grid47 Exploring patterns and algorithms
Sep 12, 2024
5 min read
Solution to LeetCode 553: Optimal Division Problem
Given an array of integers, you need to add parentheses in such a way that the division expression evaluates to the maximum possible value. The division must respect the adjacency of integers in the array.
Problem
Approach
Steps
Complexity
Input: The input is an array of integers 'nums'. The adjacent integers will perform the division operation.
Example: Input: nums = [200, 50, 10]
Constraints:
• 1 <= nums.length <= 10
• 2 <= nums[i] <= 1000
Output: Return the corresponding expression with parentheses added optimally, resulting in the maximum evaluated value.
Example: Output: '200/(50/10)'
Constraints:
• The returned expression should not have redundant parentheses.
Goal: The goal is to determine the optimal arrangement of parentheses in the given division expression to maximize the evaluated result.
Steps:
• If the array has only one element, return the number as a string.
• If the array has two elements, return them with a '/' operator.
• For arrays with more than two elements, always add parentheses around all but the first element.
Goal: The input list will have between 1 and 10 integers, and each integer is between 2 and 1000.
Steps:
• 1 <= nums.length <= 10
• 2 <= nums[i] <= 1000
Assumptions:
• The input is valid, and there is always a unique optimal expression.
• Input: Input: nums = [200, 50, 10]
• Explanation: By adding parentheses around (50/10), the expression becomes '200/(50/10)', which gives the maximum value of 40.
Approach: To maximize the result of a division expression, we need to add parentheses in a way that minimizes the divisors in the sequence. This can be done by wrapping the divisions between all but the first number in parentheses.
Observations:
• When dividing numbers, larger numbers should ideally divide smaller ones.
• The expression with the maximum value will involve minimizing the number of divisors in the sequence by grouping all but the first numbers into parentheses.
• By focusing the parentheses around the numbers that need to be divided first, we maximize the value of the expression.
Steps:
• If the array has one element, simply return that element as a string.
• If the array has two elements, return them with a division sign.
• For longer arrays, wrap all elements except the first one in parentheses.
Empty Inputs:
• There are no edge cases with empty inputs, as the array will always have at least one number.
Large Inputs:
• The solution should work efficiently for arrays of size up to 10.
Special Values:
• Ensure that the parentheses are placed optimally, without redundant brackets that do not affect the order of operations.
Constraints:
• The input size is small, and the solution should handle this easily.
string optimalDivision(vector<int>& nums) {
string s ="";
if(nums.size() ==1) return to_string(nums[0]);
if(nums.size() ==2) return to_string(nums[0]) +'/'+ to_string(nums[1]);
for(int i =0; i < nums.size(); i++) {
s += to_string(nums[i]) +"/";
if(i ==0)
s +="(";
}
s[s.size() -1] =')';
return s;
}
1 : Function Definition
string optimalDivision(vector<int>& nums) {
Defines the function `optimalDivision`, which takes a vector of integers `nums` and returns the optimal division string for them.
2 : String Initialization
string s ="";
Initializes an empty string `s` to build the division result.
3 : Edge Case Handling (Single Element)
if(nums.size() ==1) return to_string(nums[0]);
Checks if the input vector contains only one element. If true, converts the element to a string and returns it as the result.
