Prefix Sum Array in DSA C++ - Time & Space Complexity

Time Complexity: Prefix Sum Array

O(n)

Understanding Time Complexity

We want to understand how fast we can build and use a prefix sum array.

How does the time needed grow as the input array gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int n = arr.size();
vector<int> prefixSum(n);
prefixSum[0] = arr[0];
for (int i = 1; i < n; i++) {
    prefixSum[i] = prefixSum[i - 1] + arr[i];
}
// Now prefixSum[i] holds sum of arr[0] to arr[i]
    

This code builds a prefix sum array where each element stores the sum of all elements up to that index.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: A single for-loop that runs through the array once.
• How many times: Exactly n-1 times, where n is the size of the input array.

How Execution Grows With Input

As the input array grows, the number of additions grows roughly the same way.

Input Size (n)	Approx. Operations
10	9 additions
100	99 additions
1000	999 additions

Pattern observation: The operations increase linearly as the input size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to build the prefix sum array grows directly in proportion to the input size.

Common Mistake

[X] Wrong: "Building prefix sums takes quadratic time because we add many elements repeatedly."

[OK] Correct: Each element is added only once during the loop, so the total work grows linearly, not quadratically.

Interview Connect

Knowing prefix sums and their time complexity helps you quickly answer range sum queries efficiently, a common skill interviewers look for.

Self-Check

"What if we used nested loops to build the prefix sums instead? How would the time complexity change?"

Practice

1. What does a prefix sum array represent for a given array arr?

easy

A. An array where each element is the difference between the current and previous element in arr

B. An array where each element is the product of all elements from the start up to that index in arr

C. An array where each element is the sum of all elements from the start up to that index in arr

D. An array where each element is the maximum value found in arr up to that index

Solution

1. Step 1: Understand prefix sum array definition

   A prefix sum array stores cumulative sums, where each element at index i is the sum of all elements from index 0 to i in the original array.

2. Step 2: Compare options with definition

   Only An array where each element is the sum of all elements from the start up to that index in arr correctly describes this cumulative sum property. Other options describe different operations.

3. Final Answer:

   An array where each element is the sum of all elements from the start up to that index in arr -> Option C

4. Quick Check:

   Prefix sum = cumulative sum [OK]

Hint: Prefix sum means adding elements from start to current index [OK]

Common Mistakes:

• Confusing prefix sum with product or max operations
• Thinking prefix sum stores differences
• Assuming prefix sum resets at each index

2. Which of the following C++ code snippets correctly initializes a prefix sum array prefix from an integer array arr of size n?

easy

A. prefix[0] = 0; for(int i = 1; i <= n; i++) prefix[i] = prefix[i-1] + arr[i];

B. prefix[0] = arr[0]; for(int i = 1; i < n; i++) prefix[i] = prefix[i-1] + arr[i];

C. prefix[0] = arr[0]; for(int i = 0; i < n; i++) prefix[i] = prefix[i-1] + arr[i];

D. prefix[0] = arr[0]; for(int i = 1; i < n; i++) prefix[i] = arr[i] - prefix[i-1];

Solution

1. Step 1: Check initialization of prefix array

   Prefix sum starts with prefix[0] = arr[0], which is correct in options B, C, and D.

2. Step 2: Verify loop and summation logic

   prefix[0] = arr[0]; for(int i = 1; i < n; i++) prefix[i] = prefix[i-1] + arr[i]; correctly loops from i=1 to n-1 and adds prefix[i-1] + arr[i]. prefix[0] = 0; for(int i = 1; i <= n; i++) prefix[i] = prefix[i-1] + arr[i]; has wrong indexing and loop range. prefix[0] = arr[0]; for(int i = 0; i < n; i++) prefix[i] = prefix[i-1] + arr[i]; uses prefix[i-1] at i=0 which is invalid. prefix[0] = arr[0]; for(int i = 1; i < n; i++) prefix[i] = arr[i] - prefix[i-1]; subtracts instead of adding.

