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Find the Index of the Large Integer

Medium

Asked by:

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Topics:

Arrays

You are given an integer array accounts where accounts[i][j] is the amount of money the i^th customer has in the j^th bank. Return the wealth that the richest customer has.

A customer's wealth is the amount of money they have in all their bank accounts. The richest customer is the customer that has the maximum wealth.

Example 1:

Input: accounts = [[1,2,3],[3,2,1]]
Output: 6
Explanation:
1st customer has wealth = 1 + 2 + 3 = 6
2nd customer has wealth = 3 + 2 + 1 = 6
Both customers are considered the richest with a wealth of 6, so return 6.

Example 2:

Input: accounts = [[1,5],[7,3],[3,5]]
Output: 10
Explanation: 
1st customer has wealth = 6
2nd customer has wealth = 10 
3rd customer has wealth = 8
The 2nd customer is the richest with a wealth of 10.

Example 3:

Input: accounts = [[2,8,7],[7,1,3],[1,9,5]]
Output: 17

Constraints:

m == accounts.length
n == accounts[i].length
1 <= m, n <= 50
1 <= accounts[i][j] <= 100

Try coding it in LeetCode

Find the Index of the Large Integer

Medium

Asked by:

12 views

Topics:

Arrays

You are given an integer array accounts where accounts[i][j] is the amount of money the i^th customer has in the j^th bank. Return the wealth that the richest customer has.

A customer's wealth is the amount of money they have in all their bank accounts. The richest customer is the customer that has the maximum wealth.

Example 1:

Input: accounts = [[1,2,3],[3,2,1]]
Output: 6
Explanation:
1st customer has wealth = 1 + 2 + 3 = 6
2nd customer has wealth = 3 + 2 + 1 = 6
Both customers are considered the richest with a wealth of 6, so return 6.

Example 2:

Input: accounts = [[1,5],[7,3],[3,5]]
Output: 10
Explanation: 
1st customer has wealth = 6
2nd customer has wealth = 10 
3rd customer has wealth = 8
The 2nd customer is the richest with a wealth of 10.

Example 3:

Input: accounts = [[2,8,7],[7,1,3],[1,9,5]]
Output: 17

Constraints:

m == accounts.length
n == accounts[i].length
1 <= m, n <= 50
1 <= accounts[i][j] <= 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 data type of the numbers in the array, and can they be floating-point numbers or only integers?
Can the array be empty, or contain null or undefined values? What should I return in those cases?
If there are multiple largest integers (duplicates), should I return the index of the first occurrence, or is any index acceptable?
What is the expected range of the integer values in the array? Are there any limits or constraints?
If the input is valid (non-empty array with numbers), is it guaranteed that at least one element will exist, or is it possible the concept of 'largest' is undefined (e.g., array contains NaN)? What should I return in such a case?

Brute Force Solution

Approach

The brute force approach to finding the largest number in a collection involves examining each number individually. We start by assuming the first number is the largest, then compare it against every other number in the collection. If we find a bigger number, we update our assumption of which number is the largest.

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

First, consider the very first number you see as the biggest one so far.
Then, look at the second number and compare it to your current 'biggest' number.
If the second number is actually bigger, remember that number instead.
Next, compare the third number to the 'biggest' number you're remembering.
Again, if the third number is bigger, update your memory.
Keep doing this comparison for every single number in the collection.
After checking all the numbers, the one you are remembering is the biggest.

Code Implementation

def find_index_of_the_large_integer(list_of_numbers):
    if not list_of_numbers:
        return -1

    # Assume the first element is the largest to start.
    index_of_largest_number_seen_so_far = 0

    for current_index in range(1, len(list_of_numbers)):

        # If we find a larger number, update the index.
        if list_of_numbers[current_index] > list_of_numbers[index_of_largest_number_seen_so_far]:

            index_of_largest_number_seen_so_far = current_index

            # Update largest number

    return index_of_largest_number_seen_so_far

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through each element of the input array once to find the largest element. For an input array of size n, the algorithm performs a single comparison for each element to the current largest, resulting in n comparisons. Therefore, the time complexity is directly proportional to the size of the input, which gives us a Big O notation of O(n).

