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Pairs | HackerRank Solutions. Problem Statement : Given an array of integers and a target value, determine the number of pairs of array elements that have a difference equal to the target value.
Oct 16, 2020 · Efficient Solution. An efficient way to solve the problem requires some reverse thought process. Instead of finding a pair, we can find the complimentary integer to an element that would make the difference equal to the target value. (Element\ in\ array) - (Number\ to\ find) = target\ value.
Given an array of integers and a target value, determine the number of pairs of array elements that have a difference equal to the target value. Example. There are three values that differ by : , , and . Return . Function Description. Complete the pairs function below. pairs has the following parameter(s): int k: an integer, the target difference
Jun 28, 2020 · Hackerrank - Pairs Solution. You will be given an array of integers and a target value. Determine the number of pairs of array elements that have a difference equal to a target value. For example, given an array of [1, 2, 3, 4] and a target value of 1, we have three values meeting the condition: , , and . Function Description.
Sep 2, 2019 · 1. You will be given an array of integers and a target value k. Determine the number of pairs of array elements that have a difference equal to k. We will create a dictionary to save the...
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Missing Numbers is a programming challenge on HackerRank. You are given an array of unique integers and a target value. You need to find out the number of pa...
Aug 31, 2015 · HackerRank 'Pairs' Solution – Martin Kysel – Coding Challenges and More. Martin Kysel · August 31, 2015. coding-challenge hackerrank python. Short Problem Definition: Given N integers, count the number of pairs of integers whose difference is K. Link. Pairs. Complexity: time complexity is O(N\*log(N)) space complexity is O(N) Execution:
def pairs (k, arr): # Convert the list to a set for O(1) lookups arr_set = set (arr) count = 0 # Check for each element if there's a pair with difference k for x in arr: if x + k in arr_set: count += 1 if x-k in arr_set: count += 1 # Since each pair is counted twice, we return half the count return count // 2
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