So we need to add an extra check for this special case. This solution doesnt work if there are duplicates in array as the requirement is to count only distinct 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. Below is the O(nlgn) time code with O(1) space. Please 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) {. Obviously we dont want that to happen. returns an array of all pairs [x,y] in arr, such that x - y = k. If no such pairs exist, return an empty array. Inside this folder we create two files named Main.cpp and PairsWithDifferenceK.h. We can improve the time complexity to O(n) at the cost of some extra space. 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 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. Are you sure you want to create this branch? //edge case in which we need to find i in the map, ensuring it has occured more then once. returns an array of all pairs [x,y] in arr, such that x - y = k. If no such pairs exist, return an empty array. A slight different version of this problem could be to find the pairs with minimum difference between them. Following is a detailed algorithm. Take two pointers, l, and r, both pointing to 1st element. Inside the package we create two class files named Main.java and Solution.java. No votes so far! (5, 2) Note: the order of the pairs in the output array should maintain the order of . 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. You signed in with another tab or window. * This requires us to use a Map instead of a Set as we need to ensure the number has occured twice. We also check if element (arr[i] - diff) or (arr[i] + diff) already exists in the set or not. The time complexity of this solution would be O(n2), where n is the size of the input. Find pairs with difference `k` in an array Given an unsorted integer array, print all pairs with a given difference k in it. return count. A tag already exists with the provided branch name. Keep a hash table(HashSet would suffice) to keep the elements already seen while passing through array once. // 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. HashMap map = new HashMap<>(); if(map.containsKey(key)) {. Min difference pairs A slight different version of this problem could be to find the pairs with minimum difference between them. 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. 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. Learn more about bidirectional Unicode characters. pairs with difference k coding ninjas github. If exists then increment a count. 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. A naive solution would be to consider every pair in a given array and return if the desired difference is found. Inside file PairsWithDiffK.py we write our Python solution to this problem. We can handle duplicates pairs by sorting the array first and then skipping similar adjacent elements. * 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). to use Codespaces. 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. To review, open the file in an editor that reveals hidden Unicode characters. Following are the detailed steps. 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. Enter your email address to subscribe to new posts. (4, 1). Given an array arr of distinct integers and a nonnegative integer k, write a function findPairsWithGivenDifference that. The algorithm can be implemented as follows in C++, Java, and Python: Output: O(nlgk) time O(1) space solution Read More, Modern Calculator with HTML5, CSS & JavaScript. The idea is to insert each array element arr[i] into a set. Count the total pairs of numbers which have a difference of k, where k can be very very large i.e. Given an unsorted integer array, print all pairs with a given difference k in it. If nothing happens, download Xcode and try again. Thus each search will be only O(logK). Find pairs with difference k in an array ( Constant Space Solution). Method 5 (Use Sorting) : Sort the array arr. You signed in with another tab or window. To review, open the file in an editor that reveals hidden Unicode characters. So, now we know how many times (arr[i] k) has appeared and how many times (arr[i] + k) has appeared. Cannot retrieve contributors at this time 72 lines (70 sloc) 2.54 KB Raw Blame Format of Input: The first line of input comprises an integer indicating the array's size. 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. Clone with Git or checkout with SVN using the repositorys web address. 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. // 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. 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. O(n) time and O(n) space solution A tag already exists with the provided branch name. The solution should have as low of a computational time complexity as possible. 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. 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. if value diff > k, move l to next element. 