Given an array and two numbers x and k. Find the number of different ordered pairs of indexes (i, j) such that a[j] >= a[i] and there are exactly k integers num such that num is divisible by x and num is in range a[i]-a[j].
Examples:
Input : arr[] = {1, 3, 5, 7} x = 2, k = 1 Output : 3 Explanation: The pairs (1, 3), (3, 5) and (5, 7) have k (which is 1) integers i.e., 2, 4, 6 respectively for every pair in between them. Input : arr[] = {5, 3, 1, 7} x = 2, k = 0 Output : 4 Explanation: The pairs with indexes (1, 1), (2, 2), (3, 3), (4, 4) have k = 0 integers that are divisible by 2 in between them.
A naive approach is to traverse through all pairs possible and count the number of pairs that have k integers in between them which are divisible by x.
Time complexity: O(n^2), as we will be using nested loops for traversing n*n times.
An efficient approach is to sort the array and use binary search to find out the right and left boundaries of numbers(use lower_bound function inbuilt function to do it) which satisfy the condition and which do not. We have to sort the array as it is given every pair should be a[j] >= a[i] irrespective of value of i and j. After sorting we traverse through n elements, and find the number with whose multiplication with x gives a[i]-1, so that we can find k number by adding k to d = a[i]-1/x. So we binary search for the value (d+k)*x to get the multiple with which we can make a pair of a[i] as it will have exactly k integers in between a[i] and a[j]. In this way we get the left boundary for a[j] using binary search in O(log n), and for all other pairs possible with a[i], we need to find out the right-most boundary by searching the number equal to or greater than (d+k+1)*x where we will get k+1 multiples and we get the no of pairs as (right-left) boundary [index-wise].
Implementation:
// C++ program to calculate the number // pairs satisfying the condition #include <bits/stdc++.h> using namespace std;
// function to calculate the number of pairs int countPairs( int a[], int n, int x, int k)
{ sort(a, a + n);
// traverse through all elements
int ans = 0;
for ( int i = 0; i < n; i++) {
// current number's divisor
int d = (a[i] - 1) / x;
// use binary search to find the element
// after k multiples of x
int it1 = lower_bound(a, a + n,
max((d + k) * x, a[i])) - a;
// use binary search to find the element
// after k+1 multiples of x so that we get
// the answer by subtracting
int it2 = lower_bound(a, a + n,
max((d + k + 1) * x, a[i])) - a;
// the difference of index will be the answer
ans += it2 - it1;
}
return ans;
} // driver code to check the above function int main()
{ int a[] = { 1, 3, 5, 7 };
int n = sizeof (a) / sizeof (a[0]);
int x = 2, k = 1;
// function call to get the number of pairs
cout << countPairs(a, n, x, k);
return 0;
} |
// Java program to calculate the number // pairs satisfying the condition import java.util.*;
class GFG
{ // function to calculate the number of pairs static int countPairs( int a[], int n, int x, int k)
{ Arrays.sort(a);
// traverse through all elements
int ans = 0 ;
for ( int i = 0 ; i < n; i++)
{
// current number's divisor
int d = (a[i] - 1 ) / x;
// use binary search to find the element
// after k multiples of x
int it1 = Arrays.binarySearch(a,
Math.max((d + k) * x, a[i]));
// use binary search to find the element
// after k+1 multiples of x so that we get
// the answer by subtracting
int it2 = Arrays.binarySearch(a,
Math.max((d + k + 1 ) * x, a[i])) ;
// the difference of index will be the answer
ans += it1 - it2;
}
return ans;
} // Driver code public static void main(String[] args)
{ int []a = { 1 , 3 , 5 , 7 };
int n = a.length;
int x = 2 , k = 1 ;
// function call to get the number of pairs
System.out.println(countPairs(a, n, x, k));
} } // This code is contributed by Rajput-Ji |
# Python3 program to calculate the number # pairs satisfying the condition import bisect
# function to calculate the number of pairs def countPairs(a, n, x, k):
a.sort()
# traverse through all elements
ans = 0
for i in range (n):
# current number's divisor
d = (a[i] - 1 ) / / x
# use binary search to find the element
# after k multiples of x
it1 = bisect.bisect_left(a, max ((d + k) * x, a[i]))
# use binary search to find the element
# after k+1 multiples of x so that we get
# the answer by subtracting
it2 = bisect.bisect_left(a, max ((d + k + 1 ) * x, a[i]))
# the difference of index will be the answer
ans + = it2 - it1
return ans
# Driver code if __name__ = = "__main__" :
a = [ 1 , 3 , 5 , 7 ]
n = len (a)
x = 2
k = 1
# function call to get the number of pairs
print (countPairs(a, n, x, k))
# This code is contributed by # sanjeev2552 |
// C# program to calculate the number // pairs satisfying the condition using System;
class GFG{
// Function to calculate the number of pairs static int countPairs( int [] a, int n,
int x, int k)
{ Array.Sort(a);
// Traverse through all elements
int ans = 0;
for ( int i = 0; i < n; i++)
{
// Current number's divisor
int d = (a[i] - 1) / x;
// Use binary search to find the element
// after k multiples of x
int a1 = Math.Max((d + k) * x, a[i]);
int it1 = Array.BinarySearch(a, a1);
// Use binary search to find the element
// after k+1 multiples of x so that we get
// the answer by subtracting
int a2 = Math.Max((d + k + 1) * x, a[i]);
int it2 = Array.BinarySearch(a, a2);
// The difference of index will
// be the answer
ans += Math.Abs(it2 - it1);
}
return ans;
} // Driver Code static void Main()
{ int [] a = { 1, 3, 5, 7 };
int n = a.Length;
int x = 2, k = 1;
// Function call to get the number of pairs
Console.Write(countPairs(a, n, x, k));
} } // This code is contributed by SoumikMondal. |
// Javascript program to calculate the number // pairs satisfying the condition function lower_bound(arr, N, X)
{ let mid;
let low = 0;
let high = N;
while (low < high) {
mid = low + (high - low) / 2;
if (X <= arr[mid]) {
high = mid;
}
else {
low = mid + 1;
}
}
if (low < N && arr[low] < X) {
low++;
}
return low;
} // function to calculate the number of pairs function countPairs(a, n, x, k)
{ a.sort();
// traverse through all elements
let ans = 0;
for (let i = 0; i < n; i++) {
// current number's divisor
let d = (a[i] - 1) / x;
// use binary search to find the element
// after k multiples of x
let it1 = lower_bound(a, n,
Math.max((d + k) * x, a[i]));
// use binary search to find the element
// after k+1 multiples of x so that we get
// the answer by subtracting
let it2 = lower_bound(
a, n, Math.max((d + k + 1) * x, a[i]));
// the difference of index will be the answer
ans += it2 - it1;
}
return ans;
} // driver code to check the above function let a = [ 1, 3, 5, 7 ]; let n = 4; let x = 2, k = 1; // function call to get the number of pairs console.log(countPairs(a, n, x, k)); // This code is contributed by garg28harsh. |
3
Time Complexity: O(N*logN), as we are using sort function which will cost N*logN.
Auxiliary Space: O(1), as we are not using any extra space.