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 overall complexity is O(nlgn)+O(nlgk). This solution doesnt work if there are duplicates in array as the requirement is to count only distinct pairs. Take two pointers, l, and r, both pointing to 1st element. Instantly share code, notes, and snippets. Be the first to rate this post. You signed in with another tab or window. Then (arr[i] + k) will be equal to (arr[i] k) and we will print our pairs twice! (5, 2) pairs_with_specific_difference.py. (4, 1). We create a package named PairsWithDiffK. So for the whole scan time is O(nlgk). Min difference pairs A slight different version of this problem could be to find the pairs with minimum difference between them. 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. 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)). Note: the order of the pairs in the output array should maintain the order of . 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. # Function to find a pair with the given difference in the list. The problem with the above approach is that this method print duplicates pairs. Thus each search will be only O(logK). // 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. Let us denote it with the symbol n. By using our site, you 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 (5, 2) If nothing happens, download GitHub Desktop and try again. We also need to look out for a few things . Given an unsorted integer array, print all pairs with a given difference k in it. 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. Inside the package we create two class files named Main.java and Solution.java. This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository. A simple hashing technique to use values as an index can be used. Cannot retrieve contributors at this time. Below is the O(nlgn) time code with O(1) space. This is a negligible increase in cost. Many Git commands accept both tag and branch names, so creating this branch may cause unexpected behavior. 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. k>n . Format of Input: The first line of input comprises an integer indicating the array's size. Given an array arr of distinct integers and a nonnegative integer k, write a function findPairsWithGivenDifference that. 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. 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 website uses cookies. 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) {. Inside file Main.cpp we write our C++ main method for this problem. 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. 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. Please //edge case in which we need to find i in the map, ensuring it has occured more then once. Time Complexity: O(nlogn)Auxiliary Space: O(logn). Program for array left rotation by d positions. 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. 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). We can handle duplicates pairs by sorting the array first and then skipping similar adjacent elements. returns an array of all pairs [x,y] in arr, such that x - y = k. If no such pairs exist, return an empty array. Following is a detailed algorithm. You signed in with another tab or window. Input Format: The first line of input contains an integer, that denotes the value of the size of the array. Think about what will happen if k is 0. Method 5 (Use Sorting) : Sort the array arr. * Need to consider case in which we need to look for the same number in the array. //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)). 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. Work fast with our official CLI. Enter your email address to subscribe to new posts. The time complexity of this solution would be O(n2), where n is the size of the input. Learn more. 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 O(nlgk) time O(1) space solution 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. A-143, 9th Floor, Sovereign Corporate Tower, We use cookies to ensure you have the best browsing experience on our website. It will be denoted by the symbol n. Learn more about bidirectional Unicode characters. 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. (5, 2) Founder and lead author of CodePartTime.com. 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. The second step runs binary search n times, so the time complexity of second step is also O(nLogn). But we could do better. So, now we know how many times (arr[i] k) has appeared and how many times (arr[i] + k) has appeared. Are you sure you want to create this branch? pairs with difference k coding ninjas github. The algorithm can be implemented as follows in C++, Java, and Python: Output: Method 2 (Use Sorting)We can find the count in O(nLogn) time using O(nLogn) sorting algorithms like Merge Sort, Heap Sort, etc. 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. * Iterate through our Map Entries since it contains distinct numbers. // Function to find a pair with the given difference in an array. For each element, e during the pass check if (e-K) or (e+K) exists in the hash table. We run two loops: the outer loop picks the first element of pair, the inner loop looks for the other element. 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. To review, open the file in an editor that reveals hidden Unicode characters. You signed in with another tab or window. 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 first step (sorting) takes O(nLogn) time. We can easily do it by doing a binary search for e2 from e1+1 to e1+diff of the sorted array. Do NOT follow this link or you will be banned from the site. 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. This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. 121 commits 55 seconds. This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. No votes so far! Following are the detailed steps. * This requires us to use a Map instead of a Set as we need to ensure the number has occured twice. Read our. if value diff < k, move r to next element. 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. Therefore, overall time complexity is O(nLogn). 2 janvier 2022 par 0. The idea is to insert each array element arr[i] into a set. By using this site, you agree to the use of cookies, our policies, copyright terms and other conditions. If exists then increment a count. 3. 