pairs with difference k coding ninjas github

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. 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) {. 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. We also check if element (arr[i] - diff) or (arr[i] + diff) already exists in the set or not. Read More, Modern Calculator with HTML5, CSS & JavaScript. If its equal to k, we print it else we move to the next iteration. Note: the order of the pairs in the output array should maintain the order of the y element in the original array. He's highly interested in Programming and building real-time programs and bots with many use-cases. Follow me on all Networking Sites: LinkedIn : https://www.linkedin.com/in/brian-danGitHub : https://github.com/BRIAN-THOMAS-02Instagram : https://www.instagram.com/_b_r_i_a_n_#pairsum #codingninjas #competitveprogramming #competitve #programming #education #interviewproblem #interview #problem #brianthomas #coding #crackingproblem #solution Let us denote it with the symbol n. For example, in the following implementation, the range of numbers is assumed to be 0 to 99999. Each of the team f5 ltm. //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)). 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. 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(5, 2) output: [[1, 0], [0, -1], [-1, -2], [2, 1]], input: arr = [1, 7, 5, 3, 32, 17, 12], k = 17. To review, open the file in an editor that reveals hidden Unicode characters. 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. Time Complexity: O(n)Auxiliary Space: O(n), Time Complexity: O(nlogn)Auxiliary Space: O(1). For this, we can use a HashMap. For example, in A=[-1, 15, 8, 5, 2, -14, 6, 7] min diff pairs are={(5,6), (6,7), (7,8)}. 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 can easily do it by doing a binary search for e2 from e1+1 to e1+diff of the sorted array. A simple hashing technique to use values as an index can be used. 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. Learn more about bidirectional Unicode characters. Founder and lead author of CodePartTime.com. 2. This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository. Method 5 (Use Sorting) : Sort the array arr. Following is a detailed algorithm. A-143, 9th Floor, Sovereign Corporate Tower, We use cookies to ensure you have the best browsing experience on our website. For each element, e during the pass check if (e-K) or (e+K) exists in the hash table. Input Format: The first line of input contains an integer, that denotes the value of the size of the array. 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. This is O(n^2) solution. To review, open the file in an. 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. 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. Coding-Ninjas-JAVA-Data-Structures-Hashmaps, Cannot retrieve contributors at this time. We can improve the time complexity to O(n) at the cost of some extra space. Inside file PairsWithDifferenceK.h we write our C++ solution. Learn more. If exists then increment a count. Below is the O(nlgn) time code with O(1) space. The problem with the above approach is that this method print duplicates pairs. 1. Work fast with our official CLI. Program for array left rotation by d positions. We create a package named PairsWithDiffK. The first line of input contains an integer, that denotes the value of the size of the array. 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 The first step (sorting) takes O(nLogn) time. The time complexity of the above solution is O(n) and requires O(n) extra space. This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository. 2 janvier 2022 par 0. You signed in with another tab or window. The solution should have as low of a computational time complexity as possible. A very simple case where hashing works in O(n) time is the case where a range of values is very small. This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository. A tag already exists with the provided branch name. In file Main.java we write our main method . Obviously we dont want that to happen. Many Git commands accept both tag and branch names, so creating this branch may cause unexpected behavior. Take two pointers, l, and r, both pointing to 1st element. 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. Add the scanned element in the hash table. Given n numbers , n is very large. Be the first to rate this post. Many Git commands accept both tag and branch names, so creating this branch may cause unexpected behavior. To review, open the file in an editor that reveals hidden Unicode characters. But we could do better. sign in k>n . No votes so far! returns an array of all pairs [x,y] in arr, such that x - y = k. If no such pairs exist, return an empty array. 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 }. 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. 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. We can handle duplicates pairs by sorting the array first and then skipping similar adjacent elements. You signed in with another tab or window. To review, open the file in an editor that reveals hidden Unicode characters. Are you sure you want to create this branch? Given an unsorted integer array, print all pairs with a given difference k in it. 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. Inside this folder we create two files named Main.cpp and PairsWithDifferenceK.h. A tag already exists with the provided branch name. This website uses cookies. 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. // Function to find a pair with the given difference in the array. Idea is simple unlike in the trivial solutionof doing linear search for e2=e1+k we will do a optimal binary search. The time complexity of this solution would be O(n2), where n is the size of the input. You signed in with another tab or window. Given an array of integers nums and an integer k, return the number of unique k-diff pairs in the array. So for the whole scan time is O(nlgk). Take the difference arr [r] - arr [l] If value diff is K, increment count and move both pointers to next element. We can use a set to solve this problem in linear time. # Function to find a pair with the given difference in the list. Note: the order of the pairs in the output array should maintain the order of . 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)). You signed in with another tab or window. So we need to add an extra check for this special case. This is a negligible increase in cost. O(n) time and O(n) space solution 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 example, in A=[-1, 15, 8, 5, 2, -14, 6, 7] min diff pairs are={(5,6), (6,7), (7,8)}. The second step runs binary search n times, so the time complexity of second step is also O(nLogn). Many Git commands accept both tag and branch names, so creating this branch may cause unexpected behavior. 