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Leetcode 553. Optimal Division

Problem Statement

Leetcode Problem 553: Optimal Division

You are given an integer array nums. The optimal division of this array is the expression that results in the largest possible value after performing division operations between its elements. Specifically, we need to find a way to parenthesize the array such that the resulting expression has the maximum value.

For example, for the array [1000, 100, 10, 2], the optimal division is:

( \frac{1000}{(100 / (10 / 2))} )

The goal is to return this expression as a string.

Example

Input: nums = [1000, 100, 10, 2]
Output: "1000/(100/10/2)"

Clarifying Questions

1. What is the minimum and maximum length of the nums array?
   • The length can be anywhere from 1 to 10.
2. What is the range of the elements in nums?
   • The elements in the array can be any values between 1 and 1000.
3. Should we consider integer division or can we treat division as floating-point?
   • Division should be treated as floating-point to ensure the maximum value.

Strategy

The problem is essentially about where to place the parentheses to maximize the division result.

Key Insight:

• When you have more than two numbers in the division sequence, the maximum value is always obtained by dividing the first number by the result of dividing the rest of the numbers in sequence.

  For example, given numbers [a, b, c, d], the optimal division is: [ a / \frac{b}{\frac{c}{d}} ] which can be represented as a/(b/c/d).

Plan:

1. If nums has only one element, return that element as a string.
2. If nums has only two elements, return a/b.
3. For more elements, construct the optimal division as described.

Code

import java.util.Arrays;

public class OptimalDivision {
    public String optimalDivision(int[] nums) {
        int n = nums.length;
        
        // Handle case with only one number.
        if (n == 1) return Integer.toString(nums[0]);
        
        // Handle case with only two numbers.
        if (n == 2) return nums[0] + "/" + nums[1];
        
        // Handle case with more than two numbers.
        // Form the string for optimal division.
        StringBuilder sb = new StringBuilder();
        sb.append(nums[0]).append("/(").append(nums[1]);
        
        for (int i = 2; i < n; i++) {
            sb.append("/").append(nums[i]);
        }
        
        sb.append(")");
        
        return sb.toString();
    }
    
    // Example usage:
    public static void main(String[] args) {
        OptimalDivision solution = new OptimalDivision();
        int[] nums = {1000, 100, 10, 2};
        System.out.println(solution.optimalDivision(nums));  // Output: "1000/(100/10/2)"
    }
}

Time Complexity

The time complexity of the 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.

The space complexity is also O(n) due to the construction of the result string which potentially holds up to 2n characters.

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