Optimal Division - Practice Coding Problems

Optimal Division - Problem

Maximize Division Expression Value

You're given an integer array nums where adjacent integers will perform float division from left to right. For example, with nums = [2,3,4], we evaluate the expression "2/3/4".

However, you can add any number of parentheses at any position to change the order of operations and maximize the final result. Your goal is to return the expression string that produces the maximum possible value.

Key Requirements:
• Add parentheses strategically to maximize the result
• Return the complete expression as a string
• Avoid redundant parentheses
• The expression should evaluate to the maximum possible value

Example: [2,3,4] → "2/(3/4)" because 2/(3/4) = 2/(0.75) = 2.667, which is greater than 2/3/4 = 0.167

Input & Output

example_1.py — Basic Case

$ Input: [2,3,4]

› Output: "2/(3/4)"

💡 Note: We can rewrite 2/3/4 as 2/(3/4). Since 3/4 = 0.75, we get 2/0.75 = 2.667, which is much larger than the original 2/3/4 = 0.167

example_2.py — Two Numbers

$ Input: [1000,100]

› Output: "1000/100"

💡 Note: With only two numbers, we cannot add parentheses to change the result, so we return the expression as is: 1000/100 = 10

example_3.py — Single Number

$ Input: [1000]

› Output: "1000"

💡 Note: With only one number, there's no division operation, so we simply return the number itself as a string

Constraints

1 ≤ nums.length ≤ 10
2 ≤ nums[i] ≤ 1000
There is only one optimal division for the given input

Visualization

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Understanding the Visualization

Initial Setup

You have 2 pizzas, and need to share with 3 friends, then 4 friends: 2/3/4

Default Sharing

Normal left-to-right: (2/3)/4 = 0.67/4 = 0.167 pizza per person

Smart Grouping

Group friends 2&3 together first: 2/(3/4) = 2/0.75 = 2.67 pizza per person

Optimal Strategy

Always group everyone except the first 'friend' together to maximize your share!

Key Takeaway

🎯 Key Insight: To maximize division results, group all numbers after the second one together using parentheses. This creates the pattern `a/(b/c/d/...)` which is always optimal!

Asked in

G Google 35 a Amazon 28 ⊞ Microsoft 22 f Facebook 18

The optimal solution uses a mathematical insight: to maximize a division chain a/b/c/d, rewrite it as a/(b/c/d). This works because dividing by a smaller fraction yields a larger result. For arrays with 3+ elements, the pattern nums[0]/(nums[1]/nums[2]/...) is always optimal, achieving O(n) time complexity.

Common Approaches

Approach	Time	Space	Notes
✓ Mathematical Insight (Optimal)	O(n)	O(n)	Use mathematical insight: maximize by putting parentheses around everything after the second number
Brute Force (Try All Parentheses)	O(3^n)	O(n²)	Generate all possible ways to add parentheses and find the maximum

Mathematical Insight (Optimal) — Algorithm Steps

Handle special cases: single number or two numbers
For 3+ numbers, use format: first/(second/third/fourth/...)
Build the expression string with proper parentheses
Return the optimal expression

Visualization

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Step-by-Step Walkthrough

Original Expression

Start with a/b/c/d (left-to-right evaluation)

Apply Insight

Transform to a/(b/c/d) to maximize result

Why It Works

Dividing by a smaller fraction gives a larger result

Build Expression

Construct the optimal string format

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char* optimalDivision(int* nums, int numsSize) {
    char* result = (char*)malloc(1000);
    
    if (numsSize == 1) {
        sprintf(result, "%d", nums[0]);
        return result;
    }
    if (numsSize == 2) {
        sprintf(result, "%d/%d", nums[0], nums[1]);
        return result;
    }
    
    // For 3+ numbers: a/(b/c/d/...)
    sprintf(result, "%d/(", nums[0]);
    
    for (int i = 1; i < numsSize; i++) {
        if (i > 1) {
            strcat(result, "/");
        }
        char temp[20];
        sprintf(temp, "%d", nums[i]);
        strcat(result, temp);
    }
    strcat(result, ")");
    
    return result;
}

int main() {
    char input[1000];
    fgets(input, sizeof(input), stdin);
    
    int nums[100];
    int count = 0;
    char* token = strtok(input + 1, ",]");
    while (token && count < 100) {
        nums[count++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    char* result = optimalDivision(nums, count);
    printf("%s\n", result);
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the array to build the result string

✓ Linear Growth

Space Complexity

O(n)

Space needed to build the result string

⚡ Linearithmic Space

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Medium Frequency

~15 min Avg. Time

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Smart Actions

💡 Explanation

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Mathematical Insight (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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