count number of pairs with absolute difference k

Given an array arr [] and an integer K, the task is to find the count of pairs (arr [i], arr [j]) from the array such that |arr [i] - arr [j]| K. Note that (arr [i], arr [j]) and arr [j], arr [i] will be counted only once. Following is a detailed algorithm. 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. 2006 Count Number of Pairs With Absolute Difference K - YouTube The first step (sorting) takes O(nLogn) time. Given an integer array and a positive integer k, count all distinct pairs with differences equal to k. Method 1 (Simple):A simple solution is to consider all pairs one by one and check difference between every pair. Time Complexity: O(n2)Auxiliary Space: O(1), since no extra space has been taken. Count Number of Pairs With Absolute Difference K - LeetCode Perform the Binary Search as per the following: Initialize left as 0 and right as N/2 + 1. Examples: Input: arr [] = {1, 2, 3, 4}, K = 2 Output: 3 All valid pairs are (1, 3), (1, 4) and (2, 4) A very simple case where hashing works in O(n) time is the case where a range of values is very small. 1) Initialize count as 0. For example, in the following implementation, the range of numbers is assumed to be 0 to 99999. Following program implements the simple solution. Count Number of Pairs With Absolute Difference K - LeetCode A simple hashing technique to use values as an index can be used. Count pairs in an array whose absolute difference is divisible by K Approach: Sort the given array. Naive Approach: The idea is to check for each pair of the array one by one and count the total number pairs whose absolute difference is divisible by K. Time Complexity: O(N2)Space Complexity:O(1), Time Complexity: O(n+k)Auxiliary Space: O(k). Efficient Approach: To optimize the above approach, the idea is to observe the fact that for two numbers a[i] and a[j], if a[i] % k = a[j] % k, then abs(a[i] a[j]) is a multiple of K. Follow the below steps to solve the problem: Below is the implementation of the above approach. Please do LIKE, SHARE, and SUBSCRIBEDiscord Link : https://discord.gg/ntMkngpNTt ABOUT EDIGNITEEdignite NGO strives to accomplish Kalam sir's mission of an I. Following are the detailed steps. Please use ide.geeksforgeeks.org, Print the value of the count after all pairs are processed. Time Complexity: O(NlogN)Auxiliary Space: O(N). We run two loops: the outer loop picks the first element of pair, the inner loop looks for the other element. The outer loop picks every element x one by one. Given an integer array nums and an integer k, return the number of pairs (i, j) where i < j such. b) Look for arr [i] - k in the hash map, if found then increment count. Method 4 (Use Hashing):We can also use hashing to achieve the average time complexity as O(n) for many cases. 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 =. I knew it. Thanks for sticking along, it's day 15 of my coding diary. Practice Problems, POTD Streak, Weekly Contests & More! Count Number of Pairs with Absolute Difference K | Leetcode DSA b) If arr[i] + k is not found, return the index of the first occurrence of the value greater than arr[i] + k. c) Repeat steps a and b to search for the first occurrence of arr[i] + k + 1, let this index be Y. It was a light bulb moment.. Let's remove the sluggish hashmap and use an array of length 200. Well, I started this journey to discipline my self and I feel, now I am literally enjoying it. Print the value of the count after all pairs are processed.Time Complexity: O(N2)Auxiliary Space: O(1). Naive approach: The easiest way is to iterate through every possible pair in the array and if the absolute difference of the numbers is a multiple of K, then increase the count by 1. Writing code in comment? Count Number of Pairs With Absolute Difference K - LeetCode Solutions LeetCode Solutions Home Preface Style Guide Problems Problems 1. Discuss (910) Submissions. The second step can be optimized to O(n), see this. I am gradually recognizing the patterns. The pairs are: (3, 3), (3, 3), (3, 3). Count all distinct pairs with difference equal to k Efficient Approach: The efficient idea is to use Binary Search to find the first occurrence having a difference of at least K. Below are the steps: Sort the given array in increasing order. Description Solution Submissions 2006. Example 3: Count all distinct pairs with difference equal to k The task is to count the total number of pairs in the array whose absolute difference is divisible by K. Examples: Input: arr [] = {1, 2, 3, 4}, K = 2 Output: 2 Explanation: Total 2 pairs exists in the array with absolute difference divisible by 2. Count pairs in an array whose absolute difference is divisible by K, Minimize sum of absolute difference between all pairs of array elements by decrementing and incrementing pairs by 1, Length of longest