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Leetcode 553. Optimal Division

Problem Statement

You are given an integer array nums. The adjacent integers in nums will perform division operations following the order of appearance. For example, given the array nums = [1000, 100, 10, 2], we have to find the optimal way to parenthesize the division to get the maximum result.

Return a string that describes the sequence of division operations that will yield the maximum result.

Clarifying Questions

  1. Q: Can nums contain negative integers?
    • A: No, nums will contain only positive integers.
  2. Q: What is the minimum length of nums?
    • A: The length of nums will be at least 1.
  3. Q: How should we handle the case when nums has only one element?
    • A: Simply return the only element in string format as no division can be performed.

Strategy

To maximize the result of the division, we need to minimize the denominator part by dividing the largest numerator by the smallest possible denominator. This means that we should divide the first number by the result of all subsequent numbers divided together.

Given nums = [a1, a2, ..., an]:

The optimal division can be represented as:

a1 / (a2 / a3 / ... / an)

If the length of nums is 1, return a1 directly. If the length of nums is 2, return the result in the form of a1 / a2.

For any n > 2, return in the form of a1 / (a2 / a3 / ... / an).

Code

#include <vector>
#include <string>

std::string optimalDivision(std::vector<int>& nums) {
    int n = nums.size();
    if (n == 1) {
        return std::to_string(nums[0]);
    }
    if (n == 2) {
        return std::to_string(nums[0]) + "/" + std::to_string(nums[1]);
    }
    
    std::string result = std::to_string(nums[0]) + "/(" + std::to_string(nums[1]);
    for (int i = 2; i < n; ++i) {
        result += "/" + std::to_string(nums[i]);
    }
    result += ")";
    
    return result;
}

Time Complexity

The time complexity of this solution is O(n), where n is the number of elements in the array nums. This is because we are iterating through the array once to construct the result string, which takes linear time.

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