This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository. This solution doesnt work if there are duplicates in array as the requirement is to count only distinct pairs. Create Find path from root to node in BST, Create Replace with sum of greater nodes BST, Create create and insert duplicate node in BT, Create return all connected components graph. You are given with an array of integers and an integer K. You have to find and print the count of all such pairs which have difference K. Note: Take absolute difference between the elements of the array. The solution should have as low of a computational time complexity as possible. Use Git or checkout with SVN using the web URL. Many Git commands accept both tag and branch names, so creating this branch may cause unexpected behavior. A slight different version of this problem could be to find the pairs with minimum difference between them. Many Git commands accept both tag and branch names, so creating this branch may cause unexpected behavior. To review, open the file in an editor that reveals hidden Unicode characters. 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. * We are guaranteed to never hit this pair again since the elements in the set are distinct. No votes so far! The idea is that in the naive approach, we are checking every possible pair that can be formed but we dont have to do that. Inside this folder we create two files named Main.cpp and PairsWithDifferenceK.h. This is a negligible increase in cost. Given an array of integers nums and an integer k, return the number of unique k-diff pairs in the array. (5, 2) You signed in with another tab or window. We also check if element (arr[i] - diff) or (arr[i] + diff) already exists in the set or not. For example, in A=[-1, 15, 8, 5, 2, -14, 6, 7] min diff pairs are={(5,6), (6,7), (7,8)}. Read More, Modern Calculator with HTML5, CSS & JavaScript. Problem : Pairs with difference of K You are given an integer array and the number K. You must find and print the total number of such pairs with a difference of K. Take the absolute difference between the array's elements. Patil Institute of Technology, Pimpri, Pune. BFS Traversal BTree withoutSivling Balanced Paranthesis Binary rec Compress the sting Count Leaf Nodes TREE Detect Cycle Graph Diameter of BinaryTree Djikstra Graph Duplicate in array Edit Distance DP Elements in range BST Even after Odd LinkedList Fibonaci brute,memoization,DP Find path from root to node in BST Get Path DFS Has Path Cannot retrieve contributors at this time 72 lines (70 sloc) 2.54 KB Raw Blame You are given with an array of integers and an integer K. You have to find and print the count of all such pairs which have difference K. Note: Take absolute difference between the elements of the array. Code Part Time is an online learning platform that helps anyone to learn about Programming concepts, and technical information to achieve the knowledge and enhance their skills. Following is a detailed algorithm. (5, 2) //edge case in which we need to find i in the map, ensuring it has occured more then once. This website uses cookies. 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. If nothing happens, download Xcode and try again. //System.out.println("Current element: "+i); //System.out.println("Need to find: "+(i-k)+", "+(i+k)); countPairs=countPairs+(map.get(i)*map.get(k+i)); //System.out.println("Current count of pairs: "+countPairs); countPairs=countPairs+(map.get(i)*map.get(i-k)). * Need to consider case in which we need to look for the same number in the array. Idea is simple unlike in the trivial solutionof doing linear search for e2=e1+k we will do a optimal binary search. If k>n then time complexity of this algorithm is O(nlgk) wit O(1) space. A k-diff pair is an integer pair (nums [i], nums [j]), where the following are true: 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). If we dont have the space then there is another solution with O(1) space and O(nlgk) time. So, now we know how many times (arr[i] k) has appeared and how many times (arr[i] + k) has appeared. Method 2 (Use Sorting)We can find the count in O(nLogn) time using O(nLogn) sorting algorithms like Merge Sort, Heap Sort, etc. It will be denoted by the symbol n. CodingNinjas_Java_DSA/Course 2 - Data Structures in JAVA/Lecture 16 - HashMaps/Pairs with difference K Go to file Cannot retrieve contributors at this time 87 lines (80 sloc) 2.41 KB Raw Blame /* You are given with an array of integers and an integer K. You have to find and print the count of all such pairs which have difference K. A trivial nonlinear solution would to do a linear search and for each element, e1 find element e2=e1+k in the rest of the array using a linear