Suppose we have a ring, which is made of few numbers from 1 to N. We also have tow numbers A and B. Now, we can stand at any place (say x) and perform the count operation with respect of the total sum of the distance (say Z = distance from X to A + distance from X to B). We have to select X such that Z is minimized. At the end return the value of Z. We have to keep in mind that X will not be same as A and B.So, if the input is like N ... Read More

Suppose we have two numbers A and B, we have to find the minimum positive number M so that M is divisible by A and the sum of the digits of M is same as B. So, if there is no such result, then return -1.So, if the input is like A = 50, B = 2, then the output will be 200 as this is divisible by 50 and sum of its digit = 2 + 0 + 0 = 2.To solve this, we will follow these steps −Define one element type container, that contains two numbers a and ... Read More

Suppose we have an array A of numbers, in A the i-th number is the position where an island is present, and another integer k is given (1 ≤ k < N). Now, a person is standing on the 0-th island and has to reach the last island, by jumping from one island to another in exactly k jumps, we have to find the minimum of the maximum length of a jump a person will make in his/her journey. We have to keep in mind that the position of all the islands are given in ascending order.So, if the input ... Read More

Suppose we have two strings P and Q of same lengths with only lower case letters, we have to count the minimum number of pre-processing moves on string P are needed to make it equal to string Q after applying below operations −Select any index i and swap characters pi and qi.Select any index i and swap characters pi and pn – i – 1.Select any index i and swap characters qi and qn – i – 1.Note − The value of i in range (0 ≤ i < n)In one move we can change a character in P with any ... Read More

Suppose we have one N X N matrix M, and this is filled with 1, 0, 2, 3, We have to find the minimum numbers of moves required to move from source cell to destination cell. While visiting through blank cells only, we can visit up, down, right and left.Cell with value 1 indicates Source.Cell with value 2 indicates Destination.Cell with value 3 indicates Blank cell.Cell with value 0 indicates Blank Wall.There will be only one source and only one destination cells. There may be more than one path to reach destination from source cell. Now, each move in matrix ... Read More

Suppose we have a given unsorted array A[0..n-1] of size n, we have to find the minimum length subarray A[s..e] so that by sorting this subarray the whole array will be sorted. So, if the array is like [2, 6, 4, 8, 10, 9, 15], then the output will be 5. The subarray will be [6, 4, 8, 10, 9].To solve this, we will follow these steps −res := sort the nums as an arrayans := 0set r as a linked listfor i in range 0 to length of resif nums[i] is not same as res[i], then insert i into ... Read More

Suppose we have two numbers p and q, we have to find the minimum difference between any terms in the shifted infinite tables of p and q, these shifts are r and s, where r, s >= 0.So, if the input is like p = 7 and q = 17, r = 6 and s = 3, then the output will be 0 as, table of 7 = [7, 14, 21, 28, 35, 42, 49, ...] and table of 17 = [17, 34, 51, 68, 85, 102, 119, ...], then shifted table of 7 will be [13, 20, 27, 34, ... Read More

Suppose there is a candy store where N different types of candies are available and the prices of all N different types of candies are given. The store also provides an attractive offer. According to this offer, we can buy a single candy from the store and get maximum of K different types of other candies for free. We have to find the minimum amount of money we have to spend to buy all the N different types of candies. We also have to find maximum amount of money we have to spend to buy all the N different types ... Read More

Suppose we have an array of size n, if the elements in the array, are in range from 0 to k-1. Where k is denoted as a positive integer and k max_val, thenmax_val := A[i]result := ireturn resultExample Let us see the following implementation to get better understanding − Live Demodef get_max_repeating(A, k):    n = len(A)    for i in range(n):       A[A[i]%k] += k    max_val = A[0]    result = 0    for i in range(1, n):       if A[i] > max_val:          max_val = A[i]          result ... Read More

Suppose we have a given number N that in range (1= j and table[i - j] != -1):             table[i] = max(table[i], table[i - j] + 1)    return table def max_summ(table, n):    if (n < max_val):       return table[n]    else:       t = int((n - max_val) / 4)+ 1       return t + table[n - 4 * t] n = 16 table = pre_calc() print(max_summ(table, n))Input16Output4