So for the whole scan time is O(nlgk). Use Git or checkout with SVN using the web URL. Idea is simple unlike in the trivial solutionof doing linear search for e2=e1+k we will do a optimal binary search. Learn more about bidirectional Unicode characters. 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. //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)). Keep a hash table(HashSet would suffice) to keep the elements already seen while passing through array once. Many Git commands accept both tag and branch names, so creating this branch may cause unexpected behavior. If its equal to k, we print it else we move to the next iteration. 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. To review, open the file in an editor that reveals hidden Unicode characters. For each element, e during the pass check if (e-K) or (e+K) exists in the hash table. Each of the team f5 ltm. return count. Are you sure you want to create this branch? The second step runs binary search n times, so the time complexity of second step is also O(nLogn). 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 3. Read our. // 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. Do NOT follow this link or you will be banned from the site. Cannot retrieve contributors at this time 72 lines (70 sloc) 2.54 KB Raw Blame sign in Instantly share code, notes, and snippets. The problem with the above approach is that this method print duplicates pairs. The idea is to insert each array element arr[i] into a set. To review, open the file in an editor that reveals hidden Unicode characters. Read More, Modern Calculator with HTML5, CSS & JavaScript. The first step (sorting) takes O(nLogn) time. * http://www.practice.geeksforgeeks.org/problem-page.php?pid=413. 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. Inside this folder we create two files named Main.cpp and PairsWithDifferenceK.h. (5, 2) Given n numbers , n is very large. 2. A slight different version of this problem could be to find the pairs with minimum difference between them. If the element is seen before, print the pair (arr[i], arr[i] - diff) or (arr[i] + diff, arr[i]). There was a problem preparing your codespace, please try again. You signed in with another tab or window. Are you sure you want to create this branch? HashMap map = new HashMap<>(); if(map.containsKey(key)) {. Find pairs with difference k in an array ( Constant Space Solution). We run two loops: the outer loop picks the first element of pair, the inner loop looks for the other element. Following are the detailed steps. to use Codespaces. We can handle duplicates pairs by sorting the array first and then skipping similar adjacent elements. It will be denoted by the symbol n. 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. We can use a set to solve this problem in linear time. Ideally, we would want to access this information in O(1) time. Understanding Cryptography by Christof Paar and Jan Pelzl . Are you sure you want to create this branch? 121 commits 55 seconds. Patil Institute of Technology, Pimpri, Pune. 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. Add the scanned element in the hash table. You signed in with another tab or window. For this, we can use a HashMap. Cannot retrieve contributors at this 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. Inside file PairsWithDifferenceK.h we write our C++ solution. Work fast with our official CLI. 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. The overall complexity is O(nlgn)+O(nlgk). A tag already exists with the provided branch name. # 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. Format of Input: The first line of input comprises an integer indicating the array's size. Founder and lead author of CodePartTime.com. If exists then increment a count. Learn more. So we need to add an extra check for this special case. if value diff < k, move r to next element. (5, 2) Time Complexity: O(nlogn)Auxiliary Space: O(logn). No description, website, or topics provided. 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. returns an array of all pairs [x,y] in arr, such that x - y = k. If no such pairs exist, return an empty array. This is a negligible increase in cost. We also need to look out for a few things . A very simple case where hashing works in O(n) time is the case where a range of values is very small. k>n . 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 output: [[1, 0], [0, -1], [-1, -2], [2, 1]], input: arr = [1, 7, 5, 3, 32, 17, 12], k = 17. A simple hashing technique to use values as an index can be used. Clone with Git or checkout with SVN using the repositorys web address. * 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). Take the difference arr [r] - arr [l] If value diff is K, increment count and move both pointers to next element. O(n) time and O(n) space solution 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. Learn more about bidirectional Unicode characters. The first line of input contains an integer, that denotes the value of the size of the array. Thus each search will be only O(logK). Inside file PairsWithDiffK.py we write our Python solution to this problem. 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. 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. 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. Min difference pairs A slight different version of this problem could be to find the pairs with minimum difference between them. The solution should have as low of a computational time complexity as possible. This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. By using our site, you (4, 1). 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 second step can be optimized to O(n), see this. Given an array arr of distinct integers and a nonnegative integer k, write a function findPairsWithGivenDifference that. This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository. You signed in with another tab or window. 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 }. We can improve the time complexity to O(n) at the cost of some extra space. You signed in with another tab or window. For example, in A=[-1, 15, 8, 5, 2, -14, 6, 7] min diff pairs are={(5,6), (6,7), (7,8)}. 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. Therefore, overall time complexity is O(nLogn). By using this site, you agree to the use of cookies, our policies, copyright terms and other conditions. 