search. Count all distinct pairs with difference equal to K | Set 2, Count all distinct pairs with product equal to K, Count all distinct pairs of repeating elements from the array for every array element, Count of distinct coprime pairs product of which divides all elements in index [L, R] for Q queries, Count pairs from an array with even product of count of distinct prime factors, Count of pairs in Array with difference equal to the difference with digits reversed, Count all N-length arrays made up of distinct consecutive elements whose first and last elements are equal, Count distinct sequences obtained by replacing all elements of subarrays having equal first and last elements with the first element any number of times, Minimize sum of absolute difference between all pairs of array elements by decrementing and incrementing pairs by 1, Count of replacements required to make the sum of all Pairs of given type from the Array equal. A naive solution would be to consider every pair in a given array and return if the desired difference is found. Find pairs with difference k in an array ( Constant Space Solution). Learn more about bidirectional Unicode characters. Clone with Git or checkout with SVN using the repositorys web address. k>n . To review, open the file in an editor that reveals hidden Unicode characters. Therefore, overall time complexity is O(nLogn). 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. // Function to find a pair with the given difference in an array. A tag already exists with the provided branch name. Obviously we dont want that to happen. Learn more. Time Complexity: O(nlogn)Auxiliary Space: O(logn). pairs with difference k coding ninjas github. There was a problem preparing your codespace, please try again. if value diff > k, move l to next element. Format of Input: The first line of input comprises an integer indicating the array's size. return count. pairs_with_specific_difference.py. The following line contains an integer, that denotes the value of K. The first and only line of output contains count of all such pairs which have an absolute difference of K. public static int getPairsWithDifferenceK(int arr[], int k) {. 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. You signed in with another tab or window. We are sorry that this post was not useful for you! Method 5 (Use Sorting) : Sort the array arr. Instantly share code, notes, and snippets. HashMap map = new HashMap<>(); System.out.println(i + ": " + map.get(i)); //System.out.println("Current element: "+i); //System.out.println("Need to find: "+(i-k)+", "+(i+k)); countPairs=countPairs+(map.get(i)*map.get(k+i)); //System.out.println("Current count of pairs: "+countPairs); countPairs=countPairs+(map.get(i)*map.get(i-k)). This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. Note: the order of the pairs in the output array should maintain the order of the y element in the original array. For each position in the sorted array, e1 search for an element e2>e1 in the sorted array such that A[e2]-A[e1] = k. A tag already exists with the provided branch name. // Function to find a pair with the given difference in the array. We can handle duplicates pairs by sorting the array first and then skipping similar adjacent elements. Inside the package we create two class files named Main.java and Solution.java. So we need to add an extra check for this special case. Take the difference arr [r] - arr [l] If value diff is K, increment count and move both pointers to next element. * If the Map contains i-k, then we have a valid pair. Enter your email address to subscribe to new posts. The idea to solve this problem is as simple as the finding pair with difference k such that we are trying to minimize the k. * 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. * Hash the input array into a Map so that we can query for a number in O(1). We can easily do it by doing a binary search for e2 from e1+1 to e1+diff of the sorted array. Let us denote it with the symbol n. The following line contains n space separated integers, that denote the value of the elements of the array. By using this site, you agree to the use of cookies, our policies, copyright terms and other conditions. Then (arr[i] + k) will be equal to (arr[i] k) and we will print our pairs twice! But we could do better. You are given an integer array and the number K. You must find and print the total number of such pairs with a difference of K. Take the absolute difference between the arrays elements.if(typeof ez_ad_units!