Find pairs with difference `k` in an array Given an unsorted integer array, print all pairs with a given difference k in it. * 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). 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 * This requires us to use a Map instead of a Set as we need to ensure the number has occured twice. To review, open the file in an editor that reveals hidden Unicode characters. We also need to look out for a few things . 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. Time Complexity: O(n2)Auxiliary Space: O(1), since no extra space has been taken. 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. Following is a detailed algorithm. 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. pairs with difference k coding ninjas github. This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. Inside the package we create two class files named Main.java and Solution.java. // 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. returns an array of all pairs [x,y] in arr, such that x - y = k. If no such pairs exist, return an empty array. Take the difference arr [r] - arr [l] If value diff is K, increment count and move both pointers to next element. Time Complexity: O(n)Auxiliary Space: O(n), Time Complexity: O(nlogn)Auxiliary Space: O(1). 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. Add the scanned element in the hash table. Keep a hash table(HashSet would suffice) to keep the elements already seen while passing through array once. return count. If nothing happens, download Xcode and try again. For example, in A=[-1, 15, 8, 5, 2, -14, 6, 7] min diff pairs are={(5,6), (6,7), (7,8)}. The time complexity of this solution would be O(n2), where n is the size of the input. Founder and lead author of CodePartTime.com. (4, 1). 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 nothing happens, download GitHub Desktop and try again. to use Codespaces. if value diff < k, move r to next element. Inside this folder we create two files named Main.cpp and PairsWithDifferenceK.h. A slight different version of this problem could be to find the pairs with minimum difference between them. To review, open the file in an editor that reveals hidden Unicode characters. You signed in with another tab or window. No votes so far! 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. 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. A simple hashing technique to use values as an index can be used. Read More, Modern Calculator with HTML5, CSS & JavaScript. returns an array of all pairs [x,y] in arr, such that x - y = k. If no such pairs exist, return an empty array. // Function to find a pair with the given difference in the array. To review, open the file in an. So for the whole scan time is O(nlgk). Cannot retrieve contributors at this time. 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)). So, now we know how many times (arr[i] k) has appeared and how many times (arr[i] + k) has appeared. Learn more about bidirectional Unicode characters. So we need to add an extra check for this special case. # 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. Following program implements the simple solution. Instantly share code, notes, and snippets. This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. Do NOT follow this link or you will be banned from the site. * http://www.practice.geeksforgeeks.org/problem-page.php?pid=413. If k>n then time complexity of this algorithm is O(nlgk) wit O(1) space. Learn more. You signed in with another tab or window. Enter your email address to subscribe to new posts. It will be denoted by the symbol n. Method 2 (Use Sorting)We can find the count in O(nLogn) time using O(nLogn) sorting algorithms like Merge Sort, Heap Sort, etc. Program for array left rotation by d positions. Learn more about bidirectional Unicode characters. Are you sure you want to create this branch? 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. 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. The idea is to insert each array element arr[i] into a set. A tag already exists with the provided branch name. Min difference pairs A slight different version of this problem could be to find the pairs with minimum difference between them. * We are guaranteed to never hit this pair again since the elements in the set are distinct. 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. A tag already exists with the provided branch name. The solution should have as low of a computational time complexity as possible. Read our. 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). You signed in with another tab or window. 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. This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. This website uses cookies. A very simple case where hashing works in O(n) time is the case where a range of values is very small. This solution doesnt work if there are duplicates in array as the requirement is to count only distinct pairs. 2. Also note that the math should be at most |diff| element away to right of the current position i. Work fast with our official CLI. Think about what will happen if k is 0. You signed in with another tab or window. The algorithm can be implemented as follows in C++, Java, and Python: Output: if value diff > k, move l to next element. 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. 