Count ways of choosing a pair with maximum difference
Given an array of n integers, we need to find the no. of ways of choosing pairs with maximum difference.
Examples:
Input : a[] = {3, 2, 1, 1, 3} Output : 4 Explanation: Here, the maximum difference you can find is 2 which is from (1, 3). No. of ways of choosing it: 1) Choosing the first and third elements, 2) Choosing the first and fourth elements, 3) Choosing the third and fifth elements, 4) Choosing the fourth and fifth elements. Hence ans is 4. Input : a[] = {2, 4, 1, 1} Output : 2 Explanation: Here, the maximum difference is 3 from (1, 4). No. of ways choosing it: 1) Choosing the second and third elements, 2) Choosing the second and fourth elements. Hence ans is 2.
Naive Approach : A Simple solution is to find the minimum element and maximum element to find the maximum difference. Then we can find the no. of ways of choosing a pair by running two loops. In the inner loop, check if the two elements(one in outer loop and other in inner loop) are making maximum difference, if yes increase the count.at last output the count.
Time Complexity: O(n^2)
Auxiliary Space: O(1)
Efficient approach:
An efficient approach will be:
 Case I (if all the elements are equal): The ans is no. of ways of choosing 2 elements from a set of n elements which is n(n1)/2.
 Case II (If all the elements are not equal) : The answer is product of count of no. of minimum elements(c1) and count of no. of maximum elements(c2), i.e., c1*c2

// CPP Code to find no. of Ways of choosing
// a pair with maximum difference
#include <bits/stdc++.h>
using
namespace
std;
int
countPairs(
int
a[],
int
n)
{
// To find minimum and maximum of
// the array
int
mn = INT_MAX;
int
mx = INT_MIN;
for
(
int
i = 0; i < n; i++) {
mn = min(mn, a[i]);
mx = max(mx, a[i]);
}
// to find the count of minimum and
// maximum elements
int
c1 = 0;
int
c2 = 0;
// Count variables
for
(
int
i = 0; i < n; i++) {
if
(a[i] == mn)
c1++;
if
(a[i] == mx)
c2++;
}
// condition for all elements equal
if
(mn == mx)
return
n*(n  1) / 2;
else
return
c1* c2;
}
// Driver code
int
main()
{
int
a[] = { 3, 2, 1, 1, 3};
int
n =
sizeof
(a) /
sizeof
(a[0]);
cout << countPairs(a, n);
return
0;
}
Output:
4
Time Complexity: Time complexity to find minimum and maximum is O(n) and Time Complexity to find count of minimum and maximum is O(n)
Overall Time complexity : O(n)
Auxiliary Space : O(1)
Disclaimer: This does not belong to TechCodeBit, its an article taken from the below
source and credits.
source and credits: http://www.geeksforgeeks.org
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