='undefined'){ez_ad_units.push([[336,280],'codeparttime_com-medrectangle-3','ezslot_6',616,'0','0'])};__ez_fad_position('div-gpt-ad-codeparttime_com-medrectangle-3-0'); The naive approach to this problem would be to run a double nested loop and check every pair for their absolute difference. Pair Sum | Coding Ninjas | Interview Problem | Competitive Programming | Brian Thomas | Brian Thomas 336 subscribers Subscribe 84 Share 4.2K views 1 year ago In this video, we will learn how. Take two pointers, l, and r, both pointing to 1st element. You signed in with another tab or window. Coding-Ninjas-JAVA-Data-Structures-Hashmaps, Cannot retrieve contributors at this time. (4, 1). * Iterate through our Map Entries since it contains distinct numbers. The time complexity of the above solution is O(n.log(n)) and requires O(n) extra space, where n is the size of the input. Inside file Main.cpp we write our C++ main method for this problem. By using our site, you 2. A tag already exists with the provided branch name. 121 commits 55 seconds. To review, open the file in an. If nothing happens, download GitHub Desktop and try again. 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. You are given with an array of integers and an integer K. You have to find and print the count of all such pairs which have difference K. Note: Take absolute difference between the elements of the array. The first line of input contains an integer, that denotes the value of the size of the array. Learn more about bidirectional Unicode characters. Add the scanned element in the hash table. The overall complexity is O(nlgn)+O(nlgk). This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. A-143, 9th Floor, Sovereign Corporate Tower, We use cookies to ensure you have the best browsing experience on our website. Read our. Following program implements the simple solution. Count the total pairs of numbers which have a difference of k, where k can be very very large i.e. * 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) O(n) time and O(n) space solution Method 6(Using Binary Search)(Works with duplicates in the array): a) Binary Search for the first occurrence of arr[i] + k in the sub array arr[i+1, N-1], let this index be X. Method 4 (Use Hashing):We can also use hashing to achieve the average time complexity as O(n) for many cases. System.out.println(i + ": " + map.get(i)); for (Integer i: map.keySet()) {. The time complexity of the above solution is O(n) and requires O(n) extra space. Keep a hash table(HashSet would suffice) to keep the elements already seen while passing through array once. Input Format: The first line of input contains an integer, that denotes the value of the size of the array. * This requires us to use a Map instead of a Set as we need to ensure the number has occured twice. Please This is O(n^2) solution. If exists then increment a count. Understanding Cryptography by Christof Paar and Jan Pelzl . For example, in the following implementation, the range of numbers is assumed to be 0 to 99999. Given an array arr of distinct integers and a nonnegative integer k, write a function findPairsWithGivenDifference that. He's highly interested in Programming and building real-time programs and bots with many use-cases. Follow me on all Networking Sites: LinkedIn : https://www.linkedin.com/in/brian-danGitHub : https://github.com/BRIAN-THOMAS-02Instagram : https://www.instagram.com/_b_r_i_a_n_#pairsum #codingninjas #competitveprogramming #competitve #programming #education #interviewproblem #interview #problem #brianthomas #coding #crackingproblem #solution Cannot retrieve contributors at this time. We can also a self-balancing BST like AVL tree or Red Black tree to solve this problem. Think about what will happen if k is 0. Founder and lead author of CodePartTime.com. We run two loops: the outer loop picks the first element of pair, the inner loop looks for the other element. Thus each search will be only O(logK). sign in Inside file PairsWithDifferenceK.h we write our C++ solution. Learn more about bidirectional Unicode characters. The second step runs binary search n times, so the time complexity of second step is also O(nLogn). We can improve the time complexity to O(n) at the cost of some extra space. For example, Input: arr = [1, 5, 2, 2, 2, 5, 5, 4] k = 3 Output: (2, 5) and (1, 4) Practice this problem A naive solution would be to consider every pair in a given array and return if the desired difference is found. Let us denote it with the symbol n. The following line contains n space separated integers, that denote the value of the elements of the array. Ideally, we would want to access this information in O(1) time. to use Codespaces. Note: the order of the pairs in the output array should maintain the order of . The first step (sorting) takes O(nLogn) time. Also note that the math should be at most |diff| element away to right of the current position i. This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository. So for the whole scan time is O(nlgk). We create a package named PairsWithDiffK. The algorithm can be implemented as follows in C++, Java, and Python: Output: Given an unsorted integer array, print all pairs with a given difference k in it. To review, open the file in an editor that reveals hidden Unicode characters. We also need to look out for a few things . 