this point equal to K
function findPoint(arr, N, K)
{
     
    // If N is odd keep left as arr[n / 2]
    // else keep left as arr[n / 2 - 1] + 1;
    var left;
 
    if (N % 2 == 1)
    {
        left = arr[parseInt(N / 2)];
    }
    else
    {
        left = arr[parseInt(N / 2) - 1] + 1;
    }
 
    // Keep right as arr[N - 1]
    var right = arr[N - 1];
 
    // Perform binary search in the
    // right half
    while (left <= right)
    {
         
        // Calculate the mid index
        // of the range
        var mid = parseInt((left + right) / 2);
 
        var temp = findSum(arr, N, mid);
 
        // If temp is equal to K
        if (temp == K)
        {
             
            // Print the value of mid
            document.write(mid);
            return;
        }
 
        // If the value of K < temp
        else if (K < temp)
        {
             
            // Update right to mid - 1
            right = mid - 1;
        }
 
        // If the value of K > temp
        else
        {
             
            // Update left to mid + 1
            left = mid + 1;
        }
    }
 
    // Update the value of left
    left = arr[0];
 
    // Update the value of right
    right = arr[parseInt(N / 2)] - 1;
 
    // Perform binary search on the
    // left half
    while (left <= right)
    {
         
        // Calculate the mid index
        // of the range
        var mid = parseInt((left + right) / 2);
 
        var temp = findSum(arr, N, mid);
 
        // If temp is equal to K
        if (temp == K)
        {
             
            // Print mid
            document.write(mid);
            return;
        }
 
        // If K > temp
        else if (K > temp)
        {
             
            // Update right to mid - 1
            right = mid - 1;
        }
 
        // If K < temp
        else
        {
             
            // Update left to mid + 1
            left = mid + 1;
        }
    }
 
    // If no such point found
    document.write("-1");
}
 
// Driver Code
var arr = [ 1, 3, 6, 7, 11 ];
var K = 18;
var N = arr.length;
 
findPoint(arr, N, K);
 
// This code is contributed by bgangwar59
 
</script>
Output: 
8

 

Time Complexity: O(N * log2(M – m)) where M is maximum value and m is minimum value of the array.
Auxiliary Space: O(1)

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