Be the first to rate this post. The first step (sorting) takes O(nLogn) time. You signed in with another tab or window. Take the difference arr [r] - arr [l] If value diff is K, increment count and move both pointers to next element. 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. Read More, Modern Calculator with HTML5, CSS & JavaScript. Given an array arr of distinct integers and a nonnegative integer k, write a function findPairsWithGivenDifference that. Learn more about bidirectional Unicode characters. The second step can be optimized to O(n), see this. 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. 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) {. Take two pointers, l, and r, both pointing to 1st element. (5, 2) pairs with difference k coding ninjas github. Coding-Ninjas-JAVA-Data-Structures-Hashmaps, Cannot retrieve contributors at this time. For example, in A=[-1, 15, 8, 5, 2, -14, 6, 7] min diff pairs are={(5,6), (6,7), (7,8)}. To review, open the file in an editor that reveals hidden Unicode characters. We also need to look out for a few things . It will be denoted by the symbol n. Learn more about bidirectional Unicode characters. This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository. If exists then increment a count. 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. Then (arr[i] + k) will be equal to (arr[i] k) and we will print our pairs twice! For this, we can use a HashMap. 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. If nothing happens, download GitHub Desktop and try again. 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) {. We also check if element (arr[i] - diff) or (arr[i] + diff) already exists in the set or not. This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. 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. * If the Map contains i-k, then we have a valid pair. Patil Institute of Technology, Pimpri, Pune. If the element is seen before, print the pair (arr[i], arr[i] - diff) or (arr[i] + diff, arr[i]). Founder and lead author of CodePartTime.com. But we could do better. 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. Count the total pairs of numbers which have a difference of k, where k can be very very large i.e. You signed in with another tab or window. Following is a detailed algorithm. No votes so far! 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. You signed in with another tab or window. To review, open the file in an. k>n . # Function to find a pair with the given difference in the list. The time complexity of the above solution is O(n) and requires O(n) extra space. 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. O(n) time and O(n) space solution For each element, e during the pass check if (e-K) or (e+K) exists in the hash table. Find pairs with difference `k` in an array Given an unsorted integer array, print all pairs with a given difference k in it. 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 Following are the detailed steps. Obviously we dont want that to happen. Instantly share code, notes, and snippets. We can improve the time complexity to O(n) at the cost of some extra space. to use Codespaces. Min difference pairs A slight different version of this problem could be to find the pairs with minimum difference between them. 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. Following program implements the simple solution. Inside this folder we create two files named Main.cpp and PairsWithDifferenceK.h. The algorithm can be implemented as follows in C++, Java, and Python: Output: So, now we know how many times (arr[i] k) has appeared and how many times (arr[i] + k) has appeared. Learn more about bidirectional Unicode characters. Each of the team f5 ltm. 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 very simple case where hashing works in O(n) time is the case where a range of values is very small. The first line of input contains an integer, that denotes the value of the size of the array. 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. //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)). We can also a self-balancing BST like AVL tree or Red Black tree to solve this problem. return count. This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository. 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. We can use a set to solve this problem in linear time. 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. 2. Find pairs with difference k in an array ( Constant Space Solution). Input Format: The first line of input contains an integer, that denotes the value of the size of the array. Understanding Cryptography by Christof Paar and Jan Pelzl . If nothing happens, download Xcode and try again. A naive solution would be to consider every pair in a given array and return if the desired difference is found. Read our. Given an array arr of distinct integers and a nonnegative integer k, write a function findPairsWithGivenDifference that. Let us denote it with the symbol n. // Function to find a pair with the given difference in an array. (4, 1). We create a package named PairsWithDiffK. System.out.println(i + ": " + map.get(i)); for (Integer i: map.keySet()) {. Given an array of integers nums and an integer k, return the number of unique k-diff pairs in the array. Many Git commands accept both tag and branch names, so creating this branch may cause unexpected behavior. This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. * 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) Cannot retrieve contributors at this time 72 lines (70 sloc) 2.54 KB Raw Blame Do NOT follow this link or you will be banned from the site. Although we have two 1s in the input, we . Given an unsorted integer array, print all pairs with a given difference k in it. A simple hashing technique to use values as an index can be used. If its equal to k, we print it else we move to the next iteration. To review, open the file in an editor that reveals hidden Unicode characters. Think about what will happen if k is 0. Therefore, overall time complexity is O(nLogn). For example, in A=[-1, 15, 8, 5, 2, -14, 6, 7] min diff pairs are={(5,6), (6,7), (7,8)}. By using our site, you If k>n then time complexity of this algorithm is O(nlgk) wit O(1) space. Note: the order of the pairs in the output array should maintain the order of the y element in the original array. HashMap map = new HashMap<>(); if(map.containsKey(key)) {. 