Server Side Programming Articles - Page 1805 of 2646

Suppose we have a set of items: the i-th item has value values[i] and label labels[i]. Then, we will take a subset S of these items, such that −|S|

Suppose we have a set of tiles, where each tile has one letter tiles[i] printed on it. Find the number of possible non-empty sequences of letters that we can make. So if the input is “AAB”, then the output will be 8. As sequences are “A”, “B”, “AA”, “AB”, “BA”, “AAB”, “ABA”, “BAA”To solve this, we will follow these steps −Define one dfs(), that will take countsum := 0for i in range 1 to 26if count[i] = 0, then go for next iteration, without checking the restdecrease count[i] by 1, and increase sum by 1sum := sum + dfs(count)increase count[i] ... Read More

Suppose we have a matrix consisting of 0s and 1s, we can choose any number of columns in the matrix and flip every cell in that column. converting a cell changes the value of that cell from 0 to 1 or from 1 to 0. we have to find the maximum number of rows that have all values equal after some number of flips. So if the matrix is like −000001110The output will be 2. This is because after converting values in the first two columns, the last two rows have equal values.To solve this, we will follow these steps ... Read More

Suppose in a warehouse, there is a row of barcodes. The i-th barcode is barcodes[i]. We have to rearrange the barcodes so that no two adjacent barcodes are same. So if the input is [1, 1, 1, 2, 2, 2] so output is [2, 1, 2, 1, 2, 1].To solve this, we will follow these steps −make one map named dstore the frequency of numbers present in the barcode array into dx := empty listinsert all key-value pairs into xi := 0res := make a list whose length is same as barcodes, and fill [0]sort x based on the frequencywhile ... Read More

Suppose we have an array A of positive integers (not necessarily unique), we have to find the lexicographically largest permutation that is smaller than A, that can be made with one swap (A swap exchanges the positions of two numbers A[i] and A[j]). If it cannot possible, then return the same array. So if the array is like [3, 2, 1], then the output will be [3, 1, 2], by swapping 2 and 1To solve this, we will follow these steps −n := size of Afor left in range n – 2 down to -1if left = -1, then return ... Read More

Suppose we have an integer array A, we have to partition the array into (contiguous) subarrays of length at most K. After partitioning, each subarray has their values changed to become the maximum value of that subarray. We have to find the largest sum of the given array after partitioning. So if input is like [1, 15, 7, 9, 2, 5, 10] and k = 3, then the output will be 84. This is because the array becomes [15, 15, 15, 9, 10, 10, 10]To solve this, we will follow these steps −make one array dp of length same as ... Read More

Suppose we have a sequence of number is called arithmetic if it consists of at least three elements and if the difference between any two consecutive elements is the same. So for example, these are arithmetic sequence: [1, 3, 5, 7, 9], [7, 7, 7, 7], [3, -1, -5, -9], But the following sequence is not arithmetic. [1, 1, 2, 5, 7]Now a zero-indexed array A consisting of N numbers is given. A slice of that given array is any pair of integers (P, Q) such that 0

Suppose we have a positive integer K, we need find the smallest positive integer N such that N is divisible by K, and N only contains the digit 1. We have to find the length of N. If there is no such N, return -1. So if the input is like 3, then the output will be 3. The smallest answer will be N = 111.To solve this, we will follow these steps −if k is even, or k is divisible by 5, then return -1set r := 0 and N = 1for i in range 1 to K + ... Read More

As we know that the factorial of a positive integer n is the product of all positive integers less than or equal to n. So factorial(10) = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1. We will try to find a clumsy factorial: using the integers in decreasing order, we swap out the multiply operations for a fixed rotation of operations: multiply (*), divide (/), add (+) and subtract (-) in this order.The clumsy factorial is like clumsy(10) = 10 * 9 / 8 + 7 - 6 * ... Read More

Suppose we have an array A of 0s and 1s, we can update up to K values from 0 to 1. We have to find the length of the longest (contiguous) subarray that contains only 1s. So if A = [1,1,1,0,0,0,1,1,1,1,0] and k = 2, then the output will be 6, So if we flip 2 0s, the array can be of like [1,1,1,0,0,1,1,1,1,1,1], the length of longest sequence of 1s is 6.To solve this, we will follow these steps −set ans := 0, j := 0 and n := size of arrayfor i in range 0 to n – 1if A[i] is 0, then decrease k by 1while j