HackerRank Mini-Max Sum Java Solution Explained

Problem Overview

Original Problem: Mini-Max Sum | HackerRank

Difficulty: Easy

Category: Arrays, Math

Description: Given an array of five positive integers, calculate the minimum and maximum sums that can be obtained by adding exactly four of the five numbers. Output these sums as two space-separated long integers on a single line.

This problem, part of HackerRank’s One Week Preparation Kit, is an excellent exercise for beginners to practice array manipulation and handling large numbers in Java, key skills for coding interviews.

Understanding the Problem

The Mini-Max Sum problem asks you to take an array of five positive integers and find two values: the smallest possible sum and the largest possible sum when adding exactly four of the five numbers. For example, if the array is [1, 2, 3, 4, 5], the minimum sum is obtained by adding the four smallest numbers (1 + 2 + 3 + 4 = 10), and the maximum sum comes from adding the four largest numbers (2 + 3 + 4 + 5 = 14). The challenge is to compute these sums efficiently and print them as space-separated long integers.

Input and Output

  • Input:

    • A single line of five space-separated positive integers.

  • Output:

    • Two space-separated long integers: the minimum sum and the maximum sum of four out of the five integers.

  • Constraints:

    • 1 ≤ arr[i] ≤ 10^9

    • The array always contains exactly five elements.

    • All elements are positive integers.

    • The output sums may exceed 32-bit integer limits.

Initial Reasoning

To solve this, we need to:

  1. Find the sum of four out of five numbers for both the minimum and maximum cases.

  2. The minimum sum excludes the largest number, and the maximum sum excludes the smallest number.

  3. Use long to handle large sums, as inputs can be up to 10^9.

  4. Compute the total sum of the array and subtract the maximum and minimum elements to get the desired sums.
    This approach ensures a single pass through the array (O(n) time) and minimal space (O(1)).

Thought Process

Here’s the step-by-step logic:

  1. Calculate Total Sum: Sum all five numbers in the array. Since the array has exactly five elements, this gives us the base sum.

  2. Find Min and Max Elements: Identify the smallest and largest numbers in the array during a single pass to avoid sorting.

  3. Compute Min and Max Sums:

    • Minimum sum = Total sum − Maximum element.

    • Maximum sum = Total sum − Minimum element.

  4. Handle Large Numbers: Use long for all calculations to prevent integer overflow, as each element can be up to 10^9, and the sum of four elements could be up to 4 × 10^9.

  5. Edge Cases:

    • If all numbers are equal (e.g., [5, 5, 5, 5, 5]), the minimum and maximum sums are the same (e.g., 5 + 5 + 5 + 5 = 20).

    • The constraints ensure all inputs are valid (positive, exactly five elements), so no additional validation is needed.

This approach is optimal, requiring only one pass through the array (O(5) = O(1) time since the array size is fixed) and constant extra space.

Pseudocode

// miniMaxSum method:
Initialize variables to store total, minimum and maximum element
For each element in the arry
	Add the element to the total
	If the element is less than the current minimum element
		Update the current minimum
	If the element is larger than the current maximum
		Update the current maximum
Calculate minimum and maximum sums and print

Solution Code

// Note this is one of many possible solutions
import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;

class Result {

    public static void miniMaxSum(List<Integer> arr) {
    
    long totalSum = 0;
    long minElement = Long.MAX_VALUE;
    long maxElement = Long.MIN_VALUE;
    
    // Calculate total sum while finding min/max elements
    for (int element : arr) {
        totalSum += element;
        if (element < minElement) {
            minElement = element;
        }
        if (element > maxElement) {
            maxElement = element;
        }
    }
    
    long minSum = totalSum - maxElement;
    long maxSum = totalSum - minElement;
    
    System.out.println(minSum + " " + maxSum);
}

}

public class Solution {
    public static void main(String[] args) throws IOException {
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));

        List arr = Stream.of(bufferedReader.readLine().replaceAll("\\s+$", "").split(" "))
            .map(Integer::parseInt)
            .collect(toList());

        Result.miniMaxSum(arr);

        bufferedReader.close();
    }
}

Step-by-Step Explanation

Let’s break down the Java solution for beginners:

  1. Input Handling:

    • The main method uses BufferedReader to read a single line of five space-separated integers.

    • The input is processed using Stream.of and map to convert the strings into a List<Integer>.

  2. Variable Initialization:

    • In miniMaxSum, we initialize:

      • totalSum (type long) to store the sum of all elements.

      • minElement to Long.MAX_VALUE to ensure any input number is smaller.

      • maxElement to Long.MIN_VALUE to ensure any input number is larger.

  3. Single-Pass Processing:

    • A for-each loop iterates through the array (five elements).

    • For each number:

      • Add it to totalSum.

      • Update minElement if the number is smaller.

      • Update maxElement if the number is larger.

    • This combines summing and finding min/max in one pass, keeping time complexity at O(1) since the array size is fixed.

  4. Calculate Sums:

    • Compute minSum = totalSum - maxElement (sum of the four smallest numbers).

    • Compute maxSum = totalSum - minElement (sum of the four largest numbers).

    • Using long ensures no overflow, as inputs up to 10^9 could produce sums up to 4 × 10^9.

  5. Output:

    • Use System.out.println(minSum + " " + maxSum) to print the two sums as space-separated long integers.

  6. Resource Cleanup:

    • Close the BufferedReader to prevent resource leaks.

Common Mistakes to Avoid

Here are common pitfalls for beginners:

  • Using int instead of long: The input numbers (up to 10^9) and their sums (up to 4 × 10^9) can exceed the 32-bit integer limit (2^31 − 1 ≈ 2.1 × 10^9). Always use long for calculations.

  • Overcomplicating with Sorting: Sorting the array to find the min/max sums (e.g., sum first four vs. last four elements) works but is less efficient (O(n log n)) and unnecessary for a fixed-size array.

  • Incorrect Output Format: Printing sums on separate lines or without a space between them fails the problem’s output format requirement. Use println with string concatenation for simplicity.

Real-World Application

The Mini-Max Sum problem mirrors scenarios in data analysis and optimization. For example, in financial modeling, you might need to calculate the minimum and maximum possible returns from four out of five investment options. In logistics, this logic could apply to selecting the best four routes out of five to minimize or maximize total distance. The problem teaches array traversal and handling large numbers, skills essential for software development and data science.

FAQs

Q: Why use long instead of int for this problem?
A: Each input number can be up to 10^9, and the sum of four numbers can reach 4 × 10^9, which exceeds the 32-bit integer limit (≈2.1 × 10^9). Using long prevents overflow.

Q: Can I solve this by sorting the array?
A: Yes, sorting allows you to sum the first four elements (min sum) and last four elements (max sum), but it’s less efficient (O(n log n)) than the single-pass O(1) solution.

Q: Why not validate the input size?
A: The problem constraints guarantee exactly five positive integers, so no validation is needed, simplifying the code.