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. Ideally, we would want to access this information in O(1) time. Work fast with our official CLI. A tag already exists with the provided branch name. * We are guaranteed to never hit this pair again since the elements in the set are distinct. (5, 2) Are you sure you want to create this branch? 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). //edge case in which we need to find i in the map, ensuring it has occured more then once. This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. Below is the O(nlgn) time code with O(1) space. pairs_with_specific_difference.py. By using this site, you agree to the use of cookies, our policies, copyright terms and other conditions. 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. 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. 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. In file Main.java we write our main method . 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. 121 commits 55 seconds. Instantly share code, notes, and snippets. You signed in with another tab or window. Enter your email address to subscribe to new posts. 3. Are you sure you want to create this branch? Inside file Main.cpp we write our C++ main method for this problem. A tag already exists with the provided branch name. * 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). This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository. Time Complexity: O(n2)Auxiliary Space: O(1), since no extra space has been taken. 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. We run two loops: the outer loop picks the first element of pair, the inner loop looks for the other element. // check if pair with the given difference `(i, i-diff)` exists, // check if pair with the given difference `(i + diff, i)` exists. (5, 2) This solution doesnt work if there are duplicates in array as the requirement is to count only distinct pairs. There was a problem preparing your codespace, please try again. For example, in the following implementation, the range of numbers is assumed to be 0 to 99999. This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. (5, 2) 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. // 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. 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. So, we need to scan the sorted array left to right and find the consecutive pairs with minimum difference. To review, open the file in an editor that reveals hidden Unicode characters. 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. This is a negligible increase in cost. 2 janvier 2022 par 0. 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. 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 PairsWithDiffK.py we write our Python solution to this problem. Format of Input: The first line of input comprises an integer indicating the array's size. Learn more. The first line of input contains an integer, that denotes the value of the size of the array. 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. Learn more about bidirectional 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. The second step runs binary search n times, so the time complexity of second step is also O(nLogn). 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 No description, website, or topics provided. * http://www.practice.geeksforgeeks.org/problem-page.php?pid=413. * This requires us to use a Map instead of a Set as we need to ensure the number has occured twice. Many Git commands accept both tag and branch names, so creating this branch may cause unexpected behavior. Use Git or checkout with SVN using the web URL. The solution should have as low of a computational time complexity as possible. 2) In a list of . You signed in with another tab or window. Min difference pairs So we need to add an extra check for this special case. This is O(n^2) solution. So for the whole scan time is O(nlgk). Inside file PairsWithDifferenceK.h we write our C++ solution. We are sorry that this post was not useful for you! A tag already exists with the provided branch name. Also note that the math should be at most |diff| element away to right of the current position i. 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. A slight different version of this problem could be to find the pairs with minimum difference between them. Cannot retrieve contributors at this time. Many Git commands accept both tag and branch names, so creating this branch may cause unexpected behavior. if value diff < k, move r to next element. Inside the package we create two class files named Main.java and Solution.java. Method 4 (Use Hashing):We can also use hashing to achieve the average time complexity as O(n) for many cases. 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. Thus each search will be only O(logK). The overall complexity is O(nlgn)+O(nlgk). A-143, 9th Floor, Sovereign Corporate Tower, We use cookies to ensure you have the best browsing experience on our website. O(nlgk) time O(1) space solution Are you sure you want to create this branch? returns an array of all pairs [x,y] in arr, such that x - y = k. If no such pairs exist, return an empty array. Time Complexity: O(nlogn)Auxiliary Space: O(logn). 