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Sum of Digits in the Minimum Number

Easy

Asked by:

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Topics:

ArraysBit Manipulation

Given an integer array nums, return the sum of the digits of the minimum number in nums.

Return 0 if the sum of the digits of the minimum number is odd, or 1 otherwise.

Example 1:


Input: nums = [34,23,1,24,75,33,54,8]
Output: 0
Explanation: 
The minimum element in nums is 1.
The sum of digits of 1 is 1, which is an odd number.
Therefore, we return 0.

Example 2:


Input: nums = [99,77,33,66,55]
Output: 1
Explanation: 
The minimum element in nums is 33.
The sum of digits of 33 is 3 + 3 = 6, which is an even number.
Therefore, we return 1.

Constraints:

1 <= nums.length <= 100
1 <= nums[i] <= 100

Try coding it in LeetCode

Sum of Digits in the Minimum Number

Easy

Asked by:

27 views

Topics:

ArraysBit Manipulation

Given an integer array nums, return the sum of the digits of the minimum number in nums.

Return 0 if the sum of the digits of the minimum number is odd, or 1 otherwise.

Example 1:


Input: nums = [34,23,1,24,75,33,54,8]
Output: 0
Explanation: 
The minimum element in nums is 1.
The sum of digits of 1 is 1, which is an odd number.
Therefore, we return 0.

Example 2:


Input: nums = [99,77,33,66,55]
Output: 1
Explanation: 
The minimum element in nums is 33.
The sum of digits of 33 is 3 + 3 = 6, which is an even number.
Therefore, we return 1.

Constraints:

1 <= nums.length <= 100
1 <= nums[i] <= 100

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values for the numbers in the input array?
Can the input array be empty or null?
If the minimum number appears multiple times in the array, should I still only calculate the sum of digits for one instance of it?
What should I return if the sum of the digits exceeds the maximum integer value?
Are all the numbers in the input array integers?

Brute Force Solution

Approach

The brute force approach to this problem involves finding the smallest number in a given set of numbers, and then computing the sum of its digits. We find the smallest number by simply comparing each number to every other number. Then, we add up the digits of that smallest number.

Here's how the algorithm would work step-by-step:

First, we need to find the smallest number from our collection of numbers.
To do this, we'll start by assuming the first number we see is the smallest.
Then, we'll compare our current smallest number with the next number in the collection.
If the next number is smaller than our current smallest, we'll say the next number is now the smallest.
We'll keep doing this comparison, number by number, until we've looked at all the numbers in the collection.
At the end of this process, we'll have identified the absolute smallest number.
Now that we know the smallest number, we need to add up all of its digits.
We can do this by looking at each digit individually and adding it to a total sum.
For example, if the smallest number is 25, we add 2 and 5 to get a sum of 7.

Code Implementation

def sum_of_digits_in_minimum(
    numbers_list):

    smallest_number = numbers_list[0]

    # Iterate to find the actual smallest number
    for number in numbers_list:

        # Update smallest if a smaller number is found
        if number < smallest_number:

            smallest_number = number

    sum_of_digits = 0

    # Calculate sum of the digits in smallest number
    for digit_char in str(smallest_number):

        digit = int(digit_char)
        sum_of_digits += digit

    return sum_of_digits

Big(O) Analysis

Time Complexity

O(n) – Finding the minimum element involves iterating through the array of n numbers once. Comparing each number to the current minimum takes constant time. After finding the minimum, we calculate the sum of its digits. The number of digits in the minimum number is independent of the size of the input array, n, and can be considered a constant, meaning summing its digits takes constant time as well. Therefore, the overall time complexity is dominated by the initial iteration to find the minimum, resulting in O(n).

Space Complexity

O(1) – The algorithm uses a variable to store the current smallest number and another to accumulate the sum of the digits of that smallest number. These variables require a constant amount of space, irrespective of the size of the input array (N). Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

The goal is to find the smallest number in a group and then add up the digits of that smallest number. We can accomplish this by first identifying the smallest number, and then adding its digits together one at a time to get the final sum.

Here's how the algorithm would work step-by-step:

First, go through all the numbers and find the smallest one.
Now, take the smallest number you found and separate it into individual digits.
Finally, add all those individual digits together to get the sum.

Code Implementation

def sum_of_digits_in_minimum_number(number_list):
    minimum_number = number_list[0]

    for number in number_list:
        if number < minimum_number:
            minimum_number = number

    # Need to convert to string to extract each digit.
    minimum_number_string = str(minimum_number)
    digit_sum = 0

    # Iterate through each digit and add.
    for digit_character in minimum_number_string:
        digit_sum += int(digit_character)

    return digit_sum

Big(O) Analysis

Time Complexity

O(n) – Finding the minimum number in the array of n numbers requires iterating through all elements once. Summing the digits of a number involves a number of operations proportional to the number of digits. Since the minimum number is bounded by the input numbers and the number of digits in any element is independent of the size of the input array, the digit summing operation is a constant time operation. Therefore, the overall time complexity is dominated by the linear time complexity of finding the minimum, resulting in O(n).

