This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. Inside this folder we create two files named Main.cpp and PairsWithDifferenceK.h. * 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) Enter your email address to subscribe to new posts. Following are the detailed steps. This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. pairs with difference k coding ninjas github. returns an array of all pairs [x,y] in arr, such that x - y = k. If no such pairs exist, return an empty 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. (5, 2) Given an array of integers nums and an integer k, return the number of unique k-diff pairs in the array. Learn more about bidirectional Unicode characters. Learn more about bidirectional Unicode characters. 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) {. The overall complexity is O(nlgn)+O(nlgk). 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. * This requires us to use a Map instead of a Set as we need to ensure the number has occured twice. We run two loops: the outer loop picks the first element of pair, the inner loop looks for the other element. // This method does not handle duplicates in the array, // check if pair with the given difference `(arr[i], arr[i]-diff)` exists, // check if pair with the given difference `(arr[i]+diff, arr[i])` exists, // insert the current element into the set. A tag already exists with the provided branch name. To review, open the file in an editor that reveals hidden Unicode characters. 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. 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 Read our. We also check if element (arr[i] - diff) or (arr[i] + diff) already exists in the set or not. HashMap approach to determine the number of Distinct Pairs who's difference equals an input k. Clone with Git or checkout with SVN using the repositorys web address. The algorithm can be implemented as follows in C++, Java, and Python: Output: Patil Institute of Technology, Pimpri, Pune. Following program implements the simple solution. O(n) time and O(n) space solution Given an array arr of distinct integers and a nonnegative integer k, write a function findPairsWithGivenDifference that. Ideally, we would want to access this information in O(1) time. The solution should have as low of a computational time complexity as possible. If exists then increment a count. Keep a hash table(HashSet would suffice) to keep the elements already seen while passing through array once. A slight different version of this problem could be to find the pairs with minimum difference between them. System.out.println(i + ": " + map.get(i)); for (Integer i: map.keySet()) {. This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository. 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. (5, 2) Count the total pairs of numbers which have a difference of k, where k can be very very large i.e. Take the difference arr [r] - arr [l] If value diff is K, increment count and move both pointers to next element. Min difference pairs Do NOT follow this link or you will be banned from the site. Below is the O(nlgn) time code with O(1) space. 3. * Iterate through our Map Entries since it contains distinct numbers. // Function to find a pair with the given difference in the array. (5, 2) You signed in with another tab or window. 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. Cannot retrieve contributors at this time 72 lines (70 sloc) 2.54 KB Raw Blame // check if pair with the given difference `(i, i-diff)` exists, // check if pair with the given difference `(i + diff, i)` exists. Many Git commands accept both tag and branch names, so creating this branch may cause unexpected behavior. Format of Input: The first line of input comprises an integer indicating the array's size. Note: the order of the pairs in the output array should maintain the order of the y element in the original array. This solution doesnt work if there are duplicates in array as the requirement is to count only distinct pairs. A naive solution would be to consider every pair in a given array and return if the desired difference is found. Are you sure you want to create this branch? Note: the order of the pairs in the output array should maintain the order of . Pairs with difference K - Coding Ninjas Codestudio Topic list MEDIUM 13 upvotes Arrays (Covered in this problem) Solve problems & track your progress Become Sensei in DSA topics Open the topic and solve more problems associated with it to improve your skills Check out the skill meter for every topic Think about what will happen if k is 0. Are you sure you want to create this branch? If the element is seen before, print the pair (arr[i], arr[i] - diff) or (arr[i] + diff, arr[i]). Time Complexity: O(n2)Auxiliary Space: O(1), since no extra space has been taken. Thus each search will be only O(logK). Are you sure you want to create this branch? Inside file Main.cpp we write our C++ main method for this problem. Be the first to rate this post. Find pairs with difference `k` in an array Given an unsorted integer array, print all pairs with a given difference k in it. (5, 2) 2 janvier 2022 par 0. Method 2 (Use Sorting)We can find the count in O(nLogn) time using O(nLogn) sorting algorithms like Merge Sort, Heap Sort, etc. 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 }. Therefore, overall time complexity is O(nLogn). Given an unsorted integer array, print all pairs with a given difference k in it. We can handle duplicates pairs by sorting the array first and then skipping similar adjacent elements. 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. No description, website, or topics provided. This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository. 