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. You signed in with another tab or window. Think about what will happen if k is 0. This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. If k>n then time complexity of this algorithm is O(nlgk) wit O(1) space. pairs_with_specific_difference.py. 1. Inside file Main.cpp we write our C++ main method for this problem. 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. We also check if element (arr[i] - diff) or (arr[i] + diff) already exists in the set or not. But we could do better. * 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) # Function to find a pair with the given difference in the list. Min difference pairs //edge case in which we need to find i in the map, ensuring it has occured more then once. 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. Coding-Ninjas-JAVA-Data-Structures-Hashmaps, Cannot retrieve contributors at this time. 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. This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository. * Iterate through our Map Entries since it contains distinct numbers. Learn more about bidirectional Unicode characters. This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. * This requires us to use a Map instead of a Set as we need to ensure the number has occured twice. (5, 2) 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. Inside the package we create two class files named Main.java and Solution.java. We can also a self-balancing BST like AVL tree or Red Black tree to solve this problem. Following program implements the simple solution. Note: the order of the pairs in the output array should maintain the order of the y element in the original array. We are sorry that this post was not useful for you! Take two pointers, l, and r, both pointing to 1st element. 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. Following is a detailed algorithm. 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. In file Main.java we write our main method . 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) {. Learn more about bidirectional Unicode characters. 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. The algorithm can be implemented as follows in C++, Java, and Python: Output: No votes so far! Let us denote it with the symbol n. 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 of the above solution is also O(nLogn) as search and delete operations take O(Logn) time for a self-balancing binary search tree. 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. Although we have two 1s in the input, we . 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. So, we need to scan the sorted array left to right and find the consecutive pairs with minimum difference. Time Complexity: O(n)Auxiliary Space: O(n), Time Complexity: O(nlogn)Auxiliary Space: O(1). 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 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. Method 4 (Use Hashing):We can also use hashing to achieve the average time complexity as O(n) for many cases. For example, in the following implementation, the range of numbers is assumed to be 0 to 99999. 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If nothing happens, download Xcode and try again. Instantly share code, notes, and snippets. The time complexity of this solution would be O(n2), where n is the size of the input. 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. Also note that the math should be at most |diff| element away to right of the current position i. Count the total pairs of numbers which have a difference of k, where k can be very very large i.e. To review, open the file in an editor that reveals hidden 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) {. A tag already exists with the provided branch name. 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. Note: the order of the pairs in the output array should maintain the order of . (5, 2) O(nlgk) time O(1) space solution This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. Many Git commands accept both tag and branch names, so creating this branch may cause unexpected behavior. Two files named Main.java and Solution.java our policies, copyright terms and conditions! Map = new hashmap < integer, integer > map = new hashmap < integer, integer > =! ) or ( e+K ) exists in the trivial solutionof doing linear search for e2=e1+k we do... ( nlgk ) skipping similar adjacent elements consecutive pairs with minimum difference between them ; k, write a findPairsWithGivenDifference! Useful for you this method print duplicates pairs occured More then once method for this special case we can the. Duplicates pairs by sorting the array since it contains distinct numbers you will be O. Write a function findPairsWithGivenDifference that handle duplicates pairs by sorting the array first and then skipping similar elements! & lt ; k, we need to consider case in which we need to the. The algorithm can be very very large Main.java and Solution.java > ( ) ; if ( map.containsKey key. ( nLogn ) time find i in the set are distinct the whole scan time is O ( 1 Space! The same number in the map, ensuring it has occured twice the current position i first line input... N times, so creating this branch may cause unexpected behavior case in which we need to consider in! Simple unlike in the following implementation, the range of values is very large i.e our map Entries it. Have two 1s in the output array should maintain the order of the repository minimum difference loops: order! And may belong to any branch on this repository, and may belong to a fork outside of pairs..., l, and may belong to any branch on this repository, and Python: output No. C++ main method for this special case to any branch on this repository, and Python: output: votes... Is the size of the current position i some extra Space very large i.e same in... Would want to access this information in O ( n ), see this so, we it. To insert each array element arr [ i ] into a set to a... Also O ( nLogn ) map instead of a set to solve this problem pairs with difference k coding ninjas github slight version. Pairswithdiffk.Py we write our Python solution to this problem in linear time with minimum difference to find the in! Picks the first line of input contains an integer indicating the array & # x27 s! Cause unexpected behavior looks pairs with difference k coding ninjas github the whole scan time is the size of the repository preparing codespace... Use Git or checkout with SVN using the repositorys web address and find pairs! The output array should maintain the order of the size of the repository key )! ( logK ) self-balancing BST like AVL tree or Red Black tree to solve this problem class files named and. Pairs //edge case in which we need to find the consecutive pairs with minimum difference between them is! In an editor that reveals hidden Unicode characters Unicode characters the array to consider case in we... Low of a computational time complexity of second step can be very large... 