Space Complexity

O(1) – The algorithm finds the minimum number and then calculates the sum of its digits. The space used is for storing the minimum number found so far and the running sum of the digits. These are constant space operations, independent of the number of input numbers, denoted as N. Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Empty input array	Return 0 as there are no numbers to process.
Array with a single element	Return the sum of digits of the single element.
Array containing only zero	The minimum number is zero, return 0 (sum of digits of 0).
Array with large numbers that could cause integer overflow in the sum of digits	Use a larger data type like long to store the sum of digits to prevent overflow.
Array contains negative numbers	Take absolute value of the numbers before comparing them to find min.
Array containing very large number of identical elements, where one of them is minimum.	The standard findMin approach correctly identifies the minimum regardless of duplicates.
Integer overflow when calculating sum of digits of a very large minimum number.	Use long to prevent integer overflow.
Input array containing MAX_INT values.	Handle potentially slow performance with a very large array by short circuiting or pre-calculating the digits sum.

Empty input array

How to Handle:

Return 0 as there are no numbers to process.

Array with a single element

How to Handle:

Return the sum of digits of the single element.

Array containing only zero

How to Handle:

The minimum number is zero, return 0 (sum of digits of 0).

Array with large numbers that could cause integer overflow in the sum of digits

How to Handle:

Use a larger data type like long to store the sum of digits to prevent overflow.

Array contains negative numbers

How to Handle:

Take absolute value of the numbers before comparing them to find min.

Array containing very large number of identical elements, where one of them is minimum.

How to Handle:

The standard findMin approach correctly identifies the minimum regardless of duplicates.

Integer overflow when calculating sum of digits of a very large minimum number.

How to Handle:

Use long to prevent integer overflow.

Input array containing MAX_INT values.

How to Handle:

Handle potentially slow performance with a very large array by short circuiting or pre-calculating the digits sum.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values for the numbers in the input array?
Can the input array be empty or null?
If the minimum number appears multiple times in the array, should I still only calculate the sum of digits for one instance of it?
What should I return if the sum of the digits exceeds the maximum integer value?
Are all the numbers in the input array integers?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

First, we need to find the smallest number from our collection of numbers.
To do this, we'll start by assuming the first number we see is the smallest.
Then, we'll compare our current smallest number with the next number in the collection.
If the next number is smaller than our current smallest, we'll say the next number is now the smallest.
We'll keep doing this comparison, number by number, until we've looked at all the numbers in the collection.
At the end of this process, we'll have identified the absolute smallest number.
Now that we know the smallest number, we need to add up all of its digits.
We can do this by looking at each digit individually and adding it to a total sum.
For example, if the smallest number is 25, we add 2 and 5 to get a sum of 7.

Code Implementation

def sum_of_digits_in_minimum(
    numbers_list):

    smallest_number = numbers_list[0]

    # Iterate to find the actual smallest number
    for number in numbers_list:

        # Update smallest if a smaller number is found
        if number < smallest_number:

            smallest_number = number

    sum_of_digits = 0

    # Calculate sum of the digits in smallest number
    for digit_char in str(smallest_number):

        digit = int(digit_char)
        sum_of_digits += digit

    return sum_of_digits

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

First, go through all the numbers and find the smallest one.
Now, take the smallest number you found and separate it into individual digits.
Finally, add all those individual digits together to get the sum.

Code Implementation

def sum_of_digits_in_minimum_number(number_list):
    minimum_number = number_list[0]

    for number in number_list:
        if number < minimum_number:
            minimum_number = number

    # Need to convert to string to extract each digit.
    minimum_number_string = str(minimum_number)
    digit_sum = 0

    # Iterate through each digit and add.
    for digit_character in minimum_number_string:
        digit_sum += int(digit_character)

    return digit_sum

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty input array	Return 0 as there are no numbers to process.
Array with a single element	Return the sum of digits of the single element.
Array containing only zero	The minimum number is zero, return 0 (sum of digits of 0).
Array with large numbers that could cause integer overflow in the sum of digits	Use a larger data type like long to store the sum of digits to prevent overflow.
Array contains negative numbers	Take absolute value of the numbers before comparing them to find min.
Array containing very large number of identical elements, where one of them is minimum.	The standard findMin approach correctly identifies the minimum regardless of duplicates.
Integer overflow when calculating sum of digits of a very large minimum number.	Use long to prevent integer overflow.
Input array containing MAX_INT values.	Handle potentially slow performance with a very large array by short circuiting or pre-calculating the digits sum.

Empty input array

How to Handle:

Return 0 as there are no numbers to process.

Array with a single element

How to Handle:

Return the sum of digits of the single element.

Array containing only zero

How to Handle:

The minimum number is zero, return 0 (sum of digits of 0).

Array with large numbers that could cause integer overflow in the sum of digits

How to Handle:

Use a larger data type like long to store the sum of digits to prevent overflow.

Array contains negative numbers

How to Handle:

Take absolute value of the numbers before comparing them to find min.

Array containing very large number of identical elements, where one of them is minimum.

How to Handle:

The standard findMin approach correctly identifies the minimum regardless of duplicates.

Integer overflow when calculating sum of digits of a very large minimum number.

How to Handle:

Use long to prevent integer overflow.

Input array containing MAX_INT values.

How to Handle:

Handle potentially slow performance with a very large array by short circuiting or pre-calculating the digits sum.