subsequence having absolute difference of all pairs divisible by K, Count pairs in Array whose product is divisible by K, Count of pairs in Array whose product is divisible by K, Count pairs in array whose sum is divisible by 4, Count pairs in array whose sum is divisible by K, Count maximum elements of an array whose absolute difference does not exceed K, Count the number of pairs (i, j) such that either arr[i] is divisible by arr[j] or arr[j] is divisible by arr[i], Count of pairs from 1 to a and 1 to b whose sum is divisible by N, Count of all pairs in an Array with minimum absolute difference, Count all disjoint pairs having absolute difference at least K from a given array, Count Non-Repeating array elements after inserting absolute difference between all possible pairs, Count pairs from an array with absolute difference not less than the minimum element in the pair, Count pairs in an array such that the absolute difference between them is K, Count of subarrays of size K having at least one pair with absolute difference divisible by K-1, Count N-digit numbers whose digits does not exceed absolute difference of the two previous digits, Count of elements whose absolute difference with the sum of all the other elements is greater than k, Count of N-digit numbers whose absolute difference between adjacent digits is non-increasing, Check if an array can be divided into pairs whose sum is divisible by k, Searching in a map using std::map functions in C++, Maximize count of pairs whose Bitwise AND exceeds Bitwise XOR by replacing such pairs with their Bitwise AND, Maximize count of pairs whose bitwise XOR is even by replacing such pairs with their Bitwise XOR, Find K elements whose absolute difference with median of array is maximum, Complete Interview Preparation- Self Paced Course, Data Structures & Algorithms- Self Paced Course. a) Look for arr [i] + k in the hash map, if found then increment count. Initialize cnt to 0 which will store the count of all possible pairs. The inner loop considers all elements after x and checks if the difference is within limits or not. You might like previous editions of my coding diary, Count Number of Pairs With Absolute Difference K. Day #14 - Minimum Number of Operations to Move All Balls to Each Box. The following algorithm takes O(N) time and O(N) space - based on a hash table. Example 1: Input: nums = [1,2,2,1], k = 1 Output: 4 Count Number of Pairs With Absolute Difference K (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. -x if x < 0. -x if x < 0. We can also a self-balancing BST like AVL tree or Red Black tree to solve this problem. Although we have two 1s in the input, we . Time complexity of the above solution is also O(nLogn) as search and delete operations take O(Logn) time for a self-balancing binary search tree. Please use ide.geeksforgeeks.org, Time Complexity: O (N2) Auxiliary Space: O (1) -x if x < 0. By using our site, you Input: N = 3, K = 3, arr[] = {3, 3, 3}Output: 3Explanation:Total 3 pairs exists in this array with absolute difference divisible by 3. As expected, the worst approach. Given an array of integers nums and an integer k, return the number of unique k-diff pairs in the array. Pankaj Tanwar - CS Engineer, writer & creator. I again read the problem statement. Writing code in comment? Two Sum 2. By using our site, you New. 2006 - Count Number of Pairs With Absolute Difference K Problem of the day - Count Number of Pairs With Absolute Difference K. 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. Just after reading the problem statement carefully, like any other dev, a brute force, O(n2), the slowest approach came to my mind and I just started typing without wasting a second. A-143, 9th Floor, Sovereign Corporate Tower, We use cookies to ensure you have the best browsing experience on our website. 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(I found 3 approaches for the problem), Easy problems might have interesting hidden challenges, For each element, reduce the count for the current value first, and check the count of, return the final value, that's the answer. By using our site, you Practice Problems, POTD Streak, Weekly Contests & More! Frequency Array Approach: The approach to solving this problem using a frequency array is discussed in Set-1 of this article. A-143, 9th Floor, Sovereign Corporate Tower, We use cookies to ensure you have the best browsing experience on our website. generate link and share the link here. WTH, I thought I cracked it. Count distinct pairs with difference k Try It! Element of pair, the range of numbers is assumed to be to. Tanwar - CS Engineer, writer & creator every element x one by.. Look for arr [ I ] - k in the hash map, if found then increment count &.. Is assumed to be 0 to 99999 example, in the following takes! 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