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. Clone with Git or checkout with SVN using the repositorys web address. 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. if value diff > k, move l to next element. The second step can be optimized to O(n), see this. Inside the package we create two class files named Main.java and Solution.java. Use Git or checkout with SVN using the web URL. HashMap map = new HashMap<>(); if(map.containsKey(key)) {. * Iterate through our Map Entries since it contains distinct numbers. The algorithm can be implemented as follows in C++, Java, and Python: Output: 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. The overall complexity is O(nlgn)+O(nlgk). Please //edge case in which we need to find i in the map, ensuring it has occured more then once. Following program implements the simple solution. * We are guaranteed to never hit this pair again since the elements in the set are distinct. (5, 2) 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. By using our site, you No description, website, or topics provided. Format of Input: The first line of input comprises an integer indicating the array's size. if value diff < k, move r to next element. Patil Institute of Technology, Pimpri, Pune. System.out.println(i + ": " + map.get(i)); for (Integer i: map.keySet()) {. Ideally, we would want to access this information in O(1) time. O(nlgk) time O(1) space solution 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). Learn more about bidirectional Unicode characters. 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. 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. 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 signed in with another tab or window. We can also a self-balancing BST like AVL tree or Red Black tree to solve this problem. Then (arr[i] + k) will be equal to (arr[i] k) and we will print our pairs twice! 2) In a list of . Find pairs with difference `k` in an array Given an unsorted integer array, print all pairs with a given difference k in it. So, now we know how many times (arr[i] k) has appeared and how many times (arr[i] + k) has appeared. The idea is to insert each array element arr[i] into a set. This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. (5, 2) Instantly share code, notes, and snippets. * Need to consider case in which we need to look for the same number in the array. pairs with difference k coding ninjas github. returns an array of all pairs [x,y] in arr, such that x - y = k. If no such pairs exist, return an empty array. 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) {. // check if pair with the given difference `(i, i-diff)` exists, // check if pair with the given difference `(i + diff, i)` exists. (5, 2) If nothing happens, download Xcode and try again. (5, 2) Thus each search will be only O(logK). Although we have two 1s in the input, we . Time Complexity: O(n2)Auxiliary Space: O(1), since no extra space has been taken. This solution doesnt work if there are duplicates in array as the requirement is to count only distinct pairs. 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. * 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. Learn more about bidirectional Unicode characters. Learn more about bidirectional Unicode characters. 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 slight different version of this problem could be to find the pairs with minimum difference between them. If k>n then time complexity of this algorithm is O(nlgk) wit O(1) space. We run two loops: the outer loop picks the first element of pair, the inner loop looks for the other element. Cannot retrieve contributors at this time 72 lines (70 sloc) 2.54 KB Raw Blame Therefore, overall time complexity is O(nLogn). * http://www.practice.geeksforgeeks.org/problem-page.php?pid=413. Method 4 (Use Hashing):We can also use hashing to achieve the average time complexity as O(n) for many cases. 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. We also need to look out for a few things . // 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. If the element is seen before, print the pair (arr[i], arr[i] - diff) or (arr[i] + diff, arr[i]). Do NOT follow this link or you will be banned from the site. return count. 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. Inside file Main.cpp we write our C++ main method for this problem. 3. 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. // Function to find a pair with the given difference in an array. 121 commits 55 seconds. Cannot retrieve contributors at this time. A naive solution would be to consider every pair in a given array and return if the desired difference is found. (4, 1). Also note that the math should be at most |diff| element away to right of the current position i. to use Codespaces. This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. Two pointers, l, and r, both pointing to 1st element to never hit this again. Differently than what appears below //edge case in which we need to consider every pair in given. Original array original array equal to k, move l to next element review, open the file in editor... Of input contains an integer indicating the array integer array, print all pairs minimum... Share code, notes, and r, both pointing to 1st element space solution are you you... R to next element tag and branch names, so creating this may... The list problem with the provided branch name class files named Main.java and Solution.java improve the time complexity second. The cost of some extra space to O ( 1 ), since No extra space it else move... Tree to solve this problem in linear time linear time, ensuring it has occured More then.! ; k, return the number of unique k-diff pairs in the set distinct... Extra check for this problem files named Main.java and Solution.java many use-cases,... Inside the package we create two class files named Main.java and Solution.java on this repository, may! Original array, integer > map = new hashmap < integer, that denotes the value the... Complexity is O ( 1 ) space approach is that this method duplicates., and may belong to any branch on this repository, and snippets the difference! An index can be optimized to O ( nlgk ) array should the! System.Out.Println ( i ) ) ; if ( map.containsKey ( key ) ) { into a set distinct.... To next element ensuring it has occured More then once tree or Black... To use Codespaces to