k-th smallest absolute difference of two elements in an array

We are given an array of size n containing positive integers. The absolute difference between values at indices i and j is |a[i] – a[j]|. There are n*(n-1)/2 such pairs and we are asked to print the kth (1 <= k <= n*(n-1)/2) as the smallest absolute difference among all these pairs.

Examples:

Input  : a[] = {1, 2, 3, 4}
         k = 3
Output : 1
The possible absolute differences are :
{1, 2, 3, 1, 2, 1}.
The 3rd smallest value among these is 1.

Input : n = 2
        a[] = {10, 10}
        k = 1
Output : 0

Naive Method is to find all the n*(n-1)/2 possible absolute differences in O(n^2) and store them in an array. Then sort this array and print the kth minimum value from this array. This will take time O(n^2 + n^2 * log(n^2)) = O(n^2 + 2*n^2*log(n)).

The naive method won’t be efficient for large values of n, say n = 10^5.

An Efficient Solution is based on Binary Search. 

1) Sort the given array a[].
2) We can easily find the least possible absolute
   difference in O(n) after sorting. The largest
   possible difference will be a[n-1] - a[0] after
   sorting the array. Let low = minimum_difference
   and high = maximum_difference.
3) while low < high:
4)     mid = (low + high)/2
5)     if ((number of pairs with absolute difference
                                <= mid) < k):
6)        low = mid + 1
7)     else:
8)        high = mid
9) return low

We need a function that will tell us the number of pairs with a difference <= mid efficiently. Since our array is sorted, this part can be done like this: 

1) result = 0
2) for i = 0 to n-1:
3)     result = result + (upper_bound(a+i, a+n, a[i] + mid) - (a+i+1))
4) return result

Here upper_bound is a variant of binary search that returns a pointer to the first element from a[i] to a[n-1] which is greater than a[i] + mid. Let the pointer returned be j. Then a[i] + mid < a[j]. Thus, subtracting (a+i+1) from this will give us the number of values whose difference with a[i] is <= mid. We sum this up for all indices from 0 to n-1 and get the answer for the current mid.

Flowchart is as follows: 

Flowchart

Implementation:

C++

// C++ program to find k-th absolute difference
// between two elements
#include<bits/stdc++.h>
using namespace std;

// returns number of pairs with absolute difference
// less than or equal to mid.
int countPairs(int *a, int n, int mid)
{
    int res = 0;
    for (int i = 0; i < n; ++i)

        // Upper bound returns pointer to position
        // of next higher number than a[i]+mid in
        // a[i..n-1]. We subtract (a + i + 1) from
        // this position to count
        res += upper_bound(a+i, a+n, a[i] + mid) -
                                    (a + i + 1);
    return res;
}

// Returns k-th absolute difference
int kthDiff(int a[], int n, int k)
{
    // Sort array
    sort(a, a+n);

    // Minimum absolute difference
    int low = a[1] - a[0];
    for (int i = 1; i <= n-2; ++i)
        low = min(low, a[i+1] - a[i]);

    // Maximum absolute difference
    int high = a[n-1] - a[0];

    // Do binary search for k-th absolute difference
    while (low < high)
    {
        int mid = (low+high)>>1;
        if (countPairs(a, n, mid) < k)
            low = mid + 1;
        else
            high = mid;
    }

    return low;
}

// Driver code
int main()
{
    int k = 3;
    int a[] = {1, 2, 3, 4};
    int n = sizeof(a)/sizeof(a[0]);
    cout << kthDiff(a, n, k);
    return 0;
}

Java

// Java program to find k-th absolute difference
// between two elements
import java.util.Scanner;
import java.util.Arrays;

class GFG
{
    // returns number of pairs with absolute
    // difference less than or equal to mid
    static int countPairs(int[] a, int n, int mid)
    {
        int res = 0, value;
        for(int i = 0; i < n; i++)
        {
            // Upper bound returns pointer to position
            // of next higher number than a[i]+mid in
            // a[i..n-1]. We subtract (ub + i + 1) from
            // this position to count
            if(a[i]+mid>a[n-1])
              res+=(n-(i+1));
            else
            {
             int ub = upperbound(a, n, a[i]+mid);
             res += (ub- (i+1));
            }
        }
        return res;
    }

    // returns the upper bound
    static int upperbound(int a[], int n, int value)
    {
        int low = 0;
        int high = n;
        while(low < high)
        {
            final int mid = (low + high)/2;
            if(value >= a[mid])
                low = mid + 1;
            else
                high = mid;
        }

    return low;
    }

    // Returns k-th absolute difference
    static int kthDiff(int a[], int n, int k)
    {
        // Sort array
        Arrays.sort(a);

        // Minimum absolute difference
        int low = a[1] - a[0];
        for (int i = 1; i <= n-2; ++i)
            low = Math.min(low, a[i+1] - a[i]);

        // Maximum absolute difference
        int high = a[n-1] - a[0];

        // Do binary search for k-th absolute difference
        while (low < high)
        {
            int mid = (low + high) >> 1;
            if (countPairs(a, n, mid) < k)
                low = mid + 1;
            else
                high = mid;
        }

        return low;
    }

    // Driver function to check the above functions
    public static void main(String args[])
    {
        Scanner s = new Scanner(System.in);
        int k = 3;
        int a[] = {1,2,3,4};
        int n = a.length;
        System.out.println(kthDiff(a, n, k));
    }

