We run two loops: the outer loop picks the first element of pair, the inner loop looks for the other element. This website uses cookies. By using this site, you agree to the use of cookies, our policies, copyright terms and other conditions. // 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. Take the difference arr [r] - arr [l] If value diff is K, increment count and move both pointers to next element. 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. There was a problem preparing your codespace, please try again. return count. Inside file PairsWithDifferenceK.h we write our C++ solution. Input Format: The first line of input contains an integer, that denotes the value of the size of the 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) {. Work fast with our official CLI. We can improve the time complexity to O(n) at the cost of some extra space. // Function to find a pair with the given difference in an array. (5, 2) This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. The first step (sorting) takes O(nLogn) time. 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) {. 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). Use Git or checkout with SVN using the web URL. Many Git commands accept both tag and branch names, so creating this branch may cause unexpected behavior. You signed in with another tab or window. A tag already exists with the provided branch name. We are sorry that this post was not useful for you! // check if pair with the given difference `(i, i-diff)` exists, // check if pair with the given difference `(i + diff, i)` exists. 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. (5, 2) Read More, Modern Calculator with HTML5, CSS & JavaScript. Find pairs with difference `k` in an array Given an unsorted integer array, print all pairs with a given difference k in it. 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 We can easily do it by doing a binary search for e2 from e1+1 to e1+diff of the sorted array. It will be denoted by the symbol n. //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)). HashMap
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