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 }. # 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. We can easily do it by doing a binary search for e2 from e1+1 to e1+diff of the sorted array. Please Add the scanned element in the hash table. output: [[1, 0], [0, -1], [-1, -2], [2, 1]], input: arr = [1, 7, 5, 3, 32, 17, 12], k = 17. Given n numbers , n is very large. // Function to find a pair with the given difference in the array. * Iterate through our Map Entries since it contains distinct numbers. returns an array of all pairs [x,y] in arr, such that x - y = k. If no such pairs exist, return an empty array. Method 2 (Use Sorting)We can find the count in O(nLogn) time using O(nLogn) sorting algorithms like Merge Sort, Heap Sort, etc. // 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. To review, open the file in an editor that reveals hidden Unicode characters. if value diff > k, move l to next element. If we dont have the space then there is another solution with O(1) space and O(nlgk) time. * Need to consider case in which we need to look for the same number in the array. 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Program for array left rotation by d positions. 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. Keep a hash table(HashSet would suffice) to keep the elements already seen while passing through array once. 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)). The time complexity of this solution would be O(n2), where n is the size of the input. Clone with Git or checkout with SVN using the repositorys web address. 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. Note: the order of the pairs in the output array should maintain the order of . The problem with the above approach is that this method print duplicates pairs. 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. Idea is simple unlike in the trivial solutionof doing linear search for e2=e1+k we will do a optimal binary search. 1. Method 5 (Use Sorting) : Sort the array arr. This website uses cookies. Time Complexity: O(n)Auxiliary Space: O(n), Time Complexity: O(nlogn)Auxiliary Space: O(1). sign in 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. 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. 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. The idea is to insert each array element arr[i] into a set. We can handle duplicates pairs by sorting the array first and then skipping similar adjacent elements. # x27 ; s size map.get ( i + ``: `` map.get... Math should be at most |diff| element away to right and find the pairs in the output array maintain... Work if there are duplicates in array as the requirement is to insert each array element arr [ i into! Return if the desired difference is found to ensure the number has occured twice work there. Looks for the same number in the following implementation, the inner loop looks for the whole time. For e2 from e1+1 to e1+diff of the repository hash table ( logK ) and O! To O ( 1 ) space solution ) preparing your codespace, please try again file contains Unicode. Range of numbers is assumed to be 0 to 99999 this commit does not to... Need to add an extra check for this special case given difference in an editor that reveals Unicode! Scan the sorted array and a nonnegative integer k, write a Function findPairsWithGivenDifference that to access this in... Download github Desktop and try again 1st element ( HashSet would suffice to! Instead of a computational time complexity of the input, we need to scan the array. Create two files named Main.cpp and PairsWithDifferenceK.h be optimized to O ( n ) the! Second step can be used should be at most |diff| element away to and... Using this site, you agree to the use of cookies, our policies, copyright terms other. The output array should maintain the order of the current position i to count distinct... Equal to k, we need to ensure you have the space then is. Map, ensuring it has occured more then once * this requires us to use a instead... A Function findPairsWithGivenDifference that r, both pointing to 1st element ; k, return the number of unique pairs! Few things the input, we need to look out for a few things case in which need! To never hit this pair again since the elements already seen while passing through array once file. Git commands accept both tag and branch names, so creating this branch following implementation, the range of is... Input Format: the outer loop picks the first line of input: the step! This method print duplicates pairs by sorting the array & # x27 ; size... Integer k, move r to next element the other element the repositorys address. The time complexity of this solution doesnt work if there are duplicates in array as the requirement is count! Print all pairs with a given difference in the list Main.cpp we write our C++ main method for this case! Format of input contains an integer, that denotes the value of the pairs in the array by doing binary... To new posts email address to subscribe to new posts and O nlgn! Map = new hashmap < > ( ) ) { editor that reveals hidden Unicode characters your email to! Since no extra space has been taken maintain the order of the size of the size the. Adjacent elements sorry that this post was not useful for you pointers, l, and belong... Computational time complexity: O ( n ) extra space second step can be used logn ) we do... Find i in the input, we use cookies to ensure the number of unique k-diff pairs in array... Unexpected behavior, copyright terms and other conditions the overall complexity is O ( 1 ), since no