Optimal Division solution and explanation
3 more videos on this problem
The problem
You are given an integer array nums. The adjacent integers in nums will perform the float division.
- For example, for
nums = [2,3,4], we will evaluate the expression"2/3/4".
However, you can add any number of parenthesis at any position to change the priority of operations. You want to add these parentheses such the value of the expression after the evaluation is maximum.
Return the corresponding expression that has the maximum value in string format.
Note: your expression should not contain redundant parenthesis.
Example 1:
Input: nums = [1000,100,10,2] Output: "1000/(100/10/2)" Explanation: 1000/(100/10/2) = 1000/((100/10)/2) = 200 However, the bold parenthesis in "1000/((100/10)/2)" are redundant since they do not influence the operation priority. So you should return "1000/(100/10/2)". Other cases: 1000/(100/10)/2 = 50 1000/(100/(10/2)) = 50 1000/100/10/2 = 0.5 1000/100/(10/2) = 2
Example 2:
Input: nums = [2,3,4] Output: "2/(3/4)" Explanation: (2/(3/4)) = 8/3 = 2.667 It can be shown that after trying all possibilities, we cannot get an expression with evaluation greater than 2.667
Constraints:
1 <= nums.length <= 102 <= nums[i] <= 1000- There is only one optimal division for the given input.
Approach overview
Problem Overview: You get an array nums representing a chain of divisions: nums[0] / nums[1] / nums[2] / ... / nums[n-1]. You can insert parentheses anywhere to change evaluation order. The goal is to return a string expression that produces the maximum possible value.
Approach 1: Brute Force Parenthesization with Recursion/DP (Time: O(n^3), Space: O(n^2))
This problem resembles the classic matrix-chain multiplication structure. For every subarray [i, j], compute the minimum and maximum value obtainable by placing parentheses in different ways. Try every split point k where the expression becomes (i..k) / (k+1..j). Because division is not associative, both min and max values must be tracked for each interval. Dynamic programming reduces repeated work by storing results in a table. This guarantees correctness but requires evaluating many partitions and storing intermediate expressions, which makes it heavier than needed for this specific problem.
Approach 2: Greedy Observation (Time: O(n), Space: O(1))
The key insight comes from the behavior of division. For a / b / c / d, the result equals ((a / b) / c) / d by default. However, placing parentheses as a / (b / c / d) makes the denominator smaller because you divide b by multiple numbers. A smaller denominator increases the overall value. Therefore, to maximize the result, keep the first number outside and group all remaining numbers inside a single denominator.
If the array length is 1, return the number. If it is 2, simply return a/b. When the array has three or more elements, construct the expression nums[0] / (nums[1] / nums[2] / ... / nums[n-1]). This structure guarantees the largest value without exploring all parenthesis placements. Implementation is straightforward: iterate through the array and build the string while inserting one opening parenthesis after the first division and a closing parenthesis at the end.
This greedy rule works because every additional division inside the denominator shrinks it, which increases the overall quotient. The solution relies on simple math properties rather than heavy dynamic programming.
Recommended for interviews: Interviewers expect the greedy observation. Starting with the brute-force or DP reasoning shows you understand the search space of parenthesization. Recognizing that all numbers after the first should be grouped into a single denominator demonstrates the optimization insight and leads to the O(n) solution.
Approach 1: Greedy Approach
The key observation is that to maximize the result of the division, we should minimize the denominator. We achieve this by grouping as many elements as possible in the denominator so that the result becomes maximum. Hence, we should divide the first number by all the numbers after the first number treated as one single division.
def optimalDivision(nums):
if len(nums) == 1:
return str(nums[0])
if len(nums) == 2:
return f"{nums[0]}/{nums[1]}"
return f"{nums[0]}/({"/".join(map(str, nums[1:]))})"public class Solution {
public String optimalDivision(int[] nums) {
if (nums.length == 1) return String.valueOf(nums[0]);
if (nums.length == 2) return nums[0] + "/" + nums[1];
StringBuilder result = new StringBuilder();
result.append(nums[0]).append("/(");
for (int i = 1; i < nums.length; i++) {
result.append(nums[i]);
if (i != nums.length - 1) result.append("/");
}
result.append(")");
return result.toString();
}
}#include <iostream>
#include <string>
#include <vector>
#include <sstream>
using namespace std;
string optimalDivision(vector<int>& nums) {
if (nums.size() == 1) return to_string(nums[0]);
if (nums.size() == 2) return to_string(nums[0]) + "/" + to_string(nums[1]);
ostringstream result;
result << nums[0] << "/(";
for (size_t i = 1; i < nums.size(); ++i) {
result << nums[i];
if (i != nums.size() - 1) result << "/";
}
result << ")";
return result.str();
}function optimalDivision(nums) {
if (nums.length === 1) return nums[0].toString();
if (nums.length === 2) return nums[0] + '/' + nums[1];
return nums[0] + '/(' + nums.slice(1).join('/') + ')';
}#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char* optimalDivision(int* nums, int numsSize) {
if (numsSize == 1) {
char* result = (char*)malloc(24 * sizeof(char));
sprintf(result, "%d", nums[0]);
return result;
}
if (numsSize == 2) {
char* result = (char*)malloc(48 * sizeof(char));
sprintf(result, "%d/%d", nums[0], nums[1]);
return result;
}
char* result = (char*)malloc(256 * sizeof(char));
char temp[16];
sprintf(result, "%d/(", nums[0]);
for (int i = 1; i < numsSize; ++i) {
sprintf(temp, "%d", nums[i]);
strcat(result, temp);
if (i < numsSize - 1) strcat(result, "/");
}
strcat(result, ")");
return result;
}public class Solution {
public string OptimalDivision(int[] nums) {
if (nums.Length == 1) return nums[0].ToString();
if (nums.Length == 2) return nums[0] + "/" + nums[1];
var result = new System.Text.StringBuilder();
result.Append(nums[0]).Append("/(");
for (int i = 1; i < nums.Length; i++) {
result.Append(nums[i]);
if (i != nums.Length - 1) result.Append("/");
}
result.Append(")");
return result.ToString();
}
}The solution starts by checking if the length of the array is 1 or 2. For these cases, the return is straightforward as no extra parentheses are needed. For more than two numbers, the first number is divided by the expression of the rest, enclosed in parentheses to ensure that division occurs from left to right within the parentheses, thus maximizing the result.
