And this can be easily evaluated with the help of the lower_bound function in O(log(n)) time for each element. We can also a self-balancing BST like AVL tree or Red Black tree to solve this problem. Input: nums = [3,1,4,1,5], k = 2 Output: 2 Explanation: There are two 2-diff pairs in the array, (1, 3) and (3, 5). pairs has the following parameter (s): int k: an integer, the target difference int arr [n]: an array of integers Returns int: the number of pairs that satisfy the criterion Input Format The first line contains two space-separated integers and , the size of and the target value. listA = [5, 3, 7, 2, 9] k = 2 count = 0 # Elements of the list for i in range(0, len(listA)): # Make pairs for j in range(i + 1, len(listA)): if listA[i] - listA[j] == k or listA[j] - listA[i] == k: count += 1 print("Required Pairs: ",count) Output Running the above code gives us the following result Required Pairs: 3 Using While Loop If k>n then time complexity of this algorithm is O(nlgk) wit O(1) space. Example 1: Input: nums = [1,2,2,1], k = 1 Output: 4 Below is the java implementation and sample output. Answers related to "print pairs with difference k using hashmap c++". Given an array of integers nums and an integer k, return the number of unique k-diff pairs in the array. Although we have two 1s in the input, we should only return the number of unique pairs. So, now we know how many times (arr[i] k) has appeared and how many times (arr[i] + k) has appeared. * Given an integer array and a non-negative integer k, count all distinct pairs with difference equal to k, i.e., A[ i ] - A[ j ] = k. * * @param input integer array * @param k * @return number of pairs * * Approach: * Hash the input array into a Map so that we can query for a number in O(1) Practice this problem. Method 3 (Use Self-balancing BST) 2nd line contains N numbers of the set. Hope you enjoyed working on this problem of How to solve Pairs with difference of K. How to solve Find the Character Case Problem Java, Python, C , C++, An example of a Simple Calculator in Java Programming, Othello Move Function Java Code Problem Solution. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above, This article is attributed to GeeksforGeeks.org. Given an array A of size N, how do I count the number of pairs(A[i], A[j]) such that the absolute difference between them is less than or equal to K where K is any positive natural number? For example, in A=[-1, 15, 8, 5, 2, -14, 6, 7] min diff pairs are={(5,6), (6,7), (7,8)}. // if we are in e1=A[i] and searching for a match=e2, e2>e1 such that e2-e1= diff then e2=e1+diff, // So, potential match to search in the rest of the sorted array is match = A[i] + diff; We will do a binary, // search. // perform a binary search on element `A[i]-diff`. By using our site, you consent to our Cookies Policy. (1, 2) and (4, 2) Input : a [] = {1, 8, 7} K = 7 Output : 2 Pairs with difference less than 7 are (1, 7) and (8, 7) Recommended Problem Idea is simple unlike in the trivial solutionof doing linear search for e2=e1+k we will do a optimal binary search. Be the first to rate this post. Determine the number of pairs of array elements that have a difference equal to a target value. Given a sorted array and a number k. print all pairs in the array with difference between them is k. Solution 1: Using Hash Map. Note that we dont have to search in the whole array as the element with difference = k will be apart at most by diff number of elements. Iterate through each element and store a[i]+k in val for index i. We also need to look out for a few things . Input Format: The first line of input contains an integer, that denotes the value of the size of the array. Input: nums = [3,2,1,5,4], k = 2 Output: 3 Explanation: The pairs with an absolute difference of 2 are: - [ 3 ,2, 1 ,5,4] - [ 3 ,2,1, 5 ,4] - [3, 2 ,1,5, 4 ] Constraints: 1 <= nums.length <= 200 1 <= nums [i] <= 100 1 <= k <= 99 Accepted 73.5K Submissions 89.5K Companies Related Topics Array Hash Table Counting Similar Questions Given an array of n integers, We need to find all pairs with a difference less than k Examples : Input : a [] = {1, 10, 4, 2} K = 3 Output : 2 We can make only two pairs with difference less than 3. Step to solve this question: Step 1: Create a Set of all the elements. 