Find Pivot Index

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Given an array of integers nums, calculate the pivot index of this array.

The pivot index is the index where the sum of all the numbers strictly to the left of the index is equal to the sum of all the numbers strictly to the index's right.

If the index is on the left edge of the array, then the left sum is 0 because there are no elements to the left. This also applies to the right edge of the array.

Return the leftmost pivot index. If no such index exists, return -1.

For example:

nums = [1,7,3,6,5,6] should return 3. The left sum at index 3 is 1 + 7 + 3 = 11, and the right sum is 5 + 6 = 11.
nums = [1,2,3] should return -1. There is no index that satisfies the conditions.
nums = [2,1,-1] should return 0. The left sum at index 0 is 0, and the right sum is 1 + -1 = 0.

Write a function to find the pivot index of a given array of integers. Your function should take an array of integers as input and return an integer representing the pivot index, or -1 if no such index exists. Optimize for time complexity.

Find Pivot Index

Easy

Google

2 views

Topics:

Arrays

Given an array of integers nums, calculate the pivot index of this array.

The pivot index is the index where the sum of all the numbers strictly to the left of the index is equal to the sum of all the numbers strictly to the index's right.

If the index is on the left edge of the array, then the left sum is 0 because there are no elements to the left. This also applies to the right edge of the array.

Return the leftmost pivot index. If no such index exists, return -1.

For example:

nums = [1,7,3,6,5,6] should return 3. The left sum at index 3 is 1 + 7 + 3 = 11, and the right sum is 5 + 6 = 11.
nums = [1,2,3] should return -1. There is no index that satisfies the conditions.
nums = [2,1,-1] should return 0. The left sum at index 0 is 0, and the right sum is 1 + -1 = 0.

Write a function to find the pivot index of a given array of integers. Your function should take an array of integers as input and return an integer representing the pivot index, or -1 if no such index exists. Optimize for time complexity.

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 expected return value if no pivot index exists?
What are the possible range of values for the integers in the input array? Can they be negative, zero, or only positive?
What is the maximum possible length of the input array 'nums'?
If multiple pivot indices exist, should I return the first (leftmost) one?
Can I assume the input 'nums' is always a valid array and not null or empty?

Brute Force Solution

Approach

The brute force strategy involves checking every single possible location to see if it satisfies the conditions. We consider each possible location one at a time and calculate the sums of the values to its left and right. We stop when we find a location where those sums are equal.

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

Start by considering the first location in the sequence.
Calculate the sum of all the values to the left of this location. Since it's the first location, this sum will be zero.
Calculate the sum of all the values to the right of this location.
Compare the two sums. If they are equal, you've found the desired location, and you're done.
If the sums aren't equal, move to the next location in the sequence.
Repeat the process of calculating the sums to the left and right of the current location and comparing them.
Continue until you either find a location where the sums are equal or you've checked every location in the sequence.

Code Implementation

def find_pivot_index_brute_force(numbers):
    for current_index in range(len(numbers)):
        # Calculate left sum for the current index
        left_sum = 0
        for left_index in range(current_index):
            left_sum += numbers[left_index]

        # Calculate right sum for the current index
        right_sum = 0
        for right_index in range(current_index + 1, len(numbers)):
            right_sum += numbers[right_index]

        # Check if current index is the pivot index
        if left_sum == right_sum:
            return current_index

    # No pivot index found
    return -1

Big(O) Analysis

Time Complexity

O(n²) – The brute force approach iterates through each of the n elements of the array as a potential pivot index. For each element, it calculates the sum of the elements to its left and the sum of the elements to its right. Calculating each of these sums requires iterating through a portion of the array, which, in the worst case, could be almost the entire array. Thus, for each of the n elements, we potentially perform up to n operations to calculate the sums, resulting in roughly n * n operations which simplifies to O(n²).

Space Complexity

O(1) – The brute force approach calculates left and right sums for each potential pivot index. The plain English description doesn't suggest any auxiliary data structures like lists or dictionaries are created to store intermediate sums or track visited indices. The algorithm uses only a few variables to store the current index, left sum, and right sum. Therefore, the space used remains constant irrespective of the input array's size (N), leading to O(1) space complexity.

Optimal Solution

Approach

To find that special spot, we avoid recalculating the entire sum every time. Instead, we'll cleverly track the sums on either side as we go, making the process much faster.

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

First, find the total sum of all the numbers.
Start from the beginning, keeping track of the sum of the numbers to the left of your current spot.
To find the sum of the numbers to the right, subtract the current number and the left sum from the total sum.
If the left sum and the right sum are the same, you've found the special spot! Return its location.
If you reach the end without finding equal sums, there's no special spot, so indicate that.

Code Implementation

def find_pivot_index(numbers) -> int:
    total_sum = sum(numbers)

    left_sum = 0

    for i, number in enumerate(numbers):
        # Calculate right sum dynamically
        right_sum = total_sum - number - left_sum

        # Check if sums are equal, indicating pivot
        if left_sum == right_sum:
            return i

        # Accumulate left sum for next iteration
        left_sum += number

    # If no pivot index is found
    return -1

Big(O) Analysis

Time Complexity

O(n) – The algorithm first calculates the total sum of the array, which takes O(n) time where n is the size of the input array. Then, it iterates through the array once, keeping track of the left sum and calculating the right sum in constant time by subtracting from the total sum. The comparison of left and right sums also takes constant time within the loop. Since the dominant operation is the single pass through the array, the overall time complexity is O(n).

Space Complexity

O(1) – The provided solution calculates the total sum and then iterates through the input array, maintaining only a left sum. It doesn't use any additional data structures that scale with the input size N (the number of elements in the array). The space used for the total sum, the left sum, and the current element's index remains constant regardless of N. Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Empty input array	Return -1 immediately as there can be no pivot index in an empty array.
Array with a single element	Return 0, as the single element trivially satisfies the condition (left and right sums are both 0).
Array with two elements	Calculate the left and right sums (which will always be one element each) and return 0 if the right is 0, otherwise -1 if not
Array with all elements being zero	Return 0, as the first index will always be a valid pivot.
Array with only negative numbers	The solution should correctly handle negative sums and potentially no pivot exists, returning -1.
Integer overflow if calculating total sum naively	Use long data type or incremental calculation to avoid overflow when computing sums.
Large array with a pivot index near the end	The solution should avoid recalculating sums for each potential pivot and be efficient for large arrays O(n) is optimal.
No pivot index exists in the array	After iterating through all possible indices, return -1 if no pivot is found.