In-place Prefix Sum

In-place Prefix Sum

Problem Statement:

Given an array A of size N, modify the array in-place to create a prefix sum array where each element A[i] contains the sum of all elements from index 0 to i.

You must solve this problem without using any extra space (O(1) space complexity).

Examples:

Example 1:

Input: A = 1, 2, 3, 4, 5

Output: 1, 3, 6, 10, 15

Explanation:

• A0 = 1 (remains same)
• A1 = 1 + 2 = 3
• A2 = 1 + 2 + 3 = 6
• A3 = 1 + 2 + 3 + 4 = 10
• A4 = 1 + 2 + 3 + 4 + 5 = 15

Example 2:

Input: A = 2, -1, 3, -4, 5

Output: 2, 1, 4, 0, 5

Explanation:

• A0 = 2
• A1 = 2 + (-1) = 1
• A2 = 2 + (-1) + 3 = 4
• A3 = 2 + (-1) + 3 + (-4) = 0
• A4 = 2 + (-1) + 3 + (-4) + 5 = 5

Constraints:

• 1 ≤ N ≤ 10^5
• -10^9 ≤ A[i] ≤ 10^9
• Must modify the array in-place (O(1) extra space)

Important Points to Understand:

1. In-place modification: We update the original array itself, no extra array needed
2. Left-to-right traversal: Process elements from index 1 to N-1
3. Current element dependency: Each Ai depends only on Ai-1 (already computed) and original Ai
4. Space optimization: Trading the ability to access original values for O(1) space
5. Formula: A[i] = A[i-1] + A[i] for all i from 1 to N-1

Approach:

Step 1: Keep A0 as is (it's already the prefix sum for index 0)

Step 2: Iterate from index 1 to N-1:

• Add previous element to current: A[i] = A[i-1] + A[i]

Step 3: Array is now modified to contain prefix sums

Time Complexity:

• Time Complexity: O(N) - Single pass through the array
• Space Complexity: O(1) - No extra space used, only modifying input array in-place

Space Complexity:

• O(1) - Only using a loop variable, no additional data structures

Dry Run:

Input: A = [3, 1, 2, 4]

Initial state: [3, 1, 2, 4]

Step 1: i = 1
  A[1] = A[0] + A[1] = 3 + 1 = 4
  Array: [3, 4, 2, 4]

Step 2: i = 2
  A[2] = A[1] + A[2] = 4 + 2 = 6
  Array: [3, 4, 6, 4]

Step 3: i = 3
  A[3] = A[2] + A[3] = 6 + 4 = 10
  Array: [3, 4, 6, 10]

Final Output: [3, 4, 6, 10]

Brute Force Approach:

Approach: Create a new array and compute each prefix sum from scratch.

Algorithm:

1. Create new array prefix of size N
2. For each index i (0 to N-1):
   • Sum elements from 0 to i
   • Store in prefixi
3. Copy prefix array back to A

Time Complexity: O(N²) - Nested loops Space Complexity: O(N) - Extra array needed

Why it's not optimal:

• Recalculates sums repeatedly
• Uses extra space
• Much slower for large arrays

Visualization:

Original Array:     [3,    1,    2,    4]
                     ↓
Step 1 (i=1):      [3,    4,    2,    4]
                     ↑-----|
                     
Step 2 (i=2):      [3,    4,    6,    4]
                           ↑-----|
                           
Step 3 (i=3):      [3,    4,    6,   10]
                                 ↑-----|
                                 
Prefix Sum Result: [3,    4,    6,   10]

Formula at each step: A[i] = A[i-1] + A[i]

Multiple Optimized Approaches:

Approach 1: In-place Modification (Optimal - Current Solution)

Time: O(N), Space: O(1)

function inplacePrefixSum(A) {
    for (let i = 1; i < A.length; i++) {
        A[i] = A[i-1] + A[i];
    }
    return A;
}

Pros: Minimal space, simple implementation Cons: Destroys original array

Approach 2: Using Extra Array

Time: O(N), Space: O(N)

function prefixSumWithExtraSpace(A) {
    const prefix = [A[0]];
    for (let i = 1; i < A.length; i++) {
        prefix[i] = prefix[i-1] + A[i];
    }
    return prefix;
}

Pros: Preserves original array Cons: Uses O(N) extra space

Approach 3: Functional Programming Style

Time: O(N), Space: O(N)

function prefixSumFunctional(A) {
    return A.reduce((acc, curr) => {
        acc.push((acc[acc.length-1] || 0) + curr);
        return acc;
    }, []);
}

Pros: Declarative, clean code Cons: Less efficient, uses extra space

Edge Cases to Consider:

1. Single Element Array:
   • Input: 5
   • Output: 5
   • The element remains unchanged
2. All Zeros:
   • Input: 0, 0, 0, 0
   • Output: 0, 0, 0, 0
   • All prefix sums are 0
3. Negative Numbers:
   • Input: -1, -2, -3
   • Output: -1, -3, -6
   • Prefix sum decreases
4. Mixed Positive and Negative:
   • Input: 5, -3, 2, -1
   • Output: 5, 2, 4, 3
   • Sum can increase or decrease
5. Large Numbers:
   • Input: 10^9, 10^9
   • Consider integer overflow in some languages
   • JavaScript handles with Number type
6. Alternating Signs:
   • Input: 1, -1, 1, -1
   • Output: 1, 0, 1, 0
   • Sum oscillates

JavaScript Code:

function inplacePrefixSum(A) {
    const N = A.length;
    
    // Start from index 1 since A[0] is already its own prefix sum
    for (let i = 1; i < N; i++) {
        A[i] = A[i-1] + A[i];
    }
    
    return A;
}

// Example usage:
const arr1 = [1, 2, 3, 4, 5];
console.log(inplacePrefixSum(arr1)); // [1, 3, 6, 10, 15]

const arr2 = [2, -1, 3, -4, 5];
console.log(inplacePrefixSum(arr2)); // [2, 1, 4, 0, 5]

Key Takeaways:

1. In-place modification achieves O(1) space complexity - critical for memory-constrained environments
2. Dependency pattern: Each element only depends on previous element, enabling single-pass solution
3. Trade-off understanding: Sacrificing original array access for optimal space complexity
4. Foundation for advanced techniques: This pattern appears in dynamic programming and optimization problems
5. Interview importance: Common follow-up question after regular prefix sum problems
6. Real-world applications:
   • Running totals in financial calculations
   • Cumulative statistics in data analysis
   • Memory-efficient range query preprocessing
7. Common mistakes to avoid:
   • Starting loop from index 0 instead of 1
   • Not considering negative numbers
   • Forgetting that original array is lost
8. Performance characteristics:
   • Single pass: O(N) time
   • No extra allocation: O(1) space
   • Cache-friendly: sequential access pattern
9. When to use this approach:
   • Space is limited
   • Original array not needed afterward
   • Simple prefix sum queries required
10. Related concepts:
    • Cumulative frequency arrays
    • Running sums
    • Prefix product arrays
    • Difference arrays (inverse operation)