1. For those who are not familiar with JS, this is just an ordinary function that we define under the Array object's prototype (so that now every array has access to this function like arr.combinations(n)) But let's use a more expressive language and refactor the above combinations array . We are sorry that this post was not useful for you! Enter your email address to subscribe to new posts. Below is the O(nlgn) time code with O(1) space. This website uses cookies. The idea is somewhat similar to finding a pair with the given sum in the array. A simple hashing technique to use values as index can be used. Count all distinct pairs with difference equal to k in C++, Program to find out the k-th smallest difference between all element pairs in an array in C++, Find pair of numbers whose difference is an input value 'k' in an unsorted array, Count maximum elements of an array whose absolute difference does not exceed K in C++ For example, in the following implementation, range of numbers is assumed to be 0 to 99999. The inputs are an array of integers, numbers, and an integer k. The output should be an integer that is the count of all distinct pairs (a, b) where a + k = b, or k = a b. Step 2: Iterate over each element in set Check whether the number (element value + diff) is in HashSet. The outer loop picks every element x one by one. Here are some. You can have below code from GIT. Method 1 (Simple): A simple solution is to consider all pairs one by one and check difference between every pair. Example 3: i) Here we will use a binary search to find the element. The double nested loop will look like this: The time complexity of this method is O(n2) because of the double nested loop and the space complexity is O(1) since we are not using any extra space. Traverse the array again and look for value k + arr [i] in hashmap. k>n . All the N numbers are assured to be distinct. Note that in the worst case this also takes O(n2) time but works much better in general. We run two loops: the outer loop picks the first element of pair, the inner loop looks for the other element. Rearrange an array in order smallest, largest, 2nd smallest, 2nd largest, .. Pairs with difference K the number of pairs of array elements that have a difference equal to a target value You will be given an array of integers and a target value. If its equal to k, we print it else we move to the next iteration. Read More, Modern Calculator with HTML5, CSS & JavaScript. Two ways are considered different if they consist of different pairs. Therefore count is 2. // This method handles duplicates in the array, // to avoid printing duplicates (skip adjacent duplicates), // perform a binary search on element `arr[i]-diff`. DSA Editorial, Solution and Code. Create a hashmap and then traverse the array. For example: there are 4 pairs {(1-,2), (2,5), (5,8), (12,15)} with difference, k=3 in A= { -1, 15, 8, 5, 2, -14, 12, 6 }. Example 2: Input: nums = [1,2,3,4,5], k = 1 Output: 4 Explanation: There are four 1-diff pairs in the array, (1, 2), (2, 3), (3, 4) and (4, 5). Example 1: Input: nums = [1,2,2,1], k = 1 Output: 4 Explanation: The pairs with an absolute difference of 1 are: - [ 1, 2 ,2,1] - [ 1 ,2, 2 ,1] - [1, 2 ,2, 1 ] - [1,2, 2, 1 ] Example 2: Input: nums = [1,3], k = 3 Output: 0 Explanation: There are no pairs with an absolute difference of 3. See your article appearing on the GeeksforGeeks main page and help other Geeks. For example, Input: arr = [1, 5, 2, 2, 2, 5, 5, 4] k = 3 Output: (2, 5) and (1, 4) Practice this problem We have discussed a linear time solution in the previous post that takes O (n) extra space for an input containing n items. Then (arr[i] + k) will be equal to (arr[i] k) and we will print our pairs twice! Problem of the day - Count Number of Pairs With Absolute Difference K Tag - Easy Given an integer array nums and an integer k, return the number of pairs (i, j) where i < j such that |nums [i] - nums [j]| == k. The value of |x| is defined as: x if x >= 0. Results indicated that: (a) PL outperforms IL in enhancing girls' learning engagement; (b) PL and IL have the same effect on the girls' learning attitude and robotics works; (c) Mixed-gender pairing is more beneficial for girls to complete robotics works than single-gender pairing; and (d) Mixed-gender pairing and single-gender pairing have . Count distinct pairs with difference k Try It! Keep a hash table(HashSet would suffice) to keep the elements already seen while passing through array once. 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