QUESTION
Count pairs in a sorted array whose sum is less than x.
“TESTCASE_1”: “5 4\n1 6 7 8 9\n###—###SEPERATOR—###—\n0”, “TESTCASE_2”: “6 8\n4 4 5 3 1 10\n###—###SEPERATOR—###—\n10”, “TESTCASE_3”: “8 4\n2 3 8 7 9 1 0 12\n###—###SEPERATOR—###—\n11”, “TESTCASE_4”: “4 8\n1 2 3 5\n###—###SEPERATOR—###—\n5”, “TESTCASE_5”: “0\n###—###SEPERATOR—###—\n0
ANSWER
#include <iostream>
using namespace std;
int find(int arr[],int n,int x)
{
int l = 0, r = n-1;
int result = 0;
while (l < r)
{
if (arr[l] + arr[r] < x){
result += (r - l);
l++;
}
else
r--;
}
return result;
}
int main()
{
int n,x;
cin>>n>>x;
int a[n];
for(int i=0;i<n;i++)
cin>>a[i];
cout<<find(a,n,x);
return 0;
}