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)

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 then (d+k+1)*x where we will get k+1 multiples and we get the no of pairs as (right-left) boundary [index-wise].

## C++

`// cpp program to calculate the number` `// pairs satisfying th 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 bu 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 fucntion` `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

`// Java program to calculate the number` `// pairs satisfying th 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 bu 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

`# Python program to calculate the number` `# pairs satisfying th 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 bu 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` |

**Output:**

3

**Time complexity:** O(n log n)