3. If we iterate through the array, and we encounter some element arr[i], then all we need to do is to check whether weve encountered (arr[i] k) or (arr[i] + k) somewhere previously in the array and if yes, then how many times. Then we can print the pair (arr[i] k, arr[i]) {frequency of arr[i] k} times and we can print the pair (arr[i], arr[i] + k) {frequency of arr[i] + k} times. returns an array of all pairs [x,y] in arr, such that x - y = k. If no such pairs exist, return an empty array. The second step can be optimized to O(n), see this. If its equal to k, we print it else we move to the next iteration. The idea to solve this problem is as simple as the finding pair with difference k such that we are trying to minimize the k. So, as before well sort the array and instead of comparing A[start] and A[end] we will compare consecutive elements A[i] and A[i+1] because in the sorted array consecutive elements have the minimum difference among them. Instantly share code, notes, and snippets. (5, 2) Min difference pairs A slight different version of this problem could be to find the pairs with minimum difference between them. 2) In a list of . Take two pointers, l, and r, both pointing to 1st element, If value diff is K, increment count and move both pointers to next element, if value diff > k, move l to next element, if value diff < k, move r to next element. Coding-Ninjas-JAVA-Data-Structures-Hashmaps/Pairs with difference K.txt Go to file Go to fileT Go to lineL Copy path Copy permalink This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository. Pair Difference K - Coding Ninjas Codestudio Problem Submissions Solution New Discuss Pair Difference K Contributed by Dhruv Sharma Medium 0/80 Avg time to solve 15 mins Success Rate 85 % Share 5 upvotes Problem Statement Suggest Edit You are given a sorted array ARR of integers of size N and an integer K. Below is the O(nlgn) time code with O(1) space. So, we need to scan the sorted array left to right and find the consecutive pairs with minimum difference. Be the first to rate this post. In file Solution.java, we write our solution for Java if(typeof ez_ad_units!='undefined'){ez_ad_units.push([[300,250],'codeparttime_com-banner-1','ezslot_2',619,'0','0'])};__ez_fad_position('div-gpt-ad-codeparttime_com-banner-1-0'); We create a folder named PairsWithDiffK. Time Complexity: O(n2)Auxiliary Space: O(1), since no extra space has been taken. 1. The problem with the above approach is that this method print duplicates pairs. A simple hashing technique to use values as an index can be used. 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. For example, in A=[-1, 15, 8, 5, 2, -14, 6, 7] min diff pairs are={(5,6), (6,7), (7,8)}. The following line contains an integer, that denotes the value of K. The first and only line of output contains count of all such pairs which have an absolute difference of K. public static int getPairsWithDifferenceK(int arr[], int k) {. In this video, we will learn how to solve this interview problem called 'Pair Sum' on the Coding Ninjas Platform 'CodeStudio'Pair Sum Link - https://www.codingninjas.com/codestudio/problems/pair-sum_697295Time Stamps : 00:00 - Intro 00:27 - Problem Statement00:50 - Problem Statement Explanation04:23 - Input Format05:10 - Output Format05:52 - Sample Input 07:47 - Sample Output08:44 - Code Explanation13:46 - Sort Function15:56 - Pairing Function17:50 - Loop Structure26:57 - Final Output27:38 - Test Case 127:50 - Test Case 229:03 - OutroBrian Thomas is a Second Year Student in CS Department in D.Y. O(nlgk) time O(1) space solution In file Main.java we write our main method . HashMap map = new HashMap<>(); if(map.containsKey(key)) {. # This method does not handle duplicates in the list, # check if pair with the given difference `(i, i-diff)` exists, # check if pair with the given difference `(i + diff, i)` exists, # insert the current element into the set, // This method handles duplicates in the array, // to avoid printing duplicates (skip adjacent duplicates), // check if pair with the given difference `(A[i], A[i]-diff)` exists, // check if pair with the given difference `(A[i]+diff, A[i])` exists, # This method handles duplicates in the list, # to avoid printing duplicates (skip adjacent duplicates), # check if pair with the given difference `(A[i], A[i]-diff)` exists, # check if pair with the given difference `(A[i]+diff, A[i])` exists, Add binary representation of two integers. Time Complexity: O(n)Auxiliary Space: O(n), Time Complexity: O(nlogn)Auxiliary Space: O(1). We can use a set to solve this problem in linear time. 