One of the common problems presented in coding tests is finding a way to reduce a given integer to 1. In this tutorial, we will explain the algorithms, approaches, and provide actual coding examples needed to solve this problem in detail. Our goal is to perform the minimum number of operations until the given integer becomes 1.<\/p>\n
Given an integer For example, when Therefore, the total number of operations is 3.<\/p>\n To solve this problem efficiently, we can use Dynamic Programming (DP). DP involves breaking the problem into smaller subproblems and storing the solutions to those subproblems to reduce redundant calculations. This approach will be explained in detail in the following steps.<\/p>\n Create a DP table to store the minimum number of operations required to reduce each index By default, the value of For each integer In other words, we update the value of After filling the DP table for all integers, Below is the complete code based on the aforementioned explanations:<\/p>\nX<\/code>. The problem is to reduce this X<\/code> to 1 using the following three operations and find the minimum number of operations required:<\/p>\n\n
X - 1<\/code><\/li>\nX \/ 2<\/code> (only possible if X is divisible by 2)<\/li>\nX \/ 3<\/code> (only possible if X is divisible by 3)<\/li>\n<\/ul>\nX = 10<\/code>, it can be calculated as follows:<\/p>\n\n
Approach to Problem Solving<\/h2>\n
Step 1: Initialize DP Table<\/h3>\n
i<\/code> to 1. The size of the table is set to X + 1<\/code>.<\/p>\n\nX = int(input(\"Enter an integer: \"))\ndp = [0] * (X + 1)\n <\/pre>\n
Step 2: Set Base Case<\/h3>\n
dp[1]<\/code> is 0 because 1 is already the target, so no additional operations are needed.<\/p>\n\ndp[1] = 0 # 1 can be represented with 0 operations.\n <\/pre>\n
Step 3: Set Recurrence Relation<\/h3>\n
i<\/code>, perform all possible operations to update the value of dp[i]<\/code>.<\/p>\n\n
i<\/code>: dp[i] = dp[i-1] + 1<\/code><\/li>\ni<\/code> is divisible by 2: dp[i] = min(dp[i], dp[i \/\/ 2] + 1)<\/code><\/li>\ni<\/code> is divisible by 3: dp[i] = min(dp[i], dp[i \/\/ 3] + 1)<\/code><\/li>\n<\/ul>\ndp[i]<\/code> with the minimum operations.<\/p>\nStep 4: Derive Final Result<\/h3>\n
dp[X]<\/code> will be the desired result, that is, the minimum number of operations needed to reduce X<\/code> to 1.<\/p>\n\nfor i in range(2, X + 1):\n dp[i] = dp[i - 1] + 1\n if i % 2 == 0:\n dp[i] = min(dp[i], dp[i \/\/ 2] + 1)\n if i % 3 == 0:\n dp[i] = min(dp[i], dp[i \/\/ 3] + 1)\n\nprint(\"Minimum number of operations:\", dp[X])\n <\/pre>\n
Full Code<\/h2>\n
\ndef min_operations_to_one(X):\n dp = [0] * (X + 1)\n dp[1] = 0 # Base case: 1 can be represented with 0 operations.\n\n for i in range(2, X + 1):\n dp[i] = dp[i - 1] + 1 # Case of subtracting 1\n if i % 2 == 0:\n dp[i] = min(dp[i], dp[i \/\/ 2] + 1) # Case of dividing by 2\n if i % 3 == 0:\n dp[i] = min(dp[i], dp[i \/\/ 3] + 1) # Case of dividing by 3\n\n return dp[X]\n\n# User input\nX = int(input(\"Enter an integer: \"))\nresult = min_operations_to_one(X)\nprint(\"Minimum number of operations:\", result)\n <\/pre>\n
Complexity Analysis<\/h2>\n