{"id":9,"date":"2019-02-05T14:40:31","date_gmt":"2019-02-05T14:40:31","guid":{"rendered":"https:\/\/www.kindsonthegenius.com\/coding\/?p=9"},"modified":"2019-02-05T17:19:58","modified_gmt":"2019-02-05T17:19:58","slug":"question-1-find-pair-with-given-sum-in-an-array","status":"publish","type":"post","link":"https:\/\/www.kindsonthegenius.com\/coding\/question-1-find-pair-with-given-sum-in-an-array\/","title":{"rendered":"Question 1: Find Pair With Given Sum in an Array"},"content":{"rendered":"

Question 1<\/strong><\/h4>\n

Given an array of integers, you must find a pair within the array that sums up to a given sum. We assume the array is unsorted.<\/p>\n

For example:<\/p>\n

Input:\u00a0 <\/strong><\/p>\n

Array A = {4, 9, 7, 3, 6, 8}<\/p>\n

Sum = 11<\/p>\n

Output: <\/strong><\/p>\n

Pair found at 0 and\u00a0 2 (4 +\u00a0 7)<\/p>\n

Pair found at 3 and 5 (3 + 8)<\/p>\n

    \n
  1. Solution 1: Inefficient Solution<\/a><\/li>\n
  2. Solution 2: Start with Sorting<\/a><\/li>\n
  3. Solution 3: Use a hashmap<\/a><\/li>\n<\/ol>\n

     <\/p>\n

    Solution 1: Inefficient Solution (Quadratic time)<\/strong><\/h4>\n

    In this solution, we loop through the array and for each element, we calculate the sum with every other element and compare with the given sum. If the sum equals the given sum. We output the indexes.<\/p>\n