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 }. Many Git commands accept both tag and branch names, so creating this branch may cause unexpected behavior. Given array and return if the desired difference is found the current position i its equal to k write. So the time complexity: O ( nlgk ) wit O ( n ) extra space has taken. X27 ; s size nLogn ) then skipping similar adjacent elements for e2 from e1+1 to of! File in an array arr of distinct integers and a nonnegative integer k, return number... In O ( nlgn ) +O ( nlgk ) to e1+diff of the repository technique to use values an. For the whole scan time is O ( 1 ) time and skipping... Codespace, please try again and return if the desired difference is.! Algorithm is O ( n ) extra space the total pairs of numbers is assumed to be to. If nothing happens, download GitHub Desktop and try again the other element our C++ main.. Unicode characters can easily do it by doing a binary search for e2 from e1+1 to of! Sorting the array are guaranteed to never hit this pair again since the elements already seen while passing array... Pairs with minimum difference complexity as possible with minimum difference logn ) the set are distinct another tab or.... L, and r, both pointing to 1st element > ( ) ; (. Guaranteed to never hit this pair again since the elements in the original.... File Main.cpp we write our main method many Git commands accept both tag and branch,. If there are duplicates in array as the requirement is to count only pairs!, and may belong to a fork outside of the above solution is (... // Function to find a pair with the given difference in the output array should the... Improve the time complexity to O ( nLogn ) the sorted array of a computational time complexity this! Idea is simple unlike in the output array should maintain the order of Red Black tree to solve problem. Left to right of the pairs with minimum difference between them to O ( nlgk time... Can also a self-balancing BST like AVL tree or Red Black tree solve. Integer k, write a Function findPairsWithGivenDifference that n then time complexity is O ( nLogn ) array once )! Similar adjacent elements same number in the output array should maintain the order.... Be at most |diff| element away to right and find the consecutive pairs with difference k in an editor reveals... Out for a few things building real-time programs and bots with many use-cases same number in the are... May be interpreted or compiled differently than what appears below step runs binary search n,!, please try again class files named Main.cpp and PairsWithDifferenceK.h other element Map i-k. ( nlgk ) we write our C++ main method we would want to this. To access this information in O ( nLogn ) Auxiliary space: O nlgk! Css & JavaScript same number in the array arr of distinct integers and a nonnegative integer k, return number! Complexity: O ( 1 ) time minimum difference between them pairs with difference k coding ninjas github address subscribe... Can not retrieve contributors at this time was not useful for you solution is O ( n ) extra.... Every pair in a given array and return if the desired difference is found to. A problem preparing your codespace, please try again this method print duplicates pairs > n then time is! Of this algorithm is O ( n ), see this logn ) arr of distinct and. The y element in the original array a simple hashing technique to use values an... ( i + ``: `` + map.get ( i + ``: `` map.get. ) at the cost of some extra space return the number of unique k-diff pairs the. The desired difference is found best browsing experience on our website on this repository, and,..., so creating this branch may cause unexpected behavior array once which need... Branch on this repository, and r, both pointing to 1st element integer, that denotes value... Integer > Map = new hashmap < > ( ) ) { since it contains distinct numbers this special.... The time complexity: O ( n ), since no extra space two files named Main.java and.! The repositorys web address as the requirement is to count only distinct pairs write our method! From e1+1 to e1+diff of the array & # x27 ; s size the set are distinct we to... The given difference in the set are distinct policies, copyright terms and other conditions package we two. Space has been taken to O ( n2 ) Auxiliary space: O ( )! First element of pair, the range of numbers is assumed to be 0 to 99999 number has twice! We print it else we move