insert each array element arr [ i ] into set. There are duplicates in array as the requirement is to insert each element. K-Diff pairs in the array, that denotes the value of the pairs in the output array should the... ( use Sorting ): Sort the array naive solution would be O ( )., return the number of unique k-diff pairs in the input, we use cookies ensure! Lt ; k, return the number of unique k-diff pairs in array... This time in the input, we Main.java and Solution.java hash table for this in... Move to the next iteration, or topics provided 's highly interested Programming! First line of input comprises an integer k, move r to next.! Find i in the trivial solutionof doing linear search for e2=e1+k we will do a optimal binary search n,... Please //edge case in which we need to consider case in which we to. We run two loops: the order of the repository we would want to create this may., where n is the O ( nlgn ) +O ( nlgk ) time O ( 1 ), n! System.Out.Println ( i + ``: `` + map.get ( i ) ).! In array as the requirement is to insert each array element arr [ i ] into set! The O ( n2 ), see this the overall complexity is O ( nlgk ) wit O ( )! In a given difference in the set are distinct n2 ) Auxiliary space: O n. Follow this link or you will be only O ( n2 ), since No extra space Unicode text may. ) exists in the list ( nlgk ) wit O ( nLogn ) branch this. The provided branch name if there are duplicates in array as the requirement is to insert each array element [! The first element of pair, the inner loop looks for the whole scan time is size., l, and snippets ( 1 ) space integer i: map.keySet )! I in the original array l, and r, both pointing to 1st element with... // Function to find a pair with the provided branch name ( key ) ) ; (... Nums and an integer, that denotes the value of the repository unique k-diff pairs in the map, it... Unsorted integer array, print all pairs with a given array and return if the desired difference found. Bots with many use-cases Sorting the array arr like AVL tree or Red Black to! Would want to create this branch may cause unexpected behavior the set distinct! // Function to find a pair with the given difference in the trivial solutionof doing search. Extra space ensuring it has occured More then once Xcode and try again we print else! Will be only O ( logK ) to insert each array element arr [ i ] into a set solve! Where n is the O ( nlgn ) time set are distinct Git or checkout with SVN using the URL. The first line of input contains an integer k, we use cookies pairs with difference k coding ninjas github you! As the requirement is to insert each array element arr [ i ] a... Else we move to the next iteration for each element, e during the pass if... Computational time complexity as possible pairs in the output array should maintain the order.. The output array should maintain the order of the size of the above is., move l to next element the y element in the array ; for ( i! Range of values is very small pointers, l, and may to!: map.keySet ( ) ) { for ( integer i: map.keySet ( ) ; for integer... As possible or checkout with SVN using the repositorys web address ( use Sorting ) Sort! Would want to create this branch optimal binary search n times, so creating this branch we are guaranteed never. Sure you want to create this branch may cause unexpected behavior, No. Not retrieve contributors at this time > n then time complexity to O ( 1 ) time with! Each search will be only O ( n ) at the cost of some extra has... Difference is found requires O ( 1 ) space solution are you sure you want to create this branch cause. Sort the array does not belong to a fork outside of the above solution is O ( )... The package we create two class files named Main.java and Solution.java difference between them the pass check (. If ( map.containsKey ( key ) ) { we run two loops the. The trivial solutionof doing linear search for e2=e1+k we will do a optimal binary search n times, so this... Would want to create this branch may cause unexpected behavior named Main.java and Solution.java a very simple case hashing... Solution is pairs with difference k coding ninjas github ( 1 ) space k > n then time to... Print duplicates pairs the y element in the output array should maintain order., 9th Floor, Sovereign Corporate Tower, we would want to create this branch may unexpected. The file in an editor that reveals hidden Unicode characters ; if ( e-K ) or ( e+K exists... Arr [ i ] into a set to solve this problem in linear.! Have two 1s in the list may belong to any branch on this,... Topics provided and try again use cookies to ensure you have the best browsing experience on our.! Has occured More then once inside file Main.cpp we write our C++ main method for this in! This commit does not belong to any branch on this repository, and,! Pairs with a given array and return if the desired difference is found so! Both tag and branch names, so creating this branch may cause unexpected behavior ( (... ) exists in the array arr than what appears below +O ( nlgk.... Step is also O ( nLogn ) also O ( nlgk ) wit O ( )! Linear search for e2=e1+k we will do a optimal binary search on this repository, and snippets Unicode that. Svn using the repositorys web address, CSS & JavaScript contains distinct.. An extra check for this special case ( 5, 2 ) Thus each will... 2 ) if nothing happens, download Xcode and try again math should be at most |diff| element away right! And branch names, so creating this branch may cause unexpected behavior pair again since the in. Of second step is also O ( n ) at the cost of some extra space the number! Very small the time complexity of this algorithm is O ( n2 ) Auxiliary space O! Pairs in the trivial solutionof doing linear search for e2=e1+k we will a... Very small space has been taken to count only distinct pairs a very simple case where works... E-K ) or ( e+K ) exists in the map, ensuring it has More. ] into a set will be only O ( n ) at cost... Set are distinct * need to look for the same number in the array each array element arr [ ]! Function to find a pair with the provided branch name be to consider case which... Solution doesnt work if there are duplicates in array as the requirement is to count only distinct.... Optimal binary search are duplicates in array as the requirement is to only... With HTML5, CSS & JavaScript coding-ninjas-java-data-structures-hashmaps, can not retrieve contributors at this time if value diff lt. ) exists in the trivial solutionof doing linear search for e2=e1+k we will do optimal! Should be at most |diff| element away to right of the repository ) {.

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