Given an integer ‘N’, you need to make the maximum possible number of moves where each move consists of choosing a positive integer ‘X’ > 1, such that ‘N’ is divisible by ‘X’ and replacing ‘N’ with ‘N/X’.When ‘N’ becomes equal to 1 and there are no more possible valid moves. You need to stop and your score is equal to the number of moves made.Given ‘N’ is of the form a! / b! ( i.e. factorial of ‘a’ divided by factorial of ‘b’) for some positive integer ‘a’ and ‘b’ (a ≥ b).You need to find and return the maximum possible score you can achieve.
Question
Given an integer ‘N’, you need to make the maximum possible number of moves where each move consists of choosing a positive integer ‘X’ > 1, such that ‘N’ is divisible by ‘X’ and replacing ‘N’ with ‘N/X’.When ‘N’ becomes equal to 1 and there are no more possible valid moves. You need to stop and your score is equal to the number of moves made.Given ‘N’ is of the form a! / b! ( i.e. factorial of ‘a’ divided by factorial of ‘b’) for some positive integer ‘a’ and ‘b’ (a ≥ b).You need to find and return the maximum possible score you can achieve.
Solution 1
The problem can be solved by using the concept of prime factorization.
Step 1: First, we need to find the prime factorization of 'a!' and 'b!'.
Step 2: Then, we subtract the prime factorization of 'b!' from 'a!'. This will give us the prime factorization of 'a!/b!'.
Step 3: The maximum possible score we can achieve is the sum of the powers of the prime factors in the prime factorization of 'a!/b!'.
This is because each move consists of choosing a positive integer ‘X’ > 1, such that ‘N’ is divisible by ‘X’ and replacing ‘N’ with ‘N/X’. So, for each prime factor 'p' with power 'k' in the prime factorization of 'a!/b!', we can make 'k' moves by choosing 'p' as 'X'.
So, the maximum possible score is the sum of the powers of the prime factors in the prime factorization of 'a!/b!'.
This solution works because the prime factorization of a number is unique and the prime factors of 'a!' and 'b!' are the same, so the prime factorization of 'a!/b!' is the difference of the prime factorizations of 'a!' and 'b!'.
Also, the score is maximized when we make the maximum number of moves, which is achieved by choosing the prime factors as 'X' in each move.
Solution 2
The problem can be solved by using the concept of prime factorization.
Step 1: First, we need to find the prime factorization of 'a!' and 'b!'.
Step 2: Then, subtract the prime factorization of 'b!' from 'a!'. This will give us the prime factorization of 'N'.
Step 3: The maximum possible score will be the sum of the powers of the prime factors of 'N'.
This is because each move consists of choosing a positive integer 'X' such that 'N' is divisible by 'X' and replacing 'N' with 'N/X'. The maximum number of moves will be achieved when 'X' is a prime number.
So, the maximum possible score will be the sum of the powers of the prime factors of 'N'.
This solution works because the prime factorization of a number is unique. So, the maximum possible score will be the same regardless of the order in which the moves are made.
To implement this solution, you can use the Sieve of Eratosthenes algorithm to find all the prime numbers up to 'a'. Then, use these prime numbers to find the prime factorization of 'a!' and 'b!'. Finally, subtract the prime factorization of 'b!' from 'a!' and sum up the powers of the prime factors to find the maximum possible score.
Similar Questions
You are given an integer array nums.In one move, you can choose one element of nums and change it to any value.Return the minimum difference between the largest and smallest value of nums after performing at most three moves.
Consider a game where a player can score 3 or 5 or 10 points in a move. Given a total score n, find number of distinct combinations to reach the given score.Example 1:Inputn = 10Output2ExplanationThere are two ways {5,5} and {10}.
You are given an m x n matrix grid consisting of positive integers. You can move from a cell in the matrix to any other cell that is either to the bottom or to the right (not necessarily adjacent). The score of a move from a cell with the value c1 to a cell with the value c2 is c2 - c1.You can start at any cell, and you have to make at least one move.Return the maximum total score you can achieve. Example 1:Input: grid = [[9,5,7,3],[8,9,6,1],[6,7,14,3],[2,5,3,1]]Output: 9Explanation: We start at the cell (0, 1), and we perform the following moves:- Move from the cell (0, 1) to (2, 1) with a score of 7 - 5 = 2.- Move from the cell (2, 1) to (2, 2) with a score of 14 - 7 = 7.The total score is 2 + 7 = 9.Example 2:Input: grid = [[4,3,2],[3,2,1]]Output: -1Explanation: We start at the cell (0, 0), and we perform one move: (0, 0) to (0, 1). The score is 3 - 4 = -1. Constraints:m == grid.lengthn == grid[i].length2 <= m, n <= 10004 <= m * n <= 1051 <= grid[i][j] <= 105
There is an m x n grid with a ball. The ball is initially at the position [startRow, startColumn]. You are allowed to move the ball to one of the four adjacent cells in the grid (possibly out of the grid crossing the grid boundary). You can apply at most maxMove moves to the ball.Given the five integers m, n, maxMove, startRow, startColumn, return the number of paths to move the ball out of the grid boundary. Since the answer can be very large, return it modulo 109 + 7. Example 1:Input: m = 2, n = 2, maxMove = 2, startRow = 0, startColumn = 0Output: 6
A game is played by moving a game piece left or right along a horizontal game board. The board consists of spaces of various colors, as shown. The circle represents the initial location of the game piece.Yellow Black Green Green Red Yellow Black Black Yellow Black ● The following algorithm indicates how the game is played. The game continues until the game is either won by landing on the red space or lost when the piece moves off either end of the board.Step 1:Place a game piece on a space that is not red and set a counter to 0.Step 2:If the game piece is on a yellow space, move the game piece 3 positions to the left and go to step 3. Otherwise, if the game piece is on a black space, move the game piece 1 position to the left and go to step 3. Otherwise, if the game piece is on a green space, move the game piece 2 positions to the right and go to step 3.Step 3:Increase the value of the counter by 1.Step 4:If game piece is on the red space or moved off the end of the game board, the game is complete. Otherwise, go back to step 2.If a game is begun by placing the game piece on the rightmost black space for step 1, what will be the value of the counter at the end of the game?Responses2233445
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.