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. # Function to find a pair with the given difference in the list. The first line of input contains an integer, that denotes the value of the size of the array. A tag already exists with the provided branch name. The time complexity of the above solution is O(n) and requires O(n) extra space. Method 5 (Use Sorting) : Sort the array arr. We create a package named PairsWithDiffK. For each element, e during the pass check if (e-K) or (e+K) exists in the hash table. The problem with the above approach is that this method print duplicates pairs. HashMap map = new HashMap<>(); if(map.containsKey(key)) {. if value diff < k, move r to next element. The idea is to insert each array element arr[i] into a set. Inside file PairsWithDiffK.py we write our Python solution to this problem. 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. // Function to find a pair with the given difference in an array. * If the Map contains i-k, then we have a valid pair. Clone with Git or checkout with SVN using the repositorys web address. 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. Learn more. To review, open the file in an. 1. 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. 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. To review, open the file in an editor that reveals hidden Unicode characters. 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. Instantly share code, notes, and snippets. 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. We also need to look out for a few things . So, now we know how many times (arr[i] k) has appeared and how many times (arr[i] + k) has appeared. Each of the team f5 ltm. The first step (sorting) takes O(nLogn) time. sign in 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. It will be denoted by the symbol n. // 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. For example, in A=[-1, 15, 8, 5, 2, -14, 6, 7] min diff pairs are={(5,6), (6,7), (7,8)}. 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. This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. Given an array arr of distinct integers and a nonnegative integer k, write a function findPairsWithGivenDifference that. returns an array of all pairs [x,y] in arr, such that x - y = k. If no such pairs exist, return an empty array. Understanding Cryptography by Christof Paar and Jan Pelzl . We can use a set to solve this problem in linear time. So, we need to scan the sorted array left to right and find the consecutive pairs with minimum difference. pairs_with_specific_difference.py. //edge case in which we need to find i in the map, ensuring it has occured more then once. return count. Instantly share code, notes, and snippets. So for the whole scan time is O(nlgk). 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. Inside file PairsWithDifferenceK.h we write our C++ solution. Time Complexity: O(nlogn)Auxiliary Space: O(logn). A tag already exists with the provided branch name. Let us denote it with the symbol n. By using this site, you agree to the use of cookies, our policies, copyright terms and other conditions. Add the scanned element in the hash table. 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. Use Git or checkout with SVN using the web URL. 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. 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. if value diff > k, move l to next element. Many Git commands accept both tag and branch names, so creating this branch may cause unexpected behavior. If nothing happens, download Xcode and try again. If its equal to k, we print it else we move to the next iteration. If k>n then time complexity of this algorithm is O(nlgk) wit O(1) space. The second step can be optimized to O(n), see this. In file Main.java we write our main method . 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. 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. This is a negligible increase in cost. To review, open the file in an editor that reveals hidden Unicode characters. We can improve the time complexity to O(n) at the cost of some extra space. 2) In a list of . Many Git commands accept both tag and branch names, so creating this branch may cause unexpected behavior. to use Codespaces. 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) {. You signed in with another tab or window. A simple hashing technique to use values as an index can be used. There was a problem preparing your codespace, please try again. Obviously we dont want that to happen. output: [[1, 0], [0, -1], [-1, -2], [2, 1]], input: arr = [1, 7, 5, 3, 32, 17, 12], k = 17. If we dont have the space then there is another solution with O(1) space and O(nlgk) time. The first line of input contains an integer, that denotes the value of the size of the array. 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. Take two pointers, l, and r, both pointing to 1st element. 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. Idea is simple unlike in the trivial solutionof doing linear search for e2=e1+k we will do a optimal binary search. For this, we can use a HashMap. You signed in with another tab or window. This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. 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. O(nlgk) time O(1) space solution Time Complexity: O(n)Auxiliary Space: O(n), Time Complexity: O(nlogn)Auxiliary Space: O(1). We are sorry that this post was not useful for you! 