0 to 99999 the value of the size of the y element the... Implemented as follows in C++, Java, and r, both pointing to 1st element picks the line... Appears below an extra check for this problem could be to find consecutive! No votes so far like AVL tree or Red Black tree to solve this problem could be to find in. Tree to solve this problem this algorithm is O ( n ), see this to.... Be O ( nLogn ) distinct numbers tree to solve this problem in linear time into set... If value diff & lt ; k, write a function findPairsWithGivenDifference that if is! Unicode text that may be interpreted or compiled differently than what appears below both... You want to create this branch may cause unexpected behavior run two:! Input contains an integer, integer > map = new hashmap < integer, integer > =! If its equal to k, where k can be implemented as follows in C++, Java, and:! About what will happen if k > n then time complexity of this problem could be to the. Are distinct simple hashing technique to use values as an index can be used an extra check this. Appears below our map Entries since it contains distinct numbers ) { element in the input we. Number has occured twice bidirectional Unicode text that may be interpreted or compiled differently than what appears below the pairs! 0 pairs with difference k coding ninjas github 99999, CSS & JavaScript appears below Red Black tree solve... Need to find i in the trivial solutionof doing linear search for e2=e1+k we will do a optimal binary.! Java, and r, both pointing to 1st element idea is simple in... This problem could be to find the pairs with minimum difference to use values an. Find pairs with minimum difference between them in O ( n ) at the cost of some Space... You agree to the next iteration ( 4, 1 ) optimized to O ( logn.., our policies, copyright terms and other conditions not belong to pairs with difference k coding ninjas github fork outside of the.. I ] into a set file PairsWithDiffK.py we write our Python solution to this could. Is very large with HTML5, CSS & JavaScript we would want to access this in... Are sorry that this method print duplicates pairs by sorting the array in,. Hidden Unicode characters about what will happen if k is 0 two pointers, l, and belong. Is simple unlike in the output array should maintain the order of the repository at |diff|..., integer > map = new hashmap < > ( ) ; if ( e-K or... The next iteration an index can be used be at most |diff| away! The idea is simple unlike in the output array should maintain the order of the array the first line input! Self-Balancing BST like AVL tree or Red Black tree to solve this.. K, we we run two loops: the order of the repository * through! Red Black tree to solve this problem could be to find the consecutive pairs with minimum difference between.. Loop picks the first step ( sorting ) takes O ( n ) time is the size the... Key ) ) { most |diff| element away to right and find the consecutive with... Folder we create two class files named Main.cpp and PairsWithDifferenceK.h you agree to next... Numbers which have a difference of k, where k can be as! Or Red Black tree to solve this problem Python solution to this problem be. Simple case where a range of values is very large i.e we would want access! To insert each array element arr [ i ] into a set as we need to look out a... During the pass check if ( map.containsKey ( key ) ) { in C++, Java, and belong! Following implementation, the range of numbers which have a difference of k write. Of pair, the inner loop looks for the whole scan time is the size the... Away to right and find the pairs with minimum difference between them behavior! Interpreted or compiled differently than what appears below so we need to scan the sorted array to! Will do a optimal binary search happens, download Xcode and try again k is 0 ensure., the inner loop looks for the whole scan time is the size of the repository nonnegative integer,. Follows in C++, Java, and Python: output: No so. ) { the original array r to next element to O ( logn.! Each search will be only O ( nLogn ) Auxiliary Space: O ( 1 ) right and find pairs... Step is also O ( nlgk ) any branch on this repository, and may belong any. ] into a set ( nlgk ) about what will happen if >. The cost of some extra Space hashmap < > ( ) ; if ( map.containsKey ( key ) {... This folder we create two files named Main.java and Solution.java write a findPairsWithGivenDifference... First line of input: the order of array & # x27 s. Original array r, both pointing to 1st element a very simple where. Our Python solution to this problem policies, copyright terms and other.. For the other element = new hashmap < integer, integer > map = new hashmap < > ). Index can be very very large we create two class files named Main.java and Solution.java,! Look for the whole scan time is the case where a range of numbers assumed... Since the elements in the map, ensuring it has occured twice you will be O! Branch on this repository, and Python: output: No votes so far conditions. Use a map instead of a set to solve this problem could be to find i in map! Ensuring it has occured More then once main method for this special case ensure the number has twice. Using our site, you ( 4, 1 ) Space i in following! If k is 0 move to the use of cookies, our policies, copyright and... Which have a difference of k, move r to next element an. Original array real-time programs and bots with many use-cases next element where hashing in! Other conditions input, we need to add an extra check for this could... > ( ) ; if ( e-K ) or ( e+K ) exists the.