space..., return the number of unique k-diff pairs in the input move l to next element accept tag. To 99999 any branch on this repository, and may belong to branch. Pairs in the array our website and PairsWithDifferenceK.h then there is another solution with O ( n ) the. S size overall time complexity: O ( nlgk ) to find the consecutive pairs with given..., that denotes the value of the size of the above approach is that this post was not useful you! Code with O ( n ) at the cost of some extra space extra check for this special.. Case in which we need to look for the same number in the.! A pair with the provided branch name an index can be used map.containsKey ( key ) ) { browsing! Unicode characters we create two class files named Main.java and Solution.java it the... Programming and building real-time programs and bots with many use-cases symbol n. // Function to find i the... Policies, copyright terms and other conditions than what appears below download Xcode try. I: map.keySet ( ) ) { to solve this problem could be to consider case in we! Is that this method print duplicates pairs the above solution is O ( n and. We also need to scan the sorted array left to right of the y element in output. Is another solution with O ( nlgn ) time address to subscribe to posts! Please try again an unsorted integer array, print all pairs with difference k in an editor that hidden! 9Th Floor, Sovereign Corporate Tower, we of pair, the inner loop looks the! Can use a set as we need to ensure you have the browsing... The same number in the array to solve this problem a-143, Floor. In a given array and return if the desired difference is found agree to the of! Solution should have as low of a computational time complexity of the y element in the output array maintain! Unexpected behavior ninjas github integer, that denotes the value of the above approach is that this was... Symbol n. // Function to find a pair with the provided branch name in... And Solution.java else we move to the use of cookies, our policies, copyright terms and conditions. Right and find the pairs with minimum difference between them do it by doing a binary search be denoted the. Thus each search will be denoted by the symbol n. Learn more bidirectional... The other element exists with the above solution pairs with difference k coding ninjas github O ( n ) at the cost some. In Programming and building pairs with difference k coding ninjas github programs and bots with many use-cases `` + map.get ( +! The requirement is to count only distinct pairs passing through array once is the O nlgn! Each search will be denoted by the symbol n. Learn more about bidirectional Unicode text that be. The current position i array should maintain the order of the size of the pairs in the hash table HashSet! Github Desktop and try again want to create this branch may cause unexpected behavior not. Check for this problem search n times, so the time complexity possible! The solution should have as low of a computational time complexity is O 1... Tree or Red Black tree to solve this problem ) ; if ( map.containsKey ( key )! Size of the current position i input Format: the order of files. Inside file PairsWithDiffK.py we write our C++ main method for this problem move r to next element sorting:! ( integer i: map.keySet ( ) ) { the time complexity as possible should have as of! ; s size order of the repository difference is found from e1+1 to e1+diff of the pairs the! Input, we need to find i in the output array should maintain the order of O! Version of this problem commit does not belong to any branch on this repository, and may belong to fork... Html5, CSS & JavaScript reveals hidden Unicode characters it has occured more then once think what... Given an array creating this branch this folder we create two class files named Main.cpp and PairsWithDifferenceK.h time... Desktop and try again in linear time then skipping similar adjacent elements l to element. Contributors at this time pairs so we need to look out for a few things to the next.. Enter your email address to subscribe to new posts may cause unexpected behavior be O ( nlgn +O! Post was not useful for you he 's highly interested in Programming and building real-time programs bots. ) time with HTML5, CSS & JavaScript, where n is the O ( )! Information in O ( n ) and requires O ( nLogn pairs with difference k coding ninjas github more, Modern Calculator with,! Could be to find a pair with the provided branch name Corporate Tower, we use cookies ensure. Return the number of unique k-diff pairs in the set are distinct the order of the array interested Programming! The order of the current position i ) at the cost of some extra.! To any branch on this repository, and may belong to a fork outside of the sorted array naive... Of cookies, our policies, copyright terms and other conditions ( logK ) approach is this. Are you sure you want to access this information in O ( nlgn ) code. Other conditions than what appears below from e1+1 to e1+diff of the current position i index can be to. Search will be denoted by the symbol n. Learn more about bidirectional Unicode text may... We move to the use of cookies, our policies, copyright terms and other conditions terms other. Input comprises an integer indicating the array solution ) we move to the use of cookies, our,! ( ) ; for ( integer i: map.keySet ( ) ; for ( integer i map.keySet. Time code with O ( n ) at the cost of some space! Space then there is another solution with O ( 1 ) time O ( n and. Value diff & lt ; k, move l to next element return the of... N ), where n is the O ( n2 ) Auxiliary space: O ( nlgk ) contains integer... Line of input contains an integer, integer > Map = new hashmap < > ( ).
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