}
// This code is contributed by nishkarsh146

Python3

# Python3 program to find
# k-th absolute difference
# between two elements
from bisect import bisect as upper_bound

# returns number of pairs with
# absolute difference less than
# or equal to mid.
def countPairs(a, n, mid):
    res = 0
    for i in range(n):

        # Upper bound returns pointer to position
        # of next higher number than a[i]+mid in
        # a[i..n-1]. We subtract (a + i + 1) from
        # this position to count
        res += upper_bound(a, a[i] + mid)
    return res

# Returns k-th absolute difference
def kthDiff(a, n, k):

    # Sort array
    a = sorted(a)

    # Minimum absolute difference
    low = a[1] - a[0]
    for i in range(1, n - 1):
        low = min(low, a[i + 1] - a[i])

    # Maximum absolute difference
    high = a[n - 1] - a[0]

    # Do binary search for k-th absolute difference
    while (low < high):
        mid = (low + high) >> 1
        if (countPairs(a, n, mid) < k):
            low = mid + 1
        else:
            high = mid

    return low

# Driver code
k = 3
a = [1, 2, 3, 4]
n = len(a)
print(kthDiff(a, n, k))

# This code is contributed by Mohit Kumar

C#

// C# program to find k-th
// absolute difference
// between two elements
using System;
class GFG{

// returns number of pairs
// with absolute difference
// less than or equal to mid
static int countPairs(int[] a,
                      int n,
                      int mid)
{
  int res = 0;
  for(int i = 0; i < n; i++)
  {
    // Upper bound returns pointer
    // to position of next higher
    // number than a[i]+mid in
    // a[i..n-1]. We subtract
    // (ub + i + 1) from
    // this position to count
    int ub = upperbound(a, n,
                        a[i] + mid);
    res += (ub - (i));
  }
  return res;
}

// returns the upper bound
static int upperbound(int []a,
                      int n,
                      int value)
{
  int low = 0;
  int high = n;
  while(low < high)
  {
    int mid = (low + high)/2;
    if(value >= a[mid])
      low = mid + 1;
    else
      high = mid;
  }

  return low;
}

// Returns k-th absolute
// difference
static int kthDiff(int []a,
                   int n, int k)
{
  // Sort array
  Array.Sort(a);

  // Minimum absolute
  // difference
  int low = a[1] - a[0];
  for (int i = 1; i <= n - 2; ++i)
    low = Math.Min(low, a[i + 1] -
                   a[i]);

  // Maximum absolute
  // difference
  int high = a[n - 1] - a[0];

  // Do binary search for
  // k-th absolute difference
  while (low < high)
  {
    int mid = (low + high) >> 1;
    if (countPairs(a, n, mid) < k)
      low = mid + 1;
    else
      high = mid;
  }

  return low;
}

// Driver code
public static void Main(String []args)
{
  int k = 3;
  int []a = {1, 2, 3, 4};
  int n = a.Length;
  Console.WriteLine(kthDiff(a, n, k));
}
}

// This code is contributed by gauravrajput1

Javascript

<script>

// JavaScript program to find k-th
// absolute difference
// between two elements


// returns number of pairs
// with absolute difference
// less than or equal to mid
function countPairs(a, n, mid) {
    let res = 0;
    for (let i = 0; i < n; i++) {
        // Upper bound returns pointer
        // to position of next higher
        // number than a[i]+mid in
        // a[i..n-1]. We subtract
        // (ub + i + 1) from
        // this position to count
        let ub = upperbound(a, n,
            a[i] + mid);
        res += (ub - (i));
    }
    return res;
}

// returns the upper bound
function upperbound(a, n, value) {
    let low = 0;
    let high = n;
    while (low < high) {
        let mid = (low + high) / 2;
        if (value >= a[mid])
            low = mid + 1;
        else
            high = mid;
    }

    return low;
}

// Returns k-th absolute
// difference
function kthDiff(a, n, k) {
    // Sort array
    a.sort((a, b) => a - b);

    // Minimum absolute
    // difference
    let low = a[1] - a[0];
    for (let i = 1; i <= n - 2; ++i)
        low = Math.min(low, a[i + 1] -
            a[i]);

    // Maximum absolute
    // difference
    let high = a[n - 1] - a[0];

    // Do binary search for
    // k-th absolute difference
    while (low < high) {
        let mid = (low + high) >> 1;
        if (countPairs(a, n, mid) < k)
            low = mid + 1;
        else
            high = mid;
    }

    return low;
}

// Driver code

let k = 3;
let a = [1, 2, 3, 4];
let n = a.length;
document.write(kthDiff(a, n, k));



// This code is contributed by gfgking

</script>

Output

1

Time Complexity: O(nlogn)

Auxiliary Space: O(1)

Suppose, the maximum element in the array is, and the minimum element is a minimum element in the array is . Then time taken for the binary_search will be , and the time taken for the upper_bound function will be 

So, the time complexity of the algorithm is . Sorting takes . After that the main binary search over low and high takes  time because each call to the function countPairs takes time . 

So the Overall time complexity would be 

 This article is contributed by Hemang Sarkar. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.