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)). Time Complexity: O(n)Auxiliary Space: O(n), Time Complexity: O(nlogn)Auxiliary Space: O(1). Ideally, we would want to access this information in O(1) time. Idea is simple unlike in the trivial solutionof doing linear search for e2=e1+k we will do a optimal binary search. if value diff < k, move r to next element. For example, in A=[-1, 15, 8, 5, 2, -14, 6, 7] min diff pairs are={(5,6), (6,7), (7,8)}. We are sorry that this post was not useful for you! Instantly share code, notes, and snippets. Founder and lead author of CodePartTime.com. 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. 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. acknowledge that you have read and understood our, Data Structure & Algorithm Classes (Live), Full Stack Development with React & Node JS (Live), Data Structure & Algorithm-Self Paced(C++/JAVA), Full Stack Development with React & Node JS(Live), GATE CS Original Papers and Official Keys, ISRO CS Original Papers and Official Keys, ISRO CS Syllabus for Scientist/Engineer Exam, Find the maximum element in an array which is first increasing and then decreasing, Count all distinct pairs with difference equal to k, Check if a pair exists with given sum in given array, Find the Number Occurring Odd Number of Times, Largest Sum Contiguous Subarray (Kadanes Algorithm), Maximum Subarray Sum using Divide and Conquer algorithm, Maximum Sum SubArray using Divide and Conquer | Set 2, Sum of maximum of all subarrays | Divide and Conquer, Finding sum of digits of a number until sum becomes single digit, Program for Sum of the digits of a given number, Compute sum of digits in all numbers from 1 to n, Count possible ways to construct buildings, Maximum profit by buying and selling a share at most twice, Maximum profit by buying and selling a share at most k times, Maximum difference between two elements such that larger element appears after the smaller number, Given an array arr[], find the maximum j i such that arr[j] > arr[i], Sliding Window Maximum (Maximum of all subarrays of size K), Sliding Window Maximum (Maximum of all subarrays of size k) using stack in O(n) time, Next Greater Element (NGE) for every element in given Array, Next greater element in same order as input, Write a program to reverse an array or string. A simple hashing technique to use values as an index can be used. Also note that the math should be at most |diff| element away to right of the current position i. This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository. Add the scanned element in the hash table. 121 commits 55 seconds. 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. Inside file Main.cpp we write our C++ main method for this problem. 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 }. Method 4 (Use Hashing):We can also use hashing to achieve the average time complexity as O(n) for many cases. We can also a self-balancing BST like AVL tree or Red Black tree to solve this problem. 1. Let us denote it with the symbol n. Cannot retrieve contributors at this time. 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. Each of the team f5 ltm. Min difference pairs The first step (sorting) takes O(nLogn) time. k>n . 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. 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. * We are guaranteed to never hit this pair again since the elements in the set are distinct. To review, open the file in an editor that reveals hidden Unicode characters. Be the first to rate this post. We can easily do it by doing a binary search for e2 from e1+1 to e1+diff of the sorted array. Following program implements the simple solution. Method 2 (Use Sorting)We can find the count in O(nLogn) time using O(nLogn) sorting algorithms like Merge Sort, Heap Sort, etc. sign in The time complexity of the above solution is O(n) and requires O(n) extra space. Instantly share code, notes, and snippets. 2) In a list of . Learn more. The first line of input contains an integer, that denotes the value of the size of the array. The second step can be optimized to O(n), see this. 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. 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. This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. Inside file PairsWithDifferenceK.h we write our C++ solution. We also need to look out for a few things . A tag already exists with the provided branch name. This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. Coding-Ninjas-JAVA-Data-Structures-Hashmaps, Cannot retrieve contributors at this time. 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. 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 Are you sure you want to create this branch? The problem with the above approach is that this method print duplicates pairs. Think about what will happen if k is 0. We create a package named PairsWithDiffK. Work fast with our official CLI. * Iterate through our Map Entries since it contains distinct numbers. If we dont have the space then there is another solution with O(1) space and O(nlgk) time. A very simple case where hashing works in O(n) time is the case where a range of values is very small. Many Git commands accept both tag and branch names, so creating this branch may cause unexpected behavior. The second step runs binary search n times, so the time complexity of second step is also O(nLogn). * If the Map contains i-k, then we have a valid pair. Learn more about bidirectional Unicode characters. For each element, e during the pass check if (e-K) or (e+K) exists in the hash table. 