    The Java program is given below:<\/p>\n

    \/\/Question 1: Given an unsorted array of integers<\/span>\r\n\/\/ find a pair of integers with the given sum in it<\/span>\r\n\r\npackage<\/span> examples;<\/span>\r\n\r\npublic<\/span> class<\/span> FindPairSum1<\/span>{<\/span>\r\n\tpublic<\/span> static<\/span> void<\/span> findPair1<\/span>(<\/span>int<\/span> A[],<\/span> int<\/span> sum)<\/span> {<\/span>\r\n\t\t\r\n\t\t\/\/loop until element before the last<\/span>\r\n\t\tfor<\/span>(<\/span>int<\/span> i =<\/span> 0<\/span>;<\/span> i<<\/span> A.<\/span>length<\/span>-<\/span>1<\/span>;<\/span> i++)<\/span> {<\/span>\r\n\t\t\t\r\n\t\t\t\/\/loop until the last element<\/span>\r\n\t\t\tfor<\/span>(<\/span>int<\/span> j =<\/span> i+<\/span>1<\/span>;<\/span> j <<\/span> A.<\/span>length<\/span>;<\/span> j++)<\/span> {<\/span>\r\n\t\t\t\t\r\n\t\t\t\t\/\/if given sum is found, print the indexes<\/span>\r\n\t\t\t\tif<\/span>(<\/span>A[<\/span>i]<\/span> +<\/span> A[<\/span>j]<\/span> ==<\/span> sum)<\/span> {<\/span>\r\n\t\t\t\t\tSystem.<\/span>out<\/span>.<\/span>println<\/span>(<\/span>\"Pair found at \"<\/span> +<\/span> i +<\/span> \" and \"<\/span> +<\/span> j);<\/span>\r\n\t\t\t\t}<\/span>\r\n\t\t\t}<\/span>\r\n\t\t}<\/span>\r\n\t}<\/span>\/\/End of findPair1<\/span>\r\n\t\r\n\tpublic<\/span> static<\/span> void<\/span> main<\/span>(<\/span>String args[])<\/span> {<\/span>\r\n\t\tint<\/span> ar[]<\/span> =<\/span> {<\/span>3<\/span>,<\/span>5<\/span>,<\/span>2<\/span>,<\/span>8<\/span>,<\/span>0<\/span>,<\/span>1<\/span>,<\/span>11<\/span>,<\/span>7<\/span>};<\/span>\r\n\t\tint<\/span> sum =<\/span> 10<\/span>;<\/span>\r\n\t\tfindPair1(<\/span>ar,<\/span> sum);<\/span>\r\n\t}<\/span>\/\/End of main\t\t<\/span>\r\n}<\/span>\r\n<\/pre>\n
    Solution 1:<\/strong> Inefficient Solution – Quadratic Time<\/p>\n
     <\/p>\n
    Solution 2: Start by Sorting (nlog(n) time)<\/strong><\/h4>\n
    In this solution, we first sort the array. As such, the largest elements\u00a0 are are the end of the array while the smallest elements are at the beginning of the array. Then we use two iterators. The first counting from the beginning of the array while the second counting from the end of the array.<\/p>\n
    For each iteration,\u00a0 we calculate the sum to check if it equals the given sum<\/p>\n
    The Java program is given below.<\/p>\n
    Note: The index returned with this method is the index of the elements in the sorted array which is not the same as the indexes in the original array.<\/p>\n
    \/\/findPair2: Sorts the array. Maintain two indices of the two end-points<\/span>\r\n\/\/Complexity: O(nlogn)<\/span>\r\n\r\npackage<\/span> examples;<\/span>\r\nimport<\/span> java.util.Arrays<\/span>;<\/span>\r\npublic<\/span> class<\/span> FindPairSum2<\/span> {<\/span>\r\n\r\n\tpublic<\/span> static<\/span> void<\/span> findPair2<\/span> (<\/span>int<\/span> A[],<\/span> int<\/span> sum)<\/span> {<\/span>\r\n\t\tArrays.<\/span>sort<\/span>(<\/span>A);<\/span> \/\/First sort the array<\/span>\r\n\t\t\r\n\t\tint<\/span> low =<\/span> 0<\/span>;<\/span>\r\n\t\tint<\/span> high =<\/span> A.<\/span>length<\/span> -<\/span> 1<\/span>;<\/span>\r\n\t\t\r\n\t\twhile<\/span>(<\/span>low <<\/span> high)<\/span> {<\/span>\r\n\t\t\tif<\/span>(<\/span>A[<\/span>low]<\/span> +<\/span> A[<\/span>high]<\/span> ==<\/span> sum)<\/span> {<\/span>\r\n\t\t\t\tSystem.<\/span>out<\/span>.<\/span>println<\/span>(<\/span>\"Sum found are \"<\/span> +<\/span> A[<\/span>low]<\/span> +<\/span> \" and \"<\/span> +<\/span> A[<\/span>high]);<\/span>\r\n\t\t\t\tlow++;<\/span>\r\n\t\t\t}<\/span>\r\n\t\t\telse<\/span> if<\/span>(<\/span>A[<\/span>low]<\/span> +<\/span> A[<\/span>high]<\/span> <<\/span> sum)<\/span> {<\/span>\r\n\t\t\t\tlow =<\/span> low +<\/span> 1<\/span>;<\/span>\r\n\t\t\t}<\/span>\r\n\t\t\telse<\/span> {<\/span>\r\n\t\t\t\thigh =<\/span> high  -<\/span> 1<\/span>;<\/span>\r\n\t\t\t}<\/span>\r\n\t\t}<\/span> \/\/end of while loop<\/span>\r\n\t}<\/span>\r\n\t\t\t\t\r\n\r\n\tpublic<\/span> static<\/span> void<\/span> main<\/span>(<\/span>String[]<\/span> args)<\/span> {<\/span>\r\n\t\tint<\/span> ar[]<\/span> =<\/span> {<\/span>3<\/span>,<\/span>5<\/span>,<\/span>2<\/span>,<\/span>8<\/span>,<\/span>0<\/span>,<\/span>1<\/span>,<\/span>11<\/span>,<\/span>7<\/span>};<\/span>\r\n\t\tint<\/span> sum =<\/span> 10<\/span>;<\/span>\r\n\t\tfindPair2(<\/span>ar,<\/span> sum);<\/span>\r\n\t}<\/span>\r\n}<\/span>\r\n<\/pre>\n
