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
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[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 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<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;
} |
# 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.
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Improved By : mohit kumar 29