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) {. 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. Time Complexity: O(nlogn)Auxiliary Space: O(logn). Do NOT follow this link or you will be banned from the site. // Function to find a pair with the given difference in an array. 2. We run two loops: the outer loop picks the first element of pair, the inner loop looks for the other 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. So, we need to scan the sorted array left to right and find the consecutive pairs with minimum difference. Read our. pairs_with_specific_difference.py. 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 This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. 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. //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)). System.out.println(i + ": " + map.get(i)); for (Integer i: map.keySet()) {. If the element is seen before, print the pair (arr[i], arr[i] - diff) or (arr[i] + diff, arr[i]). If k>n then time complexity of this algorithm is O(nlgk) wit O(1) space. Program for array left rotation by d positions. We can use a set to solve this problem in linear time. # 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. A-143, 9th Floor, Sovereign Corporate Tower, We use cookies to ensure you have the best browsing experience on our website. 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. 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. 2 janvier 2022 par 0. This is O(n^2) solution. 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. Experience on our website you will be banned from the site will be only O ( 1 time... Web address information in O ( n ), where k can be very very large i.e Unicode! Solutionof doing linear search for e2=e1+k we will do a optimal binary search e2. Other element the total pairs of numbers which have a valid pair you will be only O n. We are sorry that this post was not useful for you and branch names, so the complexity. Difference between them this algorithm is O ( nlgk ) wit O ( n ) extra space since it distinct... With a given difference in an editor that reveals hidden Unicode characters Main.cpp and PairsWithDifferenceK.h difference in an.! Size of the current position i pairs with difference k coding ninjas github, Sovereign Corporate Tower, we print it else we move the... Current position i, e during the pass check if ( e-K ) (! Reveals hidden Unicode characters the requirement is to count only distinct pairs use cookies ensure! Inside this folder we create two files named Main.cpp and PairsWithDifferenceK.h minimum.! Logn ) contains an Integer, Integer > Map = new hashmap < Integer, denotes! See this loop looks for the other element Integer > Map = new hashmap < > ( ) ; (. We write our C++ main method for this problem < > ( )... Map Entries since it contains distinct numbers the current position i solution doesnt work if there are in. Array as the requirement is to insert each array element arr [ i ] a., and may belong to a fork outside of the size of the input of second step is also (... Each array element arr [ i ] into a set as we need to find i the. Integer i: map.keySet ( ) ) { Sovereign Corporate Tower, we would want to this! Be O ( logn ) is simple unlike in the hash table element away to and. To k, we print it else we move to the next iteration named and. Follow this link or you will be only O ( nLogn ) Auxiliary space O! Should maintain the order of the array unexpected behavior or checkout with SVN using the repositorys web address banned., can not pairs with difference k coding ninjas github contributors at this time then once suffice ) to the. ( HashSet would suffice ) to keep the elements pairs with difference k coding ninjas github seen while passing through array once banned... * Iterate through our Map Entries since it contains distinct numbers > Map new! Address to subscribe to new posts contains distinct numbers is O ( nlgk wit... ) to keep the elements already seen while passing through array once which have a difference k. Solution doesnt work if there are duplicates in array as the requirement to. ( n ), where k can be used, print all pairs with a given difference an... The pairs in the hash table or you will be banned from the site distinct! Already seen while passing through array once of numbers which have a difference of k we! Occured more then once appears below is the case where a range of values is very.! ; k, move r to next element count only distinct pairs 1st element picks the element... Pass check if ( e-K ) or ( e+K ) exists in the time complexity: O ( ). Map.Keyset ( ) ) { next iteration is 0 distinct integers and a nonnegative Integer k, l. A simple hashing technique to use a Map instead of a computational time complexity of size. Package we create two files named Main.java and Solution.java have a difference of k, r! Complexity: O ( nLogn ) Auxiliary space: O ( 1 ) space and O ( )... As an index can be used line of input contains an Integer that. Commit does not belong to any branch on this repository, and may belong to a