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. We can also a self-balancing BST like AVL tree or Red Black tree to solve this problem. Work fast with our official CLI. Given n numbers , n is very large. Then (arr[i] + k) will be equal to (arr[i] k) and we will print our pairs twice! //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)). No votes so far! # 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. This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository. The time complexity of this solution would be O(n2), where n is the size of the input. (4, 1). To review, open the file in an editor that reveals hidden Unicode characters. Cannot retrieve contributors at this time. You signed in with another tab or window. * http://www.practice.geeksforgeeks.org/problem-page.php?pid=413. For example, in the following implementation, the range of numbers is assumed to be 0 to 99999. Program for array left rotation by d positions. Inside the package we create two class files named Main.java and Solution.java. 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. You signed in with another tab or window. But we could do better. Learn more about bidirectional Unicode characters. 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. * Need to consider case in which we need to look for the same number in the array. Read More, Modern Calculator with HTML5, CSS & JavaScript. 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 So we need to add an extra check for this special case. By using our site, you 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. Min difference pairs A slight different version of this problem could be to find the pairs with minimum difference between them. 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). 2. Input Format: The first line of input contains an integer, that denotes the value of the size of the array. 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. This is O(n^2) solution. 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. 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. 121 commits 55 seconds. (5, 2) Founder and lead author of CodePartTime.com. 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)). Method 4 (Use Hashing):We can also use hashing to achieve the average time complexity as O(n) for many cases. If nothing happens, download GitHub Desktop and try again. * We are guaranteed to never hit this pair again since the elements in the set are distinct. For example, in A=[-1, 15, 8, 5, 2, -14, 6, 7] min diff pairs are={(5,6), (6,7), (7,8)}. Also note that the math should be at most |diff| element away to right of the current position i. Coding-Ninjas-JAVA-Data-Structures-Hashmaps, Cannot retrieve contributors at this time. This website uses cookies. 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No extra space has been taken technique to use a Map instead of set... Whole scan time is O ( n ) extra space not follow this link or will. Download GitHub Desktop and try again we would want to create this branch may cause unexpected behavior Modern. Difference k in it 9th Floor, Sovereign Corporate Tower pairs with difference k coding ninjas github we would want to access information. Calculator with HTML5, CSS & JavaScript Map Entries since it contains distinct numbers to k move... # Function to find a pair with the given difference in the are... More, Modern Calculator with HTML5, CSS & JavaScript // Function to find the pairs with minimum.. This information in O ( logK ) Desktop and try again 2 janvier 2022 par 0 dont have space., since no extra space belong to any branch on this repository, and belong. To 1st element time is O ( 1 ) space, so this. Of pair, the range of numbers is assumed to be 0 to 99999 5, 2 ) and. What appears below, in the Map, ensuring it has occured more then once Main.java Solution.java... Each element, e during the pass check if ( map.containsKey ( )! Named Main.java and Solution.java array once arr of distinct integers and a nonnegative integer,... Requires us to use values as an index can be optimized to O ( 1 ) space do optimal. Logk ) ( integer i: map.keySet ( ) ) ; for ( i! Maintain the order of should have as low of a set or you will banned! Already exists with the provided branch name r, both pointing to 1st.... A problem preparing your codespace, please try again from e1+1 to e1+diff of the input duplicates array... Lt ; k, write a Function findPairsWithGivenDifference that a hash table ( HashSet suffice. An index can be optimized to O ( nlgk ) look for the same number in array! The following implementation, the range of numbers is assumed to be 0 to 99999 print it we. In array as the requirement is to count only distinct pairs the y element in the Map contains i-k then! To solve this problem a computational time complexity of this problem in linear time the! Number in the array first and then skipping similar adjacent elements browsing experience on our website idea is simple in. Have as low of a set then there is another solution with O ( )... We will do a optimal binary search for e2=e1+k we will do a optimal binary search experience on our.... In which we need to look out for a few things in an editor that reveals Unicode... Or you will be banned from the site to ensure you have space. Elements in the set are distinct the site to keep the elements in the.... Calculator with HTML5, CSS & JavaScript or checkout with SVN using repositorys! The pairs in the set are distinct ; s size review, open the file an... Use cookies to ensure the number has occured more then once another solution with O n2. L to next element and then skipping similar adjacent elements ) you signed in with another tab or.... Branch on this repository, and may belong to a fork outside of the size of y. The next iteration HTML5, CSS & JavaScript 2022 par 0 cookies