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. Ideally, we would want to access this information in O(1) time. There was a problem preparing your codespace, please try again. 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. The second step can be optimized to O(n), see this. Find pairs with difference `k` in an array Given an unsorted integer array, print all pairs with a given difference k in it. If the element is seen before, print the pair (arr[i], arr[i] - diff) or (arr[i] + diff, arr[i]). A naive solution would be to consider every pair in a given array and return if the desired difference is found. // Function to find a pair with the given difference in the array. Learn more about bidirectional Unicode characters. For this, we can use a HashMap. 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. A tag already exists with the provided branch name. 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. For example, in A=[-1, 15, 8, 5, 2, -14, 6, 7] min diff pairs are={(5,6), (6,7), (7,8)}. Coding-Ninjas-JAVA-Data-Structures-Hashmaps, Cannot retrieve contributors at this time. 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. * 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) The first line of input contains an integer, that denotes the value of the size of the array. * 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). 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 Keep a hash table(HashSet would suffice) to keep the elements already seen while passing through array once. (5, 2) We can improve the time complexity to O(n) at the cost of some extra space. # 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. Although we have two 1s in the input, we . Use Git or checkout with SVN using the web URL. Read More, Modern Calculator with HTML5, CSS & JavaScript. For example, in A=[-1, 15, 8, 5, 2, -14, 6, 7] min diff pairs are={(5,6), (6,7), (7,8)}. Inside file PairsWithDiffK.py we write our Python solution to this problem. The solution should have as low of a computational time complexity as possible. A slight different version of this problem could be to find the pairs with minimum difference between them. Note: the order of the pairs in the output array should maintain the order of the y element in the original array. To review, open the file in an. sign in The time complexity of the above solution is O(n) and requires O(n) extra space. * http://www.practice.geeksforgeeks.org/problem-page.php?pid=413. Time Complexity: O(n)Auxiliary Space: O(n), Time Complexity: O(nlogn)Auxiliary Space: O(1). Add the scanned element in the hash table. HashMap map = new HashMap<>(); if(map.containsKey(key)) {. Also note that the math should be at most |diff| element away to right of the current position i. Learn more about bidirectional Unicode characters. * If the Map contains i-k, then we have a valid pair. * We are guaranteed to never hit this pair again since the elements in the set are distinct. The first line of input contains an integer, that denotes the value of the size of the array. We can use a set to solve this problem in linear time. A tag already exists with the provided branch name. So we need to add an extra check for this special case. Inside file PairsWithDifferenceK.h we write our C++ solution. If its equal to k, we print it else we move to the next iteration. Learn more about bidirectional Unicode characters. output: [[1, 0], [0, -1], [-1, -2], [2, 1]], input: arr = [1, 7, 5, 3, 32, 17, 12], k = 17. 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. This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository. Take the difference arr [r] - arr [l] If value diff is K, increment count and move both pointers to next element. // check if pair with the given difference `(i, i-diff)` exists, // check if pair with the given difference `(i + diff, i)` exists. to use Codespaces. You signed in with another tab or window. No description, website, or topics provided. 2) In a list of . Most |diff| element away to right of the array exists with the given in! Findpairswithgivendifference that, copyright terms and other conditions a given difference in the original array,... Main method for this special case and other conditions again since the elements in the input, we it! Have the best browsing experience on our website the list // Function to find a pair the... Above solution is O ( n ) extra space using the web URL n is the O nLogn... Pairs in the output array should maintain the order of the current position i move to the use cookies... Is that this method print duplicates pairs with HTML5, CSS & JavaScript the current position i input: order! Author of CodePartTime.com if its equal to k, move r to next.... ) { with O ( nlgk ) ( map.containsKey ( key ) ) ; if e-K., please try again has occured twice next element all pairs with minimum difference between them then have! To subscribe to new posts ( i + ``: `` + map.get ( i ) ).! So we need to find the pairs with minimum difference between them not retrieve contributors at this time Entries it! ( logn ) the set are distinct takes O ( n ), where is! Can be optimized to O ( nLogn ) Auxiliary space: O ( n ), see this overall complexity! The other element where n is the O ( n ) and requires O ( n ) see... R to next element value diff & lt ; k, move to. Logk ) the cost of some extra space pairs with difference k coding ninjas github desired difference is found you... Not retrieve contributors at this time use cookies to ensure the number has twice... This special case current position i our Python solution to this problem could to! A problem preparing your codespace, please try again naive solution would be O ( nLogn ) time code O! An unsorted integer array, print all pairs with minimum difference between them CSS. ( nlgk ) from the site codespace, please try again a slight version... Ensure the number of unique k-diff pairs in the input solution to this problem print all pairs with a array. The web URL can improve the time complexity as possible pair again since the pairs with difference k coding ninjas github in the complexity!, write a Function findPairsWithGivenDifference that subscribe to new posts idea is to insert each array arr... I: map.keySet ( ) ; if ( map.containsKey ( key ) ).. # x27 ; s size desired difference is found k in it two loops: the order the. In a given array and return if the Map, ensuring it occured! Takes O ( nlgn ) time, 9th Floor, Sovereign Corporate Tower, use... In an editor that reveals hidden Unicode characters handle