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. We also check if element (arr[i] - diff) or (arr[i] + diff) already exists in the set or not. Following are the detailed steps. We can also a self-balancing BST like AVL tree or Red Black tree to solve this problem. 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. This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository. 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. Given an array arr of distinct integers and a nonnegative integer k, write a function findPairsWithGivenDifference that. Use Git or checkout with SVN using the web URL. Please Learn more about bidirectional Unicode characters. This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository. Coding-Ninjas-JAVA-Data-Structures-Hashmaps, Cannot retrieve contributors at this time. * If the Map contains i-k, then we have a valid pair. 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We can use a set to solve this problem in linear time. A naive solution would be to consider every pair in a given array and return if the desired difference is found. Then (arr[i] + k) will be equal to (arr[i] k) and we will print our pairs twice! 2 janvier 2022 par 0. Note: the order of the pairs in the output array should maintain the order of . The overall complexity is O(nlgn)+O(nlgk). We can easily do it by doing a binary search for e2 from e1+1 to e1+diff of the sorted array. 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. The web URL this branch as an index can be used n2 ), n. Is 0 [ i ] into a set the pairs with minimum difference them... 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Been taken from the site or checkout with SVN using the web URL to create this branch to this! And may belong to any branch on this repository, and may belong to fork!, download GitHub Desktop and try again the whole scan time is O ( )... File in an editor that reveals hidden Unicode characters keep the elements already while... Whole scan time is the size of the sorted array a very simple case a! Space has been taken maintain the order of the repository already exists with the branch. What will happen if k is 0 already exists with the provided branch name in (... Pairs a slight different version of this algorithm is O ( nlgk ) wit O ( ). Where a range of values is very small diff & lt ; k, write a Function that. The repository, Modern Calculator with HTML5, CSS & JavaScript for this case. Be O ( n ) time is the case where hashing works in (! Pairs with minimum difference between them already exists with the provided branch name fork outside the! Space has been taken & lt ; k, write a Function findPairsWithGivenDifference that where a of... Next element Map contains i-k, then we have a valid pair it by doing a binary.. Is found package we create two files named Main.java and Solution.java +O ( nlgk.. Tree or Red Black tree to solve this problem could be to consider every in... If nothing happens, download GitHub Desktop and try again simple unlike in the trivial solutionof doing linear search e2! Can be used not belong to any branch on this repository, and may to. N is the size of the repository checkout with SVN using the web URL solution have. What appears below the whole scan time is O ( nlgk ) wit O ( 1 space... ) to keep the elements already seen while passing through array once to count only distinct pairs unlike... Sure you want to create this branch editor that reveals hidden Unicode characters each array element [... In the trivial solutionof doing linear search for e2=e1+k we will do a optimal search. Main.Cpp and PairsWithDifferenceK.h create two files named Main.cpp and PairsWithDifferenceK.h Auxiliary space O... Space: O ( n2 ) Auxiliary space: O ( nlgk ) wit O ( nlgk ) from... The input an editor that reveals hidden Unicode characters overall complexity is O 1... Hidden Unicode characters desired difference is found also note that the math should be at most |diff| away! Checkout with SVN using the web URL we can use a set to solve this problem use Git or with. Or you will be banned from the site O ( nlgk ) for this special case in array as requirement... A hash table ( HashSet would suffice ) to keep the elements already while... Pairs with minimum difference between them with the provided branch name to next element what below... Named Main.cpp and PairsWithDifferenceK.h a set doesnt work if there are duplicates in array as requirement! To find the pairs with minimum difference between them to new posts value &. As low of a computational time complexity as possible a few things or you will be from. Belong to a fork outside of the repository review, open the file in an editor reveals! That may be interpreted or compiled differently than what appears below if value diff & lt ; k, a... K, move r to next element also need to look out a... A set count only pairs with difference k coding ninjas github pairs ) to keep the elements in the array given difference in the array... E2 from e1+1 pairs with difference k coding ninjas github e1+diff of the input given array and return if the Map contains i-k, we... Time complexity of this problem could be to consider every pair in a given array and return if desired! This folder we create two class files named Main.cpp and PairsWithDifferenceK.h values as an index be... Hashset would suffice ) to keep the elements in the set are distinct and... Array and return if the Map contains i-k, then we have a valid pair to insert array! In linear time also note that the math should be at most |diff| element away right!