    Solution 2: Maintain two indexes<\/p>\n
     <\/p>\n
    Solution 3 – Use of HashMap (Linear time O(n))<\/strong><\/h4>\n
    In this solution, we iterate though the array just once and store each element in a hash set.. For each element, we would check if the complement of the element is already\u00a0 in the hashset. If it is there, then we have\u00a0 a sum. If it is not there then we insert it.<\/p>\n
    The Java implementation is given below.<\/p>\n
    \/\/Find pair with given sum using a hashma[<\/span>\r\npackage<\/span> examples;<\/span>\r\n\r\nimport<\/span> java.util.HashMap<\/span>;<\/span>\r\nimport<\/span> java.util.Map<\/span>;<\/span>\r\n\r\npublic<\/span> class<\/span> FindPairSum3<\/span> {<\/span>\r\n\t\r\n\tpublic<\/span> static<\/span> void<\/span> findPair<\/span>(<\/span>int<\/span> A[],<\/span> int<\/span> sum)<\/span> {<\/span>\r\n\t\t\r\n\t\tMap<<\/span>Integer,<\/span> Integer><\/span> map =<\/span> new<\/span> HashMap<>();<\/span>\r\n\t\t\r\n\t\tfor<\/span>(<\/span>int<\/span> i =<\/span> 0<\/span>;<\/span> i<<\/span> A.<\/span>length<\/span>;<\/span> i++)<\/span> {<\/span>\r\n\t\t\t\r\n\t\t\t\/\/if complement is already in map, print the pair<\/span>\r\n\t\t\tif<\/span>(<\/span>map.<\/span>containsKey<\/span>(<\/span>sum-<\/span>A[<\/span>i]))<\/span> {<\/span>\r\n\t\t\t\tSystem.<\/span>out<\/span>.<\/span>println<\/span>(<\/span>\"Pair found at \"<\/span> +<\/span> map.<\/span>get<\/span>(<\/span>sum-<\/span>A[<\/span>i])<\/span> +<\/span> \" and \"<\/span> +<\/span> i);<\/span>\r\n\t\t\t}<\/span>\r\n\t\t\t\r\n\t\t\tmap.<\/span>put<\/span>(<\/span>A[<\/span>i],<\/span> i);<\/span>\r\n\t\t\t\r\n\t\t}<\/span>\/\/end of for<\/span>\r\n\t\t\r\n\t}<\/span>\/\/end for function<\/span>\r\n\t\r\n\r\n\tpublic<\/span> static<\/span> void<\/span> main<\/span>(<\/span>String[]<\/span> args)<\/span> {<\/span>\r\n\t\tint<\/span> ar[]<\/span> =<\/span> {<\/span>3<\/span>,<\/span> 20<\/span>,<\/span> 5<\/span>,<\/span> 6<\/span>,<\/span>1<\/span>,<\/span>9<\/span>,<\/span> 7<\/span>,<\/span> 22<\/span>};<\/span>\r\n\t\tint<\/span> sum =<\/span> 29<\/span>;<\/span>\r\n\t\t\r\n         findPair(<\/span>ar,<\/span> sum);<\/span>\r\n\t}<\/span>\r\n\r\n}<\/span>\r\n<\/pre>\n
    Solution 3<\/strong> : Using a HashMap<\/p>\n","protected":false},"excerpt":{"rendered":"
    Question 1 Given an array of integers, you must find a pair within the array that sums up to a given sum. We assume the … <\/p>\n","protected":false},"author":395,"featured_media":14,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[8,2,3],"tags":[4,5],"class_list":["post-9","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-array","category-coding-challenge","category-java","tag-hackerrank","tag-java-problems-and-solutions"],"yoast_head":"\nQuestion 1: Find Pair With Given Sum in an Array - Coding Challenge<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.kindsonthegenius.com\/coding\/question-1-find-pair-with-given-sum-in-an-array\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Question 1: Find Pair With Given Sum in an Array - Coding Challenge\" \/>\n<meta property=\"og:description\" content=\"Question 1 Given an array of integers, you must find a pair within the array that sums up to a given sum. 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