Suppose we have a number N, we have to find all factors of N and return the product of four factors of N such that: The sum of the four factors is same as N. The product of the four factors is maximum. All four factors can be equal to each other to maximize the product.So, if the input is like N = 60, then the output will be All the factors are -> 1 2 3 4 5 6 10 12 15 20 30 60 and product is 50625, as we 15 has been selected four times to make ... Read More

Suppose we have a sorted doubly linked list of unique positive numbers; we have to find pairs in the doubly linked list whose product is same as a given value x. We have to keep in mind that, this will be solved without consuming any extra space.So, if the input is like L = 1 ⇔ 2 ⇔ 4 ⇔ 5 ⇔ 6 ⇔ 8 ⇔ 9 and x = 8, then the output will be (1, 8), (2, 4)To solve this, we will follow these steps −curr := head, nxt := headwhile nxt.next is not None is non-zero, donxt ... Read More

Suppose we have an array A where GCD of every possible pair of elements of another array is given, we have to find the original numbers which are used to compute the given GCD array.So, if the input is like A = [6, 1, 1, 13], then the output will be [13, 6] as gcd(13, 13) is 13, gcd(13, 6) is 1, gcd(6, 13) is 1, gcd(6, 6) is 6To solve this, we will follow these steps −n := size of Asort array A in descending orderoccurrence := an array of size A[0] and fill with 0for i in range ... Read More

Suppose we have a sequence of numbers called bn, this is represented using a recurrence relation like b1=1 and bn+1/bn=2n . We have to find the value of log2(bn) for a given n.So, if the input is like 6, then the output will be 5 as log2(bn) = (n * (n - 1)) / 2 = (6*(6-1))/2 = 15We can solve this problem by solving this relation as follows −bn+1/bn = 2nbn/bn-1 = 2n-1…b2/b1 = 21 , If we multiply all of above, we can get(bn+1/bn).(bn/bn-1)……(b2/b1) = 2n + (n-1)+……….+1So, bn+1/b1 = 2n(n+1)/2As 1 + 2 + 3 + ………. ... Read More

Suppose we have a string whose length is m, and this string is containing only lowercase letters, we have to find the n-th permutation of string lexicographically.So, if the input is like string = "pqr", n = 3, then the output will be "qpr" as all permutations are [pqr, prq, qpr, qrp, rpq, rqp], they are in sorted order.To solve this, we will follow these steps −MAX_CHAR := 26MAX_FACT := 20factorials := an array of size MAX_FACTfactorials[0] := 1for i in range 1 to MAX_FACT, dofactorials[i] := factorials[i - 1] * isize := size of stringoccurrence := an array of ... Read More

Suppose we have two integers N and K; we have to find N unique values whose bit-wise OR is same as K. If there is no such result, then return -1So, if the input is like N = 4 and K = 6, then the output will be [6, 0, 1, 2].To solve this, we will follow these steps −MAX := 32visited := a list of size MAX and fill with Falseres := a new listDefine a function add() . This will take numpoint := 0value := 0for i in range 0 to MAX, doif visited[i] is non-zero, thengo for ... Read More

Suppose we have an array A of n unique numbers, these n elements are present in the array in ascending order, but there is one missing element. We have to find the missing element.So, if the input is like A = [1, 2, 3, 4, 5, 6, 7, 9], then the output will be 8.To solve this, we will follow these steps −n := size of Aleft := 0right := n - 1mid := 0while right > left , domid := left +(right - left) / 2if A[mid] - mid is same as A[0], thenif A[mid + 1] - A[mid] ... Read More

Suppose we have an array of jobs with different time requirements, there are k different persons to assign jobs and we also have how much time t an assignee takes to do one unit of the job. We have to find the minimum time to complete all jobs with following constraints.An assignee can be assigned only the contiguous jobs.Two assignees cannot share or perform a single job.So, if the input is like k = 4, t = 5, job = {12, 6, 9, 15, 5, 9}, then the output will be 75 as we get this time by assigning [12], ... Read More

Suppose we have an array of positive numbers; we replace each element from that array array so that the difference between two adjacent elements in the array is either less than or equal to a given target. Now, we have to minimize the adjustment cost, so the sum of differences between new value and old value. More precisely, we minimize ∑|A[i] – Anew[i]| where i in range 0 to n-1, here n is denoted as size of A and Anew is the array with adjacent difference less than or equal to target.So, if the input is like [56, 78, 53, ... Read More

Suppose we have Binary Search Tree(BST), we have to find median of it. We know for even number of nodes, median = ((n/2th node + (n+1)/2th node) /2 For odd number of nodes, median = (n+1)/2th node.So, if the input is likethen the output will be 7To solve this, we will follow these steps −if root is same as None, thenreturn 0node_count := number of nodes in the treecount_curr := 0current := rootwhile current is not null, doif current.left null, thencount_curr := count_curr + 1if node_count mod 2 is not 0 and count_curr is same as(node_count + 1) /2, thenreturn ... Read More