Decision Making Pattern
What is the Decision Making Pattern?
The Decision Making Pattern in dynamic programming involves making optimal choices at each step to build the overall optimal solution.
It's characterized by having multiple options at each state, and we need to decide which option leads to the optimal result.
When to Use the Decision Making Pattern
The Decision Making Pattern is useful when:
- You have multiple choices at each step
- You need to find the optimal sequence of decisions
- The problem involves maximizing or minimizing a value
- The problem can be broken down into a series of decisions
Decision Making Template
- Define the state: What information do we need to make a decision at each step?
- Identify the decisions: What choices do we have at each step?
- Formulate the recurrence relation: How does the current state depend on previous states and the decision made?
- Implement the solution using either memoization or tabulation
Example: House Robber
Here's a solution for the House Robber problem, where you need to decide which houses to rob to maximize the amount of money:
Example: Best Time to Buy and Sell Stock
Here's a solution for the Best Time to Buy and Sell Stock problem, where you need to decide when to buy and sell a stock to maximize profit:
Example: Jump Game
Here's a solution for the Jump Game problem, where you need to decide how many steps to jump at each position to reach the end:
Key Insights
- State Definition: Clearly define what information you need to make a decision at each step
- Decision Space: Identify all possible decisions at each state
- Optimal Substructure: Ensure that the optimal solution can be built from optimal solutions to subproblems
- Greedy vs. DP: Some decision problems can be solved greedily, but many require DP to find the optimal solution
Pattern 4: Decision Making
Learn how to make optimal choices at each step to build the overall optimal solution.
What is the Decision Making Pattern?
The Decision Making Pattern in dynamic programming involves making optimal choices at each step to build the overall optimal solution.
It's characterized by having multiple options at each state, and we need to decide which option leads to the optimal result.
Key Insight: The Decision Making Pattern is about exploring all possible choices at each step and selecting the one that leads to the optimal solution. It's particularly useful for optimization problems where we need to make a series of decisions to maximize or minimize a value.
When to Use the Decision Making Pattern
Multiple Choices
When you have multiple choices at each step and need to decide which one to take to achieve the optimal result.
Optimization Problems
When the problem involves maximizing or minimizing a value, such as maximum profit, minimum cost, or optimal strategy.
Sequential Decisions
When the problem can be broken down into a series of decisions, where each decision affects the available choices for future decisions.
The Decision Making Approach
1Define State
Define what information you need to make a decision at each step. This will determine the dimensions of your DP table or the parameters of your recursive function.
2Identify Decisions
Identify all possible decisions or choices at each state. These will be the branches in your decision tree.
3Formulate Recurrence
Define how the current state depends on previous states and the decision made. This is the heart of the DP approach.
4Implement Solution
Implement the solution using either memoization (top-down) or tabulation (bottom-up), depending on the problem's characteristics.
Example: House Robber
Here's a solution for the House Robber problem, where you need to decide which houses to rob to maximize the amount of money:
12345678910111213141516171819202122232425262728function rob(nums) { // Handle edge cases if (!nums || nums.length === 0) return 0; if (nums.length === 1) return nums[0]; // Create a DP array where dp[i] represents the maximum amount // that can be robbed up to house i const dp = new Array(nums.length); // Base cases dp[0] = nums[0]; // If there's only one house, rob it dp[1] = Math.max(nums[0], nums[1]); // If there are two houses, rob the one with more money // Fill the DP array for (let i = 2; i < nums.length; i++) { // At each house, we have two choices: // 1. Rob the current house and add it to the max amount robbed up to i-2 // 2. Skip the current house and take the max amount robbed up to i-1 dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i]); } // Return the maximum amount that can be robbed return dp[nums.length - 1];} // Example usage:console.log(rob([1, 2, 3, 1])); // Output: 4 (rob house 1 and 3)console.log(rob([2, 7, 9, 3, 1])); // Output: 12 (rob house 1, 3, and 5)How it works: At each house, we have two choices: rob it or skip it. If we rob the current house, we can't rob the adjacent house, so we add the current house's value to the maximum amount robbed up to two houses ago. If we skip the current house, we take the maximum amount robbed up to the previous house. We choose the option that gives us the maximum amount.
Example: Best Time to Buy and Sell Stock
Here's a solution for the Best Time to Buy and Sell Stock problem, where you need to decide when to buy and sell a stock to maximize profit:
1234567891011121314151617181920212223function maxProfit(prices) { // Handle edge cases if (!prices || prices.length <= 1) return 0; let maxProfit = 0; let minPrice = prices[0]; // Iterate through the prices for (let i = 1; i < prices.length; i++) { // Update the minimum price seen so far minPrice = Math.min(minPrice, prices[i]); // Update the maximum profit if selling at the current price // would result in a higher profit maxProfit = Math.max(maxProfit, prices[i] - minPrice); } return maxProfit;} // Example usage:console.log(maxProfit([7, 1, 5, 3, 6, 4])); // Output: 5 (buy at 1, sell at 6)console.log(maxProfit([7, 6, 4, 3, 1])); // Output: 0 (no profit possible)How it works: We keep track of the minimum price seen so far and the maximum profit we can make. At each day, we have two decisions: update the minimum price if the current price is lower, and update the maximum profit if selling at the current price would result in a higher profit.
Example: Jump Game
Here's a solution for the Jump Game problem, where you need to decide how many steps to jump at each position to reach the end:
12345678910111213141516171819202122232425262728function canJump(nums) { // Handle edge cases if (!nums || nums.length === 0) return false; if (nums.length === 1) return true; // Initialize the maximum reachable position let maxReach = nums[0]; // Iterate through the array for (let i = 0; i < nums.length; i++) { // If we can't reach the current position, return false if (maxReach < i) return false; // Update the maximum reachable position maxReach = Math.max(maxReach, i + nums[i]); // If we can reach the last index, return true if (maxReach >= nums.length - 1) return true; } // If we've gone through the entire array and haven't reached the end, // it's not possible to reach the end return false;} // Example usage:console.log(canJump([2, 3, 1, 1, 4])); // Output: true (jump 1 step from index 0 to 1, then 3 steps to the last index)console.log(canJump([3, 2, 1, 0, 4])); // Output: false (you will always arrive at index 3 with value 0, which means you can't reach the last index)How it works: We keep track of the maximum position we can reach. At each position, we update the maximum reachable position based on the current position and the maximum jump length. If at any point we can't reach the current position, we return false. If we can reach the last position, we return true.
Key Insights for Decision Making Problems
State Definition
Clearly define what information you need to make a decision at each step. This could include the current position, the resources available, or the decisions made so far.
Decision Space
Identify all possible decisions at each state. This could be binary (e.g., take or skip an item) or involve multiple options (e.g., choose one of several paths).
Optimal Substructure
Ensure that the optimal solution can be built from optimal solutions to subproblems. This is a key requirement for dynamic programming to work.
Greedy vs. DP
Some decision problems can be solved greedily (making the locally optimal choice at each step), but many require DP to find the globally optimal solution by considering all possible decision sequences.