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Nov 24, 2013 · I was solving the problem on hackerrank. I had two approaches in my mind : input : unsorted array (a) and k. First Approach : 1) Sort the array. 2) for each array element a [i] ,find the element a [i]+K using binary search.If found increament the count and break the inner loop. Second Approach : 1) Sort the array. 2) for each array element a [i ...
Dec 11, 2016 · second line (as an array): [1, 5, 3, 4, 2]; How many pairs in the array have a difference of K ? sample answer: "There are 3 pairs of integers in the set with a difference of 2." here is something I'm working on for this, but I am missing something :
Aug 28, 2021 · Sort the array compare 2 neighbour numbers (i, i+1): if difference between them is equal to given number k, increment result if the difference is less than given number, then compare i+1 with 0,...,i-1 elements to try to find possible pair among those. if the difference is grater than given number k, then increment i and do the same for the next pair.
I've arrived at a solution I'm content with in terms of exploiting Kotlin's function programming capabilities. However, I'm not getting the expected answer... Problem summary: Given an Array: -> find and return the total number of pairs. i.e: input-> [10, 20, 20, 10, 10, 30, 50, 10, 20] number of pairs-> 3
Apr 28, 2023 · Instead of storing the data in an array of values, store in an array of deltas. There are many ways of doing this. Suppose that your array looks like: A = [a0, a1, a2, a3, a4, a5, a6, a7, a8, a9] Transform it into pairs ai-aj where j is the first previous number divisible by a higher power of 2. Like this.
I have a problem which is on Hackerrank it goes like DIVISIBLE SUM PAIRS. I will anyhow give the problem statement here: Problem Statement. Given an array, where we have to find the number of pairs divisible by the given number k, and there is one more condition to it, which is : the arr[i] < arr[j] from those pairs. Example
Feb 17, 2018 · For the 3rd or 4th case I construct a new array one element smaller in size than the input array and with a 1 as the last element. Now, the result of arr = 55 68 31 80 57 18 34 28 76 55 according to Hackerrank should be 508. But I get 564. Since it has to be 508 I guess the array should be 1 68 1 80 1 1 34 1 76 1. For other arrays I get the ...
Dec 25, 2020 · Let the input array be of B-bit numbers. Here is an easy to understand and implement solution. Eliminate all values <= k. The above image shows 5 10-bit numbers. Step 1: Adjacency Graph. Store a list of set bits. In our example, 7th bit is set for numbers at index 0,1,2,3 in the input array. Step 2: The challenge is to avoid counting the same ...
Jan 27, 2014 · Thanks a lot :) how this condition is true for array [2,3], xor of 2 and 3 is 1, but the result gives 1+2=3. @RaviSevta This would give 1*1*1 + 2*0*2 = 1 which is correct. @RaviSevta It's 1*2*1 + 1*2*2 = 6. For each bit position, the sum of number of 1-bits and 0-bits must be 3 because there are 3 numbers in the list.
Aug 2, 2019 · Give an array of integers, compute the maximum difference between any item and any lower indexed smaller item for all possible pairs. In other words, for the array arr, find the maximum value of arr[j] - arr[i] for all i, j where 0 <= i < j < n and arr[i] < arr[j]. If no item has a smaller item with a lower index, then return -1.
