Suppose there is a number line from 1 to n. At first we are at position 0, jump one step to go 1, then jump two place to reach at position 3, then jump three position to reach at 6 and so on. We have to check whether maintaining this, we can reach at position n or not.So, if the input is like n = 21, then the output will be True, because 1+2+3+4+5+6 = 21To solve this, we will follow these steps −j:= (1 + square root of (1+8*n)) / 2if |j - int part of j|

Suppose we have an undirected graph G with n nodes. Now consider the cost of a simple undirected graph is the sum of the costs of its nodes. And The cost of a node is D^k, where D is its degree. Now we have n and k values. We have to find the sum of the costs of all possible simple undirected graphs with n nodes. The result may be very large, so return the result modulo 1005060097.So, if the input is like n = 3 k = 2, then the output will be 36 because, there are eight simple ... Read More

Suppose we have a set A where all elements from 1 to n are present. And P(A) represents all permutations of elements present in A. We have to find number of elements in P(A) which satisfies the given conditionsFor all i in range [1, n], A[i] is not same as iThere exists a set of k indices {i1, i2, ... ik} such that A[ij] = ij+1 for all j < k and A[ik] = i1 (cyclic).So, if the input is like n = 3 k = 2, then the output will be 0 because -Consider the Array's are 1 indexed. ... Read More

Suppose we have a string with n number of A's and 2n number of B's. We have to find number of arrangements possible such that in each prefix and each suffix have number of B's greater than or equal to the number of A'sSo, if the input is like n = 2, then the output will be 4 because there are two A's and four B's, so possible arrangements are [BBAABB, BABABB, BBABAB, BABBAB].To solve this, we will follow these steps −Define a method solve, this will take nif n is same as 1, thenreturn 1if n is same as ... Read More

Suppose there is a stair case with N steps. One can either go step by step, or at each step, one can take a jump of at most N steps. We have to find the number of ways in which we can go to the top floor. The N value may be large we are only interested in the first and the last K digits of number of ways.So, if the input is like N = 10 k = 2, then the output will be 63 because there are 10 steps, if there are S number of ways in which ... Read More

Suppose we are at at (0, 0) position in the Cartesian plane. We want to go to the point (x, y) using only horizontal(H) and vertical(V) moves of single unit. There are more than one possible ways to reach destination. Each way comprises of few H moves and few V moves. (For example if we want to go to point (2, 2) from point (0, 0), then HVVH is one of the possible ways.) If we have another value k, we have to find the lexicographically kth smallest way of going to the destination.So, if the input is like (x, ... Read More

Suppose we have m number of letters and another value n. We have to count number of strings of length n created with letters by taking from these m letters, and string has no palindromic substring of length greater than 1. If answer is too large then mod the result by 10^9+7.So, if the input is like n = 2 m = 3, then the output will be 6 because m = 3, so if the alphabets are {x, y, z}, we can generate strings like: [xx, xy, xz, yx, yy, yz, zx, zy, zz] but [xx, yy, zz] are ... Read More

Suppose we have an array nums and a value k. We have to find number of consecutive subsequences whose sum is divisible by k.So, if the input is like k = 3 nums = [1, 2, 3, 4, 1], then the output will be 4 because the subsequences are [3], [1, 2], [1, 2, 3] and [2, 3, 4].To solve this, we will follow these steps −x := An array of size k and fill with 0x[0] := 1r:= 0, s:= 0for each elem in nums, dos :=(s + elem) mod kr := r + x[s]x[s] := x[s] + 1return ... Read More

Suppose we have an array nums. We have to find number of pairs (i, j) are there such that nums[i] = nums[j] but i is not same as j.So, if the input is like nums = [1, 3, 1, 3, 5], then the output will be 4, because the pairs are (0, 2), (2, 0), (1, 3) and (3, 1)To solve this, we will follow these steps −d := a new mapfor each c in nums, dod[c] := (d[c] + 1) when c is present in d otherwise 1res := 0for each c is in the list of elements (x ... Read More

Suppose we have n balls which are numbered by an array nums, whose size is n and nums[i] represents the number of ball i. Now we have another value k. In each turn we pick k balls from n different balls and find the difference of maximum and minimum values of k balls and store the difference in a table. Then put these k balls again into that pot and pick again until we have selected all possible selections. Finally find the sum of all differences from the table. If the answer is too large, then return result mod 10^9+7.So, ... Read More