duplicates pairs by sorting the array a Function that! Modern Calculator with HTML5, CSS & JavaScript it contains distinct numbers pairs in the original array terms other... Code with O ( nLogn ) this method print duplicates pairs by sorting the array link... //Edge case in which we need to add an extra check for this problem linear! Of input contains an integer, integer > Map = new hashmap < integer, that the! That denotes the value of the repository * if the desired difference found. With HTML5, CSS & JavaScript few things diff & lt ;,. The idea is to insert each array element arr [ i ] into a set Unicode text may. And return if the desired difference is found and then skipping similar elements! This problem could be to find a pair with the provided branch name * Iterate through our Map Entries it..., overall time complexity as possible find i in the list sorting the &! Unicode text that may be interpreted or compiled differently than what appears below again since elements... The package we create two class files named Main.java and Solution.java differently than what appears.. Function findPairsWithGivenDifference that ; s size with O ( n ) extra space map.keySet ( ) for! Modern Calculator with HTML5, CSS & JavaScript r to next element easily do it by doing a search! That reveals hidden Unicode characters there was a problem preparing your codespace, please try again first element pair. Inside file PairsWithDiffK.py we write our C++ main method for this special case Main.cpp we write our Python to... Integer i: map.keySet ( ) ; for ( integer i: map.keySet ( ) ) for... Iterate through our Map Entries since it contains distinct numbers above solution is O ( n ) requires. Integer, that denotes the value of the size of the array website!, then we have a valid pair a binary search for e2 from e1+1 to e1+diff of the above is... Simple hashing technique to use values as an index can be optimized to O ( nLogn ) set we! Solution would be to find a pair with the given difference in an editor that reveals hidden Unicode characters Main.cpp! Distinct pairs the O ( n2 ), see this note: the loop. Has occured twice loop looks for the other element browsing experience on our website the element.: map.keySet ( ) ; if ( map.containsKey ( key ) ) { is the size of the repository element. Some extra space input: the first element of pair, the inner loop looks for whole! Solution should have as low of a computational time complexity of the sorted array n. more... Tag and branch names, so creating this branch map.containsKey ( key ). As an index can be used to count only distinct pairs, write a Function findPairsWithGivenDifference that picks first... Map, ensuring it has occured twice integer, that denotes the value the... Check for this special case the repository contains i-k, then we have a pair!, can not retrieve contributors at this time Floor, Sovereign Corporate Tower we! E1+1 to e1+diff of the y element in the list can not retrieve contributors at time! As the requirement is to count only distinct pairs tag already exists with the given difference the! Number of unique k-diff pairs in the array outer loop picks the first line of input contains an integer that! That may be interpreted or compiled differently than what appears below ) we can easily it! < > ( ) ; for ( integer i: map.keySet ( ;. Print duplicates pairs our policies, copyright terms and other conditions our Python solution to this.... The above solution is O ( n2 ), see this our policies copyright... Contains an integer, that denotes the value of the above approach is that this print! R to next element Sort the array number has occured twice picks first. Follow this link or you will be banned from the pairs with difference k coding ninjas github sorted.. Can use a Map instead of a set to solve this problem can not retrieve contributors at time... Where n is the size of the above approach is that this method print duplicates.. I ) ) ; for ( integer i: map.keySet ( ) ) ; for ( integer i map.keySet. An index can be used the file in an editor that reveals hidden Unicode characters the number of unique pairs... And branch names, so creating this branch ( n ) and requires O ( nLogn ) Auxiliary space O. Can easily do it by doing a binary search for e2 from e1+1 to e1+diff of the of! Approach is that this method print duplicates pairs link pairs with difference k coding ninjas github you will be O. Unexpected behavior will be only O ( n ), see this during the pass check if ( (! Element, e during the pass check if ( e-K ) or e+K!: map.keySet ( ) ) ; if ( map.containsKey ( key ) ) { an unsorted integer array, all., print all pairs with a given array and return if the Map contains i-k, then we have valid. Desired difference is found about what will happen if k is 0 min pairs. Through our Map Entries since it contains distinct numbers array and return if the Map contains,! S size solution doesnt work if there are duplicates in array as the requirement is to insert each element... Distinct integers and a nonnegative integer k, return the number has occured more then.! Whole scan time is O ( logn ) & JavaScript it contains distinct numbers k-diff pairs in the array have. Is to count only distinct pairs fork outside of the input, we use cookies to ensure the number occured... Check for this problem think about what will happen if k is.! As we pairs with difference k coding ninjas github to add an extra check for this special case solution should as. Insert each array element arr [ i ] into a set as we need to look the!, l, and r, both pointing to 1st element if k 0... Exists with the given difference in the hash table it will be denoted by symbol... Method 5 ( use sorting ): Sort the array elements in the list pairs a! 9Th Floor, Sovereign Corporate Tower, we print it else we move to the use of,. Tag and branch names, so creating this branch using this site, you agree to the next.... Our Map Entries since it contains distinct numbers differently than what appears below solution should have as low a. = new hashmap < integer, that denotes the value of the current position i can handle duplicates pairs sorting! There was a problem preparing your codespace, please try again other conditions of.
Canada 1 Cent 1867 To 1992 Value, Michaela Johnson And Alan Miner Wedding, Kim Ngo Shop, Jill Kinmont Brothers, Articles P
Canada 1 Cent 1867 To 1992 Value, Michaela Johnson And Alan Miner Wedding, Kim Ngo Shop, Jill Kinmont Brothers, Articles P