Time Complexity: O(n), where n is the number of elements in nums. We iterate through the nums once to construct the string.
Space Complexity: O(n), for storing the joined string.
Approach 2: Default Approach
function optimalDivision(nums: number[]): string {
const n = nums.length;
const res = nums.join('/');
if (n > 2) {
const index = res.indexOf('/') + 1;
return `${res.slice(0, index)}(${res.slice(index)})`;
}
return res;
}func optimalDivision(nums []int) string {
n := len(nums)
if n == 1 {
return strconv.Itoa(nums[0])
}
if n == 2 {
return fmt.Sprintf("%d/%d", nums[0], nums[1])
}
ans := &strings.Builder{}
ans.WriteString(fmt.Sprintf("%d/(", nums[0]))
for _, num := range nums[1 : n-1] {
ans.WriteString(strconv.Itoa(num))
ans.WriteByte('/')
}
ans.WriteString(fmt.Sprintf("%d)", nums[n-1]))
return ans.String()
}impl Solution {
pub fn optimal_division(nums: Vec<i32>) -> String {
let n = nums.len();
match n {
1 => nums[0].to_string(),
2 => nums[0].to_string() + "/" + &nums[1].to_string(),
_ => {
let mut res = nums[0].to_string();
res.push_str("/(");
for i in 1..n {
res.push_str(&nums[i].to_string());
res.push('/');
}
res.pop();
res.push(')');
res
}
}
}
}Complexity comparison
| Approach | Complexity |
|---|---|
| Greedy Approach | Time Complexity: O(n), where n is the number of elements in nums. We iterate through the nums once to construct the string. |
| Default Approach | – |
Complexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Brute Force Parenthesization | Exponential | O(n) | Conceptual understanding of all possible parenthesis placements |
| Dynamic Programming (Interval DP) | O(n^3) | O(n^2) | When systematically computing min/max values for each subarray |
| Greedy Parenthesis Placement | O(n) | O(1) | Optimal production solution using the division property insight |
Frequently asked questions
Is Optimal Division easy or hard?
Optimal Division is generally rated Medium because the implementation is simple but the key greedy insight is not immediately obvious. Many candidates initially attempt dynamic programming before recognizing the mathematical pattern that leads to the O(n) solution.
How to solve Optimal Division in O(n)?
Observe that dividing by a smaller value increases the result. By placing parentheses as nums[0]/(nums[1]/nums[2]/.../nums[n-1]), the denominator becomes as small as possible. Implementation only requires iterating through the array and formatting the string with one opening parenthesis after the first division and one closing parenthesis at the end.
What is the best approach for Optimal Division?
The greedy observation is the best approach. For arrays with more than two numbers, the optimal expression is nums[0]/(nums[1]/nums[2]/.../nums[n-1]). Grouping all remaining numbers inside the denominator minimizes the denominator and maximizes the overall result. This runs in O(n) time and O(1) space.
What data structure is used in Optimal Division?
The primary structure is a simple array traversal since the input is a list of numbers. The optimal solution relies on mathematical properties of division rather than complex data structures. Dynamic programming variants use a 2D DP table to store minimum and maximum values for each subarray.
What is the time complexity of Optimal Division?
The optimal greedy solution runs in O(n) time because it simply iterates through the array once to build the expression string. Space complexity is O(1) aside from the output string. A dynamic programming alternative takes O(n^3) time due to evaluating all interval partitions.
Optimal Division Python or Java solution approach?
In Python or Java, iterate through the array and construct the expression string. If the array length is greater than two, insert an opening parenthesis after the first division and close it after the last number. This greedy formatting produces the maximum value in O(n) time.
Is Optimal Division asked at Google, Amazon, or Meta?
Optimal Division is a classic expression-optimization problem seen in coding interviews that test mathematical reasoning and greedy thinking. Variants involving expression evaluation or parenthesis placement have appeared in interviews at large tech companies including Google and Amazon.