Space Complexity

O(1) – The algorithm maintains a variable to store the index of the current largest number. No additional data structures dependent on the input size N (the number of elements in the collection) are created. Therefore, the space required remains constant regardless of the input size, resulting in a space complexity of O(1).

Optimal Solution

Approach

The efficient way to find the largest number is to avoid comparing every single number. We can use a divide and conquer strategy similar to how you would find a word in a dictionary. This narrows down the search quickly.

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

Start by looking at the middle number.
Compare the middle number with the number next to it on either side.
If the middle number is larger than both of its neighbors, you've found the largest number and are done.
If either neighbor is bigger than the middle number, then the largest number must be in the half that contains that bigger neighbor. Focus on that half and ignore the other half.
Repeat the process of checking the middle number of the remaining half and comparing it to its neighbors. Keep narrowing down the search until you find the largest number.

Code Implementation

def find_peak_element_index(numbers):
    left_index = 0
    right_index = len(numbers) - 1

    while left_index < right_index:
        middle_index = (left_index + right_index) // 2

        # Check if middle element is greater than its neighbors.
        if numbers[middle_index] > numbers[middle_index - 1] and \
           numbers[middle_index] > numbers[middle_index + 1]:

            return middle_index

        # If the right neighbor is greater, search the right half.
        if numbers[middle_index + 1] > numbers[middle_index]:
            left_index = middle_index + 1

        # Otherwise, search the left half.
        else:
            right_index = middle_index

    # When left and right indices converge, it's the peak.
    return left_index

Big(O) Analysis

Time Complexity

O(log n) – The algorithm employs a divide and conquer strategy. At each step, the search space is halved by comparing the middle element to its neighbors and focusing on the half containing the larger neighbor. This halving of the input size 'n' continues until the largest element is found. The number of steps required to halve the problem until a single element is isolated is logarithmic, hence the time complexity is O(log n).

Space Complexity

O(log N) – The algorithm uses a divide and conquer approach. The space complexity is determined by the maximum depth of the recursion stack, which is logarithmic with respect to the input size N, where N is the number of elements in the input. In each recursive call, a constant amount of memory is used for variables (middle index). Since the problem size is halved at each step, the maximum depth of the recursion is log base 2 of N. Therefore, the space complexity is O(log N).

Edge Cases

Case	How to Handle
Null or empty input array	Return -1 or throw an IllegalArgumentException, depending on requirements.
Array with only one element	Return 0 since it's the largest element or throw an exception since there is no element to compare to.
Array with all identical values	Return 0 as the index of the first element since it is equal to any other element in the array or the problem should specify what to return in case of a tie.
Array with extremely large numbers (potential integer overflow)	Use long data type to store the numbers and perform comparisons.
Array with negative numbers	The algorithm should correctly handle negative numbers by comparing them according to their numerical value.
Maximum sized input array (memory constraints)	Ensure the solution uses memory efficiently, considering in-place operations or streaming if possible.
Array containing zeros	The algorithm should correctly compare zeros with other positive and negative numbers.
Extremely skewed distribution with one very large number.	The algorithm should efficiently identify the largest number without being significantly impacted by the range of values.

Null or empty input array

How to Handle:

Return -1 or throw an IllegalArgumentException, depending on requirements.

Array with only one element

How to Handle:

Return 0 since it's the largest element or throw an exception since there is no element to compare to.

Array with all identical values

How to Handle:

Return 0 as the index of the first element since it is equal to any other element in the array or the problem should specify what to return in case of a tie.

Array with extremely large numbers (potential integer overflow)

How to Handle:

Use long data type to store the numbers and perform comparisons.

Array with negative numbers

How to Handle:

The algorithm should correctly handle negative numbers by comparing them according to their numerical value.

Maximum sized input array (memory constraints)

How to Handle:

Ensure the solution uses memory efficiently, considering in-place operations or streaming if possible.

Array containing zeros

How to Handle:

The algorithm should correctly compare zeros with other positive and negative numbers.

Extremely skewed distribution with one very large number.

How to Handle:

The algorithm should efficiently identify the largest number without being significantly impacted by the range of values.

