# 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) 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
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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 k-th 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 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 number of pairs with 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 which 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 current mid.

 `// C++ program to find k-th absolute difference ` `// between two elements ` `#include ` `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 - a; ` `    ``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; ` ` `  `    ``// 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); ` `    ``cout << kthDiff(a, n, k); ` `    ``return` `0; ` `} `

 `# 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 `

Output:
```1
```

The time complexity of the algorithm is O( n*logn + n*logn*logn). Sorting takes O(n*logn). After that the main binary search over low and high takes O(n*logn*logn) time because each call to the function int f(int c, int n, int* a) takes time O(n*logn).

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