Problem Description

You are given a 0-indexed integer array nums of length n.

A valid split exists at index i if both of these conditions are met:

1. The sum of the first i + 1 elements is greater than or equal to the sum of the last n - i - 1 elements
2. There is at least one element to the right of i (meaning 0 <= i < n - 1)

Your task is to return the total number of valid splits in the array nums.

For example, if we split at index i, we're dividing the array into two parts:

• Left part: elements from index 0 to i (inclusive)
• Right part: elements from index i + 1 to n - 1 (inclusive)

The split is valid when the sum of the left part is greater than or equal to the sum of the right part, and the right part is not empty.

Intuition

To find valid splits, we need to compare the sum of the left part with the sum of the right part at each possible split position. A naive approach would be to calculate both sums from scratch at each index, but this would be inefficient.

The key insight is that we can use the relationship between the left sum and right sum more cleverly. If we know the total sum s of the entire array, and we know the sum of the left part t at any split point, then the sum of the right part is simply s - t.

So the condition "left sum >= right sum" becomes t >= s - t, which simplifies to 2 * t >= s or t >= s - t.

As we traverse the array from left to right, we can maintain a running sum t that represents the sum of elements from index 0 to the current index. At each position i, we check if t >= s - t. If this condition is satisfied, we have found a valid split.

We only need to check positions from index 0 to n - 2 (hence nums[:-1] in the code) because we need at least one element in the right part for a valid split.

This approach allows us to solve the problem in a single pass through the array with O(n) time complexity, after calculating the total sum once at the beginning.

Learn more about Prefix Sum patterns.

Solution Approach

The solution uses a prefix sum approach to efficiently count valid splits.

Step-by-step implementation:

1. Calculate the total sum: First, we compute the total sum s of all elements in the array using sum(nums). This gives us the sum of the entire array.

2. Initialize variables:

   • ans = 0: Counter for valid splits
   • t = 0: Running prefix sum

3. Traverse and check each potential split point: We iterate through nums[:-1] (all elements except the last one) because we need at least one element in the right part:

   • For each element x, we add it to our running prefix sum: t += x
   • This t now represents the sum of all elements from index 0 to the current index (left part)
   • The sum of the right part is s - t

4. Check validity condition: At each position, we check if t >= s - t:

   • If true, we increment our answer counter: ans += t >= s - t
   • The expression t >= s - t evaluates to True (1) or False (0), which gets added to ans

5. Return the result: After checking all possible split positions, return ans.

Time Complexity: O(n) - We traverse the array twice: once to calculate the total sum and once to check each split position.

Space Complexity: O(1) - We only use a constant amount of extra space for variables s, ans, and t.

The elegance of this solution lies in avoiding redundant calculations by maintaining a running prefix sum and using the mathematical relationship between the left sum, right sum, and total sum.

Example Walkthrough

Let's walk through the solution with nums = [10, 4, -8, 7].

Step 1: Calculate total sum

• s = 10 + 4 + (-8) + 7 = 13

Step 2: Initialize variables

• ans = 0 (count of valid splits)
• t = 0 (running prefix sum)

Step 3: Check each potential split point

We iterate through nums[:-1] which is [10, 4, -8]:

Index 0 (element = 10):

• Update prefix sum: t = 0 + 10 = 10
• Left part sum = 10 (just element at index 0)
• Right part sum = s - t = 13 - 10 = 3 (elements at indices 1, 2, 3)
• Check: Is t >= s - t? Is 10 >= 3? Yes!
• This is a valid split: [10] | [4, -8, 7]
• ans = 1

Index 1 (element = 4):

• Update prefix sum: t = 10 + 4 = 14
• Left part sum = 14 (elements at indices 0, 1)
• Right part sum = s - t = 13 - 14 = -1 (elements at indices 2, 3)
• Check: Is t >= s - t? Is 14 >= -1? Yes!
• This is a valid split: [10, 4] | [-8, 7]
• ans = 2

Index 2 (element = -8):

• Update prefix sum: t = 14 + (-8) = 6
• Left part sum = 6 (elements at indices 0, 1, 2)
• Right part sum = s - t = 13 - 6 = 7 (element at index 3)
• Check: Is t >= s - t? Is 6 >= 7? No!
• This is not a valid split: [10, 4, -8] | [7]
• ans = 2 (unchanged)

Step 4: Return result

• Total number of valid splits = 2

The valid splits are at indices 0 and 1, where the left part sums (10 and 14) are greater than or equal to their corresponding right part sums (3 and -1).

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the array nums. This is because:

• Computing the initial sum s = sum(nums) takes O(n) time
• The for loop iterates through n-1 elements (all elements except the last one)
• Inside each iteration, we perform constant time operations: adding to t, comparison t >= s - t, and incrementing ans
• Overall, we have O(n) + O(n-1) = O(n)

The space complexity is O(1). The algorithm only uses a fixed amount of extra space:

• Variable s to store the total sum
• Variable ans to count valid splits
• Variable t to track the running sum of the left part
• Variable x for iteration

These variables use constant space regardless of the input size, resulting in O(1) space complexity.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Integer Overflow in Other Languages

While Python handles arbitrarily large integers automatically, implementing this solution in languages like Java or C++ can lead to integer overflow when dealing with large array values or long arrays.

Problem Example:

# In Java/C++, if nums contains large values like [10^9, 10^9, 10^9, ...]
# The sum calculation might overflow a 32-bit integer

Solution: Use 64-bit integers (long in Java, long long in C++) for sum calculations:

# Python handles this automatically, but in other languages:
# Java: long totalSum = 0L;
# C++: long long totalSum = 0LL;

2. Incorrect Loop Boundary

A frequent mistake is iterating through the entire array instead of stopping at n-2 (or using nums[:-1] in Python).

Problem Example:

# WRONG: This includes the last element as a potential split point
for i in range(len(nums)):
    left_sum += nums[i]
    # This would count an invalid split where right part is empty

Solution: Always ensure at least one element remains in the right part:

# CORRECT: Stop before the last element
for i in range(len(nums) - 1):  # or use nums[:-1]
    left_sum += nums[i]
    right_sum = total_sum - left_sum
    # Right part is guaranteed to have at least one element

3. Floating Point Precision Issues

If the array contains floating-point numbers (though the problem typically uses integers), comparison operations might fail due to precision errors.

Problem Example:

# If nums contained floats
nums = [0.1, 0.2, 0.3]  # 0.1 + 0.2 might not exactly equal 0.3 due to floating point representation

Solution: For floating-point comparisons, use an epsilon value:

EPSILON = 1e-9
valid_splits += (left_sum >= right_sum - EPSILON)

4. Misunderstanding the Split Definition

Some might incorrectly interpret the split point, thinking element at index i belongs to the right part instead of the left part.

Problem Example:

# WRONG interpretation: thinking nums[i] is the first element of the right part
for i in range(1, len(nums)):  # Starting from 1
    left_sum = sum(nums[:i])   # Excludes nums[i]
    right_sum = sum(nums[i:])   # Includes nums[i]

Solution: Remember that splitting at index i means:

• Left part: indices [0, i] inclusive
• Right part: indices [i+1, n-1] inclusive

# CORRECT: nums[i] belongs to the left part
left_sum += nums[i]  # Include current element in left
right_sum = total_sum - left_sum  # Everything else is right