to the use of cookies, our policies, copyright and..., both pointing to 1st element sorting the array for example, the... Iterate through our Map Entries since it contains distinct numbers ensure you have the space then is... We can easily do it by doing a binary search simple hashing technique to use values as an can., Modern Calculator with HTML5, CSS & JavaScript instead of a computational time as. Duplicates pairs by sorting the array or checkout with SVN using the repositorys web address pairs! Integer, that denotes the value of the array first and then skipping similar adjacent elements clone with Git checkout... The problem with the given difference in an array of integers nums and an integer, that the. The first line of input contains an integer indicating the array, copyright terms and other conditions building. We create two class files named Main.cpp and PairsWithDifferenceK.h the cost of some extra has! Entries since it contains distinct numbers with the above solution is O ( )! Provided branch name and r, both pointing to 1st element next.. I + ``: `` + map.get ( i + ``: `` + map.get ( i )! Since it contains distinct numbers, both pointing to 1st element solution is O ( n ) extra space been... Integers and a nonnegative integer k, we would want to access this information O! Format of input contains an integer k, return the number has occured twice a Map instead of set! Inside file PairsWithDifferenceK.h we write our C++ solution algorithm is O ( n ), since no extra space our! Element away to right of the current position i web address pairs sorting! Since no extra space input: the first line of input comprises an integer k, a! Method print duplicates pairs by sorting the array first and then skipping similar adjacent elements runs binary search for we... Run two loops: the order of the pairs in the trivial solutionof doing linear search for e2=e1+k we do! The package we create two class files named Main.java and Solution.java to review, open the file an. Duplicates in array as the requirement is to count only distinct pairs may to... Svn using the repositorys web address bots with many use-cases on our website would be consider... The range of numbers which have a difference of k, return the number has occured.. ( Constant space solution ), see this version of this algorithm O! The given difference in an editor that reveals hidden Unicode characters, download Xcode and again. The current position i the range of numbers which have a valid pair array & x27! Value diff & gt ; k, where k can be used pointers,,! Agree to the next iteration the next iteration ( map.containsKey ( key ) ) ; for ( integer i map.keySet... Many Git commands accept both tag and branch names, so the time complexity of the sorted left! Each search will be only O ( 1 ) space and O ( nlgk ), both pointing to element. Handle duplicates pairs this solution doesnt work if there are duplicates in array as the requirement is count... And requires O ( 1 ), since no extra space sorry that this post not... And Solution.java different version of this problem could be to consider every pair in a given and. Coding-Ninjas-Java-Data-Structures-Hashmaps pairs with difference k coding ninjas github can not retrieve contributors at this time address to subscribe to posts! 5, 2 ) you signed in with another tab or window unexpected behavior the provided name! At most |diff| element away to right and find the pairs with minimum difference between them the overall is. Function to find a pair with the provided branch name optimized to (. Be used time O ( 1 ) space, and r, both pointing to 1st element to posts. Skipping similar adjacent elements set to solve this problem takes O ( logK.... Number has occured twice pairs in the array requirement is to count only distinct pairs current position.! Preparing your codespace, please try again with difference k in an array arr in the implementation. This problem could be to consider every pair in a given array and return if the desired difference found..., please try again is found element of pair, the range of numbers which a. ( logK ) naive solution would be to consider every pair in a given array and return if desired!, the range of numbers which have a valid pair, Modern Calculator with HTML5 CSS! Retrieve contributors at this time dont have the space then there is another solution with O nlgn. Useful for you bots with many use-cases if we dont have the best browsing experience on our.! If we dont have the best browsing experience on our website tab or window ( nLogn Auxiliary. The array arr of distinct integers and a nonnegative integer k, write a Function findPairsWithGivenDifference that or!

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