to ensure the has... Be only O ( n ) at the cost of some extra space in O ( ). Html5, CSS & JavaScript ) extra space diff & lt ; k, a... Reveals hidden Unicode characters, in the output array should maintain the order of the repository >. Move to the next iteration ) +O ( nlgk ) time main for. A nonnegative integer k, move r to next element the requirement is to count distinct... Or window accept both tag and branch names, so creating this branch may unexpected! Requires O ( nLogn ) Auxiliary space: O ( 1 ) time branch,. Be optimized to O ( nLogn ) a slight different version of this problem the Map ensuring. And O ( logn ) requires O ( nLogn ) Auxiliary space: O ( nlgk ) time pointing 1st! +O ( nlgk ) problem in linear time, that denotes the value of the above is., open the file in an editor that reveals hidden Unicode characters to! Index can be used programs and bots with many use-cases simple unlike in the following implementation, range. To 1st element sorting the array pairs with difference k coding ninjas github and then skipping similar adjacent elements unexpected behavior two named. As the requirement is to insert each array element arr [ i ] into a to... Set as we need to look for the same number in the hash.. Logk ) was a problem preparing your codespace, please try again to consider every pair in a given and. Inside file PairsWithDiffK.py we write our C++ main method for this problem in linear time e-K ) (. Passing through array once do not follow this link or you will be only (... During the pass check if ( e-K ) or ( e+K ) exists in the following implementation, the of. Equal to k, move l to next element format of input contains an indicating... Integers and a nonnegative integer k, move r to next element both. The site signed in with another tab or window sorting the array to review, open the in! Run two loops: the order of to ensure you have the best browsing experience on our.. Implementation, the inner loop looks for the other element Desktop and try again ) wit (. * this requires us to use a set the package we create two class files named Main.java Solution.java... ) { work if there are duplicates in array as the requirement is to insert each array arr. Names, so creating this branch unlike in the trivial solutionof doing linear search for e2=e1+k will... Pass check if ( e-K ) or ( e+K ) exists pairs with difference k coding ninjas github the output array maintain! Is assumed to be 0 to 99999 right and find the pairs in the hash table HashSet! Index can be used similar adjacent elements difference is found in the original array array... Time is O ( nLogn ) Auxiliary space: O ( 1 ) space with. Both tag and branch names, so creating this branch may cause unexpected behavior, download Desktop! To use values as an index can be used Entries since it contains numbers... Therefore, overall time complexity: O ( nlgk ) 0 to 99999 this file contains bidirectional text! Technique to use values as an index can be optimized to O ( logK.!, l, and may belong to any branch on this repository, and may to. We use cookies to ensure you have the space then there is another solution with (... Provided branch name use Git or checkout with SVN using the web URL solution O. This requires us to use values as an index can be optimized to O ( nlgn ) +O ( ). Return pairs with difference k coding ninjas github the desired difference is found the size of the array first and then skipping adjacent! Is assumed to be 0 to 99999 SVN using the web URL browsing experience on our website input. 2022 par 0 array & # x27 ; s size not useful for you hashing technique to use a.. Main.Java and Solution.java pairs with minimum difference between them pair again since pairs with difference k coding ninjas github elements the... Loop looks for the whole scan time is O ( 1 ) space and O ( nlgn time... Problem could be to find the consecutive pairs with minimum difference between them this... Array, print all pairs with minimum difference between them // Function to find i in the hash table low... Arr [ i ] into a set pairs with minimum difference in editor! Integer indicating the array integer array, print all pairs with minimum difference between them optimized to O ( )... Our website be interpreted or compiled differently than what appears below & lt ; k write... Git commands accept both tag and branch names, so creating this branch may cause unexpected behavior the contains... Will do a optimal binary search for e2=e1+k we will do a optimal binary search difference... Hash table and building real-time programs and bots with many use-cases the same number in the Map contains i-k then! ) Auxiliary space: O ( nlgn ) time extra space element, e during pass. And r, both pointing to 1st element do a optimal binary search for e2=e1+k we do! Scan the sorted array requires O ( n ) extra space has been taken bots with many.. Inner loop looks for the whole scan time is O ( nlgk ) tag already exists the! 0 to 99999 find i in the array we need pairs with difference k coding ninjas github look out for a few things we it. Space and O ( 1 ) space and O ( n ) and requires O ( n2,... ( nlgn ) time + map.get ( i + ``: `` + map.get i... Idea is to count only distinct pairs to this problem as the requirement is to insert each array arr., print all pairs with minimum difference between them tag already exists with the given difference in the list >! Not useful for you you signed in with another tab or window set to solve problem! Checkout with SVN using the repositorys web address doesnt work if there are duplicates array! [ i ] into a set and return if the desired difference is.!
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