DSA: Prefix Sum Pattern (Real Use Cases)

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Solution to Last Week's Challenge

Last week's problem was: DSA: Detect Cycles in Graphs (DFS vs Union-Find)

The optimal solution uses dsa: detect cycles in graphs (dfs vs union-find) pattern with careful state management.

Complexity Analysis:

  • Time Complexity: O(n)
  • Space Complexity: O(1)

In the toolkit of a senior engineer, few patterns are as deceptively simple yet profoundly impactful as the Prefix Sum. At its core, it is a preprocessing technique used to reduce the time complexity of range queries from linear $O(N)$ to constant $O(1)$. While it might seem like a basic arithmetic trick, its applications span from low-level image processing (Integral Images) to high-scale telemetry systems calculating error rates over rolling windows.

In technical interviews, the Prefix Sum is rarely the final answer on its own. Instead, it serves as a foundational building block for more complex algorithms involving hash maps, sliding windows, or binary searches. Mastering this pattern demonstrates to an interviewer that you understand the trade-off between space and time—specifically, how investing $O(N)$ auxiliary space can yield massive performance gains in read-heavy systems.

Theory

The Prefix Sum pattern involves creating a new array (often called a "sum array") where each element at index i stores the cumulative sum of the original array from the start up to i. Mathematically, if $A$ is our input array, the prefix sum array $P$ is defined as: $P[i] = \sum_{k=0}^{i} A[k]$

The "magic" happens when you need to calculate the sum of a subarray from index i to j. Instead of iterating through the elements, you use the formula: $Sum(i, j) = P[j] - P[i-1]$

This transformation is the discrete version of the Fundamental Theorem of Calculus. By storing the "integral" of the array, we can find the "area" under any segment in constant time.

Complexity Analysis

  • Preprocessing Time: $O(N)$ to iterate through the array once.
  • Query Time: $O(1)$ to perform a single subtraction.
  • Space Complexity: $O(N)$ to store the prefix sums, though in some "in-place" scenarios, this can be reduced to $O(1)$ if the input array can be modified.

Pattern Visualization

Identifying when to use a Prefix Sum is key. Usually, the problem involves "subarrays," "sum of ranges," or "counting occurrences over a window." If the array is static (no updates between queries), Prefix Sum is almost always the optimal choice.

Implementation

Let's look at the evolution from a naive approach to an optimized Prefix Sum implementation in Python.

Problem: Range Sum Query

Given an array nums, handle multiple queries to find the sum of elements between indices left and right.

python
# Approach 1: Brute Force
# Time: O(N * Q) where Q is number of queries
# Space: O(1)
def range_sum_brute(nums, left, right):
    total = 0
    for i in range(left, right + 1):
        total += nums[i]
    return total

# Approach 2: Prefix Sum (Optimized)
# Time: O(N) Preprocessing, O(1) per Query
# Space: O(N)
class RangeSum:
    def __init__(self, nums):
        # We pad with a leading 0 to handle the left=0 case gracefully
        self.prefix_sums = [0] * (len(nums) + 1)
        for i in range(len(nums)):
            self.prefix_sums[i + 1] = self.prefix_sums[i] + nums[i]

    def query(self, left, right):
        # Sum(left, right) = P[right + 1] - P[left]
        return self.prefix_sums[right + 1] - self.prefix_sums[left]

# Example Usage
data = [1, 2, 3, 4, 5]
rs = RangeSum(data)
print(rs.query(1, 3)) # Output: 9 (2+3+4)

Complexity Analysis Table

ApproachPreprocessingQuery TimeSpace ComplexityBest For
Brute Force$O(1)$$O(N)$$O(1)$Single query, small data
Prefix Sum$O(N)$$O(1)$$O(N)$Multiple queries, static data
Segment Tree$O(N)$$O(\log N)$$O(N)$Frequent updates and queries

Common Patterns

1. Subarray Sum Equals K (Prefix Sum + Hash Map)

This is a classic LeetCode Medium. We use a hash map to store how many times a particular prefix sum has occurred. If current_sum - K exists in the map, it means there is a subarray ending at the current index that sums to K.

python
def subarraySum(nums, k):
    count = 0
    curr_sum = 0
    # Map stores {prefix_sum: frequency}
    prefix_map = {0: 1}
    
    for x in nums:
        curr_sum += x
        # If (curr_sum - k) exists, we found a subarray
        if curr_sum - k in prefix_map:
            count += prefix_map[curr_sum - k]
        prefix_map[curr_sum] = prefix_map.get(curr_sum, 0) + 1
    return count

2. Difference Array (Range Updates)

If you need to add a value v to all elements from index i to j, you can use a "Difference Array." You increment D[i] by v and decrement D[j+1] by v. The prefix sum of D then gives you the final values.

Practice Problems

The following chart maps common problems based on their frequency in FAANG interviews and their technical difficulty.

This Week's Interview Challenge

Now that you understand the mechanics, it is time to apply them. This week’s challenge is a frequent "Screener" question used by companies like Google and Meta.

Problem: Find the Equilibrium Index (Pivot Index)

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 right of the index.

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.

Example 1:

  • Input: nums = [1, 7, 3, 6, 5, 6]
  • Output: 3
  • Explanation: Left sum (1+7+3 = 11) == Right sum (5+6 = 11).

Example 2:

  • Input: nums = [1, 2, 3]
  • Output: -1
  • Explanation: No index satisfies the condition.

Constraints:

  • 1 <= nums.length <= 10^4
  • -1000 <= nums[i] <= 1000

Hints:

  1. Calculate the total sum of the entire array first.
  2. As you iterate through the array, keep track of the "running left sum."
  3. The "right sum" for any index i can be calculated as total_sum - left_sum - nums[i].
  4. Check if left_sum == right_sum at each step.

Template:

python
def find_pivot_index(nums: list[int]) -> int:
    # Your code here
    pass

# Submit your solution and check back next week for the detailed answer!

Conclusion

The Prefix Sum pattern is a testament to the power of preprocessing. By spending a little time (and space) upfront, we can drastically optimize the performance of our applications. When you encounter problems involving range queries or subarray sums, let Prefix Sum be your first thought. Remember to handle edge cases like empty arrays or negative numbers, and always consider if a hash map could complement your prefix sum to solve search-related variants.

References