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Finding K closest element


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In Finding K closest element problem we have given a sorted array and a value x. The problem is to find the K number of elements closest to x in the given array.

Given an array arr[] ={12, 16, 22, 30, 35, 39, 42,45, 48, 50, 53, 55, 56} and x = 35. Your task is to find the k=4 closest element to x.

Result: 30,39,42,45

Note that if x is in the array then it will be skipped in the output.

Algorithm

  1. Declare and define a binary search method.
  2. Declare a function solution in which array[], x, and k are passed as an argument from the main function.
  3. Call a binary search function from solution function, which returns the position of x to cp in solution function.
  4. Set n to array length, left to cp-1, right to cp+1, and count to 0.
  5. Iterates in the while loop, over the array till the following condition, satisfies:
    1. left>=0 and
    2. right < n and
    3. count < k
  6. Open if block in while loop, in which conditions are given as:
    1. x – array[left]
    2. array[right] – x
  7. If satisfies the condition, then print array[left] and do left —.
  8. If above condition failed then print array[right] and do right++ and do count++ inside the loop.
  9. Open if block with conditions given as if (right = = n ), open the while loop with conditions given as count < k and left >= 0 if satisfies , then print array[left] and do count++.
  10. Again open second if block with conditions given as if (left < 0 ), open the while loop with conditions given as count < k and left >= 0, if satisfies, then print array[right] and do count++.

The approach behind is we will first do a binary search to find the point nearest index to x. Then, we will search left and right for K closest element.

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Explanation for Finding K closest element

So we have our first task as we have to find the ‘k’ closest elements to x in the array which is sorted and all gonna make our task easier. For this, we can use a binary search method to find out the element position, with that our task becomes easier to find k closest elements.

So our main idea is to obtain the position of x and do some operations and find out the number with the least difference with x in the given array. For this, we are going to pass the arguments of arr[], x, and k. Arguments like arr, 0, length of the array, and the number of which we have to find the position are passed into a binary search function from which we obtain the value of the position. With this position we are going to set the value of left to just before the x and value of the right to just after the x, that is left = cp -1( just before the x) and right to cp+1 (just after the x).

Now if arr[left], means the number just before the x has a difference( x – arr[left] ) less than ( arr[right] –x ), means the number just after the x then print the value of arr[left] and do left–, for the next time, else print arr[right] and keep increasing the value od count by 1.

Example

Suppose array given:

arr={12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56};

x=35, we got the position as 4 of 35.

Here left = cp -1 = 3 and right = cp + 1 = 5

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Therefore, we compare x – arr[left] < arr[right] – x => 5 < 4, which fails the condition,

So in else block it will print arr[right] that is 39 and do right++ and count++ that is count=1.

Now left = 3 and right becomes 6

So we compare x – arr[left] < arr[right] – x => 5 < 7, which is true condition,

So in if block it will print arr[left] that is 30 and do left– and count++ that is count=2.

Finding K closest element

Now left=2 and right =6

So we compare x – arr[left] < arr[right] – x => 13 < 7, which is true condition,

So in else block it will print arr[right] that is 42 and do right++ and count++ that is count=3.

Now left=2 and right=7

So we compare x – arr[left] < arr[right] – x => 13 < 7, which fails the condition,

So in if else it will print arr[right] that is 42 and do right++ and count++ that is count=4.

Finding K closest element

And now our count condition is failed means we got our all k closest elements.

And we have the answer as [39 30 42 45].

Implementation

C++ Program for Finding K closest element

#include<stdio.h>
int Binary_Search(int arr[], int low, int high, int x)
{
    int mid = (low + high)/2;

    if (arr[high] <= x)
        return high;

    if (arr[low] > x)
        return low;

    if (arr[mid] <= x && arr[mid+1] > x)
        return mid;

    if(arr[mid] < x)
        return Binary_Search(arr, mid+1, high, x);

    return Binary_Search(arr, low, mid - 1, x);
}

void solution(int arr[], int x, int k, int num)
{
    int y = Binary_Search(arr, 0, num-1, x);
    int r = y+1;
    int count = 0;

    if (arr[y] == x) y--;
    while (y >= 0 && r < num && count < k)
    {
        if (x - arr[y] < arr[r] - x)
            printf("%d ", arr[y--]);
        else
            printf("%d ", arr[r++]);
        count++;
    }

    while (count < k && y >= 0)
        printf("%d ", arr[y--]), count++;

    while (count < k && r < num)
        printf("%d ", arr[r++]), count++;
}

int main()
{
    int arr[] = {12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56};
    int num = sizeof(arr)/sizeof(arr[0]);
    int x = 35;
    int k = 4;
    solution(arr, x, k, num);
    return 0;
}
39 30 42 45

Java Program for Finding K closest element

public class kce {

  public static int binarySearch(int[] arr,int low, int high, int x){
    int mid = 0;
    while(low<=high){
      mid = (high+low)/2;
      if(arr[mid] == x){
        return mid;
      }
      else if(x<arr[mid]){
        high = mid-1;
      }
      else{
        low = mid+1;
      }
    }

    return mid;

  }

  public static void solution(int[] arr,int x, int k){
    int cp = binarySearch(arr, 0, arr.length-1, x);
    int n = arr.length-1;
    int left = cp-1;
    int right = cp+1;
    int count = 0;


    while(left>=0 && right<n && count <k){
      if(x-arr[left]< arr[right] -x){
        System.out.print(arr[left--] + " ");
        left--;
      }
      else{
        System.out.print(arr[right++] + " ");
        right++;
      }
      count++;

    }

    if(right == n){
      while(count<k && left>=0){
        System.out.print(arr[left] + " ");
        count++;
      }
    }

    if(left < 0){
      while(count<k && left>=0){
        System.out.print(arr[right] + " ");
        count++;
      }
    }


  }

  public static void main(String[]args){

    int k = 4;
    int x = 35;
    int arr1[] = {12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56};
    solution(arr1,x,k);
  }

}
39 30 45 50

Complexity Analysis

Time Complexity

O(Logn+ K) where “n” is the size of the array and “k” is the number of the closest element we have to find.

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Space Complexity

O(k) to store the required number of the closest elements.

References

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