Optimal Division - Practice Coding Problems

Optimal Division - 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 parentheses at any position to change the priority of operations. You want to add these parentheses such that 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 parentheses.

Input & Output

Example 1 — Basic Case

$ Input: nums = [2,3,4]

› Output: "2/(3/4)"

💡 Note: We want to maximize 2/3/4. By adding parentheses as 2/(3/4), we get 2/0.75 = 2.67, which is maximum possible.

Example 2 — Two Elements

$ Input: nums = [2,3]

› Output: "2/3"

💡 Note: With only two elements, no parentheses can be added. The result is simply 2/3.

Example 3 — Single Element

$ Input: nums = [5]

› Output: "5"

💡 Note: With only one element, the expression is just the number itself: 5.

Constraints

1 ≤ nums.length ≤ 10
2 ≤ nums[i] ≤ 1000

Visualization

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Understanding the Visualization

Input Array

Array [2,3,4] represents expression 2/3/4

Add Parentheses

Strategic placement: 2/(3/4) maximizes result

Output Expression

Return the string with optimal parentheses

Key Takeaway

🎯 Key Insight: To maximize division, group everything after the first operator in parentheses to minimize the effective denominator

Asked in

G Google 12 F Facebook 8

The key insight is that to maximize a division expression a/b/c/d, we should minimize the denominator by grouping as a/(b/c/d). Best approach is the Mathematical Insight method. Time: O(n), Space: O(n)

Common Approaches

Approach	Time	Space	Notes
✓ Mathematical Insight	O(n)	O(n)	Use mathematical property: to maximize a/b/c/d, rewrite as a/(b/c/d)
Dynamic Programming - All Combinations	O(2ⁿ)	O(n²)	Try all possible ways to parenthesize the expression using recursive division

Mathematical Insight — Algorithm Steps

Step 1: Recognize that a/b/c = a/(b/c) maximizes when we minimize the denominator
Step 2: For arrays with length > 2, optimal form is nums[0]/(nums[1]/nums[2]/...)
Step 3: Build the expression string with proper parentheses
Step 4: Handle edge cases for arrays of length 1 or 2

Visualization

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Step-by-Step Walkthrough

Original Expression

2/3/4 evaluated left to right

Add Parentheses

2/(3/4) groups the denominator

Maximized Result

2/(0.75) = 2.67 is optimal

Code -

solution.c — C

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

char* solution(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 arrays with length > 2, optimal is nums[0]/(nums[1]/nums[2]/...)
    sprintf(result, "%d/(", nums[0]);
    
    for (int i = 1; i < numsSize; i++) {
        char temp[20];
        sprintf(temp, "%d", nums[i]);
        strcat(result, temp);
        if (i < numsSize - 1) {
            strcat(result, "/");
        }
    }
    
    strcat(result, ")");
    return result;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    line[strcspn(line, "\n")] = 0;
    
    int nums[1000];
    int size = 0;
    
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[size++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    char* result = solution(nums, size);
    printf("%s\n", result);
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through array to build the result string

✓ Linear Growth

Space Complexity

O(n)

Space needed to construct the output string

⚡ Linearithmic Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Mathematical Insight — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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