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Count ways of choosing a pair with maximum difference in C++
We are given with an array of numbers Arr[]. The goal is to count the number of pairs whose difference is equal to the maximum difference of all possible pairs. Count pairs (i!=j) and arr[x]- arr[y] is maximum possible.
We will do this by first finding the maximum difference where (i!=j). And store as maxdiff. Then count all those pairs that have difference=maxdiff.
Let’s understand with examples.
Input − arr[]= { 1,2,3,2,4,1,5 }
Output − No. of ways of choosing pair with maximum difference − 2
Explanation −
Here minimum no. is 1 and maximum number is 5, maximum difference =5-1=4 Pair 1 [ 1,2,3,2,4,1,5 ] → (1,5) difference=4 Pair 2 [ 1,2,3,2,4,1,5 ] → (1,5) difference=4 Number of pairs with difference which is maximum=2.
Input − arr[]= { 2,4,2,4 }
Output − No. of ways of choosing pair with maximum difference − 4
Explanation −
Here minimum no. is 2 and maximum number is 4, maximum difference =4-2=2 Pair 1 [ 2,4,2,4 ] → (2,4) difference=2 Pair 2 [ 2,4,2,4 ] → (2,4) difference=2 Pair 3 [ 2,4,2,4 ] → (4,2) difference=2 Pair 4 [ 2,4,2,4 ] → (2,4) difference=2 Number of pairs with difference which is maximum=4.
Approach used in the below program is as follows
We take an integer array Arr[] initialized with random numbers.
Take a variable N which stores the length of Arr[].
Function countPairs(int arr[],int n) takes an array, its length as input and returns ways of choosing the pairs whose difference is equal to the maximum difference.
Take the initial variable count as 0 for the number of ways.
Take the variable diff as the difference of each pair.
Take the maxdiff variable as the maximum difference of all pairs.
Find maximum and minimum values from the array and store in maxx and mini respectively
Now maxdiff will be maxx-mini.
Traverse array using two for loops for each element of the pair.
Outer loop from 0<=i<n-1, inner loop i<j<n
Calculate diff=arr[i]-arr[j] or arr[j]-arr[i] count as separate. If diff==maxdiff increment count as this pair has the maximum difference.
At the end of all loops count will have a total number of pairs that meet the condition.
Return the count as result.
Example
#include <bits/stdc++.h>
using namespace std;
int countWays(int arr[],int n){
int count = 0;
int diff;
int maxdiff=0; //making minimum as larger than any product in array
int mini,maxx;
mini=maxx=arr[0];
for (int i = 0; i < n; i++) //find minimum and maximum values{
if(arr[i]<mini)
{mini=arr[i];}
if(arr[i]>maxx)
{ maxx=arr[i]; }
}
maxdiff=maxx-mini; //this is maximum difference
//cout<<maxx<<" "<<mini;
for (int i = 0; i < n-1; i++){
for (int j = i+1; j < n; j++){
diff=arr[i]-arr[j]; //pair 1
if ( diff==maxdiff ){
count++;
//cout<<endl<<"a :"<<arr[i]<<" b :"<<arr[j]; //to print
}
diff=arr[j]-arr[i]; //pair 2
if ( diff==maxdiff ){
count++;
//cout<<endl<<"a :"<<arr[i]<<" b :"<<arr[j]; //to print
}
}
}
return count;
}
int main(){
int Arr[]={ 3, 2, 1, 1, 3 };
int N=5; //length of array
cout <<endl<< "No. of ways of choosing pair with maximum difference : "<<countWays(Arr,N);
return 0;
}Output
If we run the above code it will generate the following output −
No. of ways of choosing pair with maximum difference : 4
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