Largest Number At Least Twice of Others

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class Solution:
    def dominantIndex(self, nums: List[int]) -> int:
        max_val = max(nums)
        max_idx = nums.index(max_val)

        for num in nums:
            if num != max_val and num * 2 > max_val:
                return -1

        return max_idx

This problem can be solved by first finding the maximum element and its index, and then checking whether this maximum element is at least twice as large as all other elements in the array. If it is, then we return the index of the maximum element. Otherwise, we return -1.

Here’s the Python code:

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class Solution:
    def dominantIndex(self, nums: List[int]) -> int:
        max_val = max(nums)
        max_idx = nums.index(max_val)

        for num in nums:
            if num != max_val and num * 2 > max_val:
                return -1

        return max_idx

In this code, we first find the maximum value max_val in the list nums and its index max_idx. Then, we iterate over nums and if we find any number that is not equal to max_val and is more than half of max_val, we return -1, indicating that max_val is not at least twice as large as all other numbers. If no such number is found, we return max_idx, the index of max_val.

Identifying Problem Isomorphism

“Largest Number At Least Twice of Others” can be mapped to “Dominant Index”.

Both problems have the same essence and require the same steps for solving. They involve identifying an element in an array that is at least twice as large as every other number in the array.

In the “Largest Number At Least Twice of Others” problem, you’re asked to identify such a number, whereas in the “Dominant Index” problem, you’re asked to find the index of this number.

“Dominant Index” is a simpler as it directly asks for the index of the number which reduces one step compared to “Largest Number At Least Twice of Others” which asks for the number itself.

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class Solution:
    def dominantIndex(self, nums):
        max_val = -1
        index = -1
        second = -1
        for i in range(len(nums)):
            if nums[i] > max_val:
                second = max_val
                max_val = nums[i]
                index = i
            elif nums[i] > second:
                second = nums[i]
        return index if second * 2 <= max_val else -1

Problem Classification

It is a basic algorithmic problem where the task is to analyze the given array and find the index of the largest element given certain conditions.

‘What’ Components:

  1. An integer array, ’nums’, where the largest integer is unique.
  2. The task is to find out whether the largest element in the array is at least twice as much as every other number in the array.
  3. If the largest element is at least twice as much as every other number in the array, return the index of the largest element. Otherwise, return -1.

This problem can be classified as a ‘Search and Decision’ problem. The reason for this classification is because the problem involves searching for the largest element and its index in an array (which is a Search operation), and then making a decision based on the comparison of the largest element with all other elements in the array (which is a Decision operation).

Language Agnostic Coding Drills

  1. Dissection of the code:

This code involves several distinct concepts: - Variable Initialization: This is evident in the beginning of the method, where max_val, index, and second are initialized to -1. - Looping through an array: The for-loop iterates through the nums array. - Conditional Statements: The if and elif statements are used to check conditions for array elements. - Updating Variables: Inside the if and elif blocks, variables max_val, index, and second are updated based on certain conditions. - Ternary Operator: The last line uses a ternary operator to determine which value to return.

  1. Coding concepts listed in order of increasing difficulty:

    • Variable Initialization: This is a fundamental concept in almost all programming languages. Difficulty: 1/5
    • Looping through an array: This is another basic concept where the coder needs to understand how to iterate over elements in an array. Difficulty: 2/5
    • Conditional Statements: This requires understanding of comparison operators and logical flow. Difficulty: 2/5
    • Updating Variables: This concept involves understanding the current logical flow and updating variables accordingly. Difficulty: 3/5
    • Ternary Operator: It’s a compact form of if-else statement, used for simple decisions. Understanding when to use it can be tricky for beginners. Difficulty: 3/5

  2. Problem-solving approach:

From the problem statement, we need to find if the maximum number in an array is at least twice as large as every other number in the array. This involves finding the largest number (and its index) and the second largest number.

The approach here is to iterate over the array and maintain variables for the maximum and second maximum numbers. If we find a number larger than the current maximum, we update the second maximum with the value of the current maximum and the maximum with the current number. If the number is not larger than the maximum but is larger than the second maximum, we just update the second maximum.

At the end, we check if the maximum number is at least twice as large as the second maximum. If it is, we return the index of the maximum number, otherwise, we return -1.

Each of these coding drills contributes to the overall solution in a way that it brings us closer to the desired output as described in the problem statement. For instance, variable initialization sets up our required variables, the for-loop helps us navigate through the array, conditional statements help us make decisions on the fly while navigating, variable updating assists in maintaining a record of our findings, and finally, the ternary operator aids in deciding what to return.

Targeted Drills in Python

  1. Python-based coding drills for each concept:

    • Variable Initialization:
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    # Example of variable initialization
variable = 0
    • Looping through an array:
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    # Example of looping through an array
array = [1, 2, 3, 4, 5]
for element in array:
    print(element)
    • Conditional Statements:
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    # Example of conditional statements
value = 5
if value > 3:
    print("Value is greater than 3")
elif value == 3:
    print("Value is equal to 3")
else:
    print("Value is less than 3")
    • Updating Variables:
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    # Example of updating variables
count = 0
count += 1  # count is now 1
    • Ternary Operator:
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    # Example of ternary operator
value = 5
print("Value is greater than 3") if value > 3 else print("Value is less than or equal to 3")

  2. Problem-specific concepts:

The problem-specific concept here is finding the maximum and second maximum element in the list. This concept is essential to our problem as we are required to find the largest and second largest number to be able to determine whether the largest number is at least twice as large as every other number in the array.

  • Finding maximum and second maximum:
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# Example of finding maximum and second maximum
array = [1, 2, 3, 4, 5]
max_val = second_max = -1
for num in array:
    if num > max_val:
        second_max = max_val
        max_val = num
    elif num > second_max:
        second_max = num
print("Max:", max_val, "Second max:", second_max)
  1. Integration of drills for the final solution:

Start by initializing the variables max_val, second_max and max_index to -1. Then iterate over the array using a for-loop. For each element in the array, use conditional statements to check if the current element is greater than max_val. If it is, update second_max with the value of max_val and max_val with the current element’s value and update max_index with the current index. If it’s not, but is greater than second_max, update second_max with the current element’s value. After iterating through all the elements in the array, use a ternary operator to check if max_val is at least twice as much as second_max. If it is, return max_index, else return -1.

This way, all the individual drills contribute to forming the final solution.