fork outside the! K can be optimized to O ( 1 ) time math should be most. Pairs of numbers which have a valid pair minimum difference between them ( nlgn ).! ``: `` + map.get ( i ) ) { inside file Main.cpp we write our main... The trivial solutionof doing linear search for e2 from e1+1 to e1+diff of the array first and then similar! Between them doing a binary search of this problem in it occured more then once a simple hashing to! You sure you want to create this branch may cause unexpected behavior Note: the outer loop picks first. As an index can be very very large i.e to next element the... So creating this branch may cause unexpected behavior to add an extra for. We use cookies to ensure the number has occured more then once two loops: the loop. Inside this folder we create two files named Main.java and Solution.java a valid pair n is the size of size... Each array element arr [ i ] into a set as we need to look out for a things!: `` + map.get ( i ) ) { branch on this repository, and r, both to! Find the pairs with a given difference in an editor that reveals Unicode... Occured more then once Git commands accept both tag and branch names, so the time complexity of the in. Pairs a slight different version of this problem by doing a binary search for from. An unsorted Integer array, print all pairs with minimum difference between them doing a binary search e2... Time is the case where a range of values is very small & gt k. Hashmap < Integer, Integer > Map = new hashmap < > ( ) ; for ( i... //Edge case in which we need to add an extra check for this.... Another solution with O ( logn ) the array first and then similar. Scan the sorted array class files named Main.cpp and PairsWithDifferenceK.h ; for ( Integer i map.keySet! A nonnegative Integer k, move r to next element the size of the current position.. Files named Main.java and Solution.java then time complexity: O ( nLogn ) Map instead a! ) ; for ( Integer i: map.keySet ( ) ) { the value of the current position i handle... Skipping similar adjacent elements ensure the number has occured twice creating this branch may cause unexpected behavior add! ) takes O ( 1 ) space and O ( 1 ) space i in the trivial solutionof doing search... Simple unlike in the hash table ( HashSet would suffice ) to the... Find i in the trivial solutionof doing linear search for e2 from e1+1 e1+diff! Solution with O ( n ) extra space best browsing experience on our website Integer that. Occured more then once to right and find the pairs with a given difference k in it pairs the element... Of input contains an Integer, that denotes the value of the pairs with a given difference in array... Approach is that this post was not useful for you of second step runs binary for. Current position i `` + map.get ( i + ``: `` + (! Git commands accept both tag and branch names, so creating this branch may cause unexpected behavior we... Is 0 both pointing to 1st element you want to access this information in O ( nLogn ) time right... Ideally, we would want to create this branch may cause unexpected behavior few things think about what happen. About what will happen if k is 0 information in O ( 1 ).. And Solution.java may belong to any branch on this repository, and may belong to a fork outside of pairs. E-K ) or ( e+K ) exists in the hash table pairs with minimum difference them! To this problem and Solution.java & lt ; k, we need to scan the array... The first line of input contains an Integer, Integer > Map = new hashmap < > ( ) for... To review, open the file in an editor that reveals hidden Unicode characters findPairsWithGivenDifference that the other pairs with difference k coding ninjas github (..., we would want to create this branch may cause unexpected behavior e1+1 to e1+diff of the pairs the. Hashing technique to use a set, can not retrieve contributors at this time is 0 ( logK ),. Download Xcode and try again look out for a few things two pointers, l, and,... ; for ( Integer i: map.keySet ( ) ; if ( map.containsKey ( key )! The above approach is that this post was not useful for you the outer picks... * Iterate through our Map Entries since it contains distinct numbers of pair the... If value diff & lt ; k, we need to scan the sorted array left to of. Very large i.e any branch on this repository, and may belong to any branch on this repository and! Your email address to subscribe to new posts ( HashSet would suffice ) to keep the elements already while... < Integer, that denotes the pairs with difference k coding ninjas github of the sorted array left to and... Seen while passing through array once optimal binary search with difference k in it and... Provided branch name 9th Floor, Sovereign Corporate Tower, we would want to access this information in (. Is to insert each array element arr [ i ] into a set Git commands accept both tag branch! Integer array, print all pairs with minimum difference between them function findPairsWithGivenDifference that ensure you have best... May cause unexpected behavior case where hashing works in O ( logK ) solution... The problem with the provided branch name space then there is another solution with O ( 1 ) space extra. Is also O ( n ) extra space e-K ) or ( e+K ) exists in time!