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 data type of the numbers in the array, and can they be floating-point numbers or only integers?
Can the array be empty, or contain null or undefined values? What should I return in those cases?
If there are multiple largest integers (duplicates), should I return the index of the first occurrence, or is any index acceptable?
What is the expected range of the integer values in the array? Are there any limits or constraints?
If the input is valid (non-empty array with numbers), is it guaranteed that at least one element will exist, or is it possible the concept of 'largest' is undefined (e.g., array contains NaN)? What should I return in such a case?

Brute Force Solution

Approach

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

First, consider the very first number you see as the biggest one so far.
Then, look at the second number and compare it to your current 'biggest' number.
If the second number is actually bigger, remember that number instead.
Next, compare the third number to the 'biggest' number you're remembering.
Again, if the third number is bigger, update your memory.
Keep doing this comparison for every single number in the collection.
After checking all the numbers, the one you are remembering is the biggest.

Code Implementation

def find_index_of_the_large_integer(list_of_numbers):
    if not list_of_numbers:
        return -1

    # Assume the first element is the largest to start.
    index_of_largest_number_seen_so_far = 0

    for current_index in range(1, len(list_of_numbers)):

        # If we find a larger number, update the index.
        if list_of_numbers[current_index] > list_of_numbers[index_of_largest_number_seen_so_far]:

            index_of_largest_number_seen_so_far = current_index

            # Update largest number

    return index_of_largest_number_seen_so_far

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Start by looking at the middle number.
Compare the middle number with the number next to it on either side.
If the middle number is larger than both of its neighbors, you've found the largest number and are done.
If either neighbor is bigger than the middle number, then the largest number must be in the half that contains that bigger neighbor. Focus on that half and ignore the other half.
Repeat the process of checking the middle number of the remaining half and comparing it to its neighbors. Keep narrowing down the search until you find the largest number.

Code Implementation

def find_peak_element_index(numbers):
    left_index = 0
    right_index = len(numbers) - 1

    while left_index < right_index:
        middle_index = (left_index + right_index) // 2

        # Check if middle element is greater than its neighbors.
        if numbers[middle_index] > numbers[middle_index - 1] and \
           numbers[middle_index] > numbers[middle_index + 1]:

            return middle_index

        # If the right neighbor is greater, search the right half.
        if numbers[middle_index + 1] > numbers[middle_index]:
            left_index = middle_index + 1

        # Otherwise, search the left half.
        else:
            right_index = middle_index

    # When left and right indices converge, it's the peak.
    return left_index

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input array	Return -1 or throw an IllegalArgumentException, depending on requirements.
Array with only one element	Return 0 since it's the largest element or throw an exception since there is no element to compare to.
Array with all identical values	Return 0 as the index of the first element since it is equal to any other element in the array or the problem should specify what to return in case of a tie.
Array with extremely large numbers (potential integer overflow)	Use long data type to store the numbers and perform comparisons.
Array with negative numbers	The algorithm should correctly handle negative numbers by comparing them according to their numerical value.
Maximum sized input array (memory constraints)	Ensure the solution uses memory efficiently, considering in-place operations or streaming if possible.
Array containing zeros	The algorithm should correctly compare zeros with other positive and negative numbers.
Extremely skewed distribution with one very large number.	The algorithm should efficiently identify the largest number without being significantly impacted by the range of values.

Null or empty input array

How to Handle:

Return -1 or throw an IllegalArgumentException, depending on requirements.

Array with only one element

How to Handle:

Return 0 since it's the largest element or throw an exception since there is no element to compare to.

Array with all identical values

How to Handle:

Return 0 as the index of the first element since it is equal to any other element in the array or the problem should specify what to return in case of a tie.

Array with extremely large numbers (potential integer overflow)

How to Handle:

Use long data type to store the numbers and perform comparisons.

Array with negative numbers

How to Handle:

The algorithm should correctly handle negative numbers by comparing them according to their numerical value.

Maximum sized input array (memory constraints)

How to Handle:

Ensure the solution uses memory efficiently, considering in-place operations or streaming if possible.

Array containing zeros

How to Handle:

The algorithm should correctly compare zeros with other positive and negative numbers.

Extremely skewed distribution with one very large number.

How to Handle:

The algorithm should efficiently identify the largest number without being significantly impacted by the range of values.