3. Final Answer:

   prefix[0] = arr[0]; for(int i = 1; i < n; i++) prefix[i] = prefix[i-1] + arr[i]; -> Option B

4. Quick Check:

   Initialize first, then add previous prefix + current [OK]

Hint: Start prefix[0] = arr[0], then add previous prefix + current [OK]

Common Mistakes:

• Starting loop at 0 and accessing prefix[-1]
• Using wrong loop bounds (<= n instead of < n)
• Subtracting instead of adding elements

3. What is the output of the following C++ code?

int arr[] = {2, 4, 6, 8};
int n = 4;
int prefix[4];
prefix[0] = arr[0];
for(int i = 1; i < n; i++) {
    prefix[i] = prefix[i-1] + arr[i];
}
for(int i = 0; i < n; i++) {
    std::cout << prefix[i] << " ";
}

medium

A. 2 6 12 20

B. 2 4 6 8

C. 2 8 14 20

D. 2 6 10 18

Solution

1. Step 1: Calculate prefix sums step-by-step

   prefix[0] = 2
   prefix[1] = prefix[0] + arr[1] = 2 + 4 = 6
   prefix[2] = prefix[1] + arr[2] = 6 + 6 = 12
   prefix[3] = prefix[2] + arr[3] = 12 + 8 = 20

2. Step 2: Print prefix array elements

   Output is: 2 6 12 20

3. Final Answer:

   2 6 12 20 -> Option A

4. Quick Check:

   Sum accumulates: 2, 6, 12, 20 [OK]

Hint: Add current element to previous prefix sum each step [OK]

Common Mistakes:

• Printing original array instead of prefix sums
• Incorrect addition order
• Off-by-one errors in loop

4. Identify the error in the following prefix sum array code and select the corrected version:

int arr[] = {1, 3, 5, 7};
int n = 4;
int prefix[4];
prefix[0] = arr[0];
for(int i = 1; i <= n; i++) {
    prefix[i] = prefix[i-1] + arr[i];
}

medium

A. Change loop condition to i < n to avoid out-of-bounds access

B. Initialize prefix[0] = 0 instead of arr[0]

C. Use i = 0 as loop start and keep condition i < n

D. Replace prefix[i] with arr[i] inside loop

Solution

1. Step 1: Analyze loop bounds and array indexing

   The loop runs with i from 1 to n inclusive (i <= n). Since arrays are 0-indexed, prefix[n] and arr[n] access out-of-bounds elements causing runtime error.

2. Step 2: Correct loop condition

   Changing loop condition to i < n ensures indices stay within valid range (0 to n-1).

3. Final Answer:

   Change loop condition to i < n to avoid out-of-bounds access -> Option A

4. Quick Check:

   Loop must stay within array size [OK]

Hint: Loop index must be less than array size, not less or equal [OK]

Common Mistakes:

• Using <= n causing out-of-bounds errors
• Starting loop at 0 and accessing prefix[-1]
• Incorrect initialization of prefix[0]

5. Given an array arr of size n, how can you use a prefix sum array to efficiently find the sum of elements between indices l and r (inclusive)?

hard

A. Sum = prefix[r] / prefix[l-1] if l > 0, else prefix[r]

B. Sum = prefix[r] + prefix[l-1] if l > 0, else prefix[r]

C. Sum = prefix[r] * prefix[l-1] if l > 0, else prefix[r]

D. Sum = prefix[r] - prefix[l-1] if l > 0, else prefix[r]

Solution

1. Step 1: Understand prefix sum usage for range queries

   Prefix sum at index r gives sum from 0 to r. To get sum from l to r, subtract sum from 0 to l-1, which is prefix[l-1]. If l is 0, sum is prefix[r].

2. Step 2: Apply formula for sum between l and r

   Sum = prefix[r] - prefix[l-1] if l > 0, else prefix[r]. Other operations like addition, multiplication, or division are incorrect.

3. Final Answer:

   Sum = prefix[r] - prefix[l-1] if l > 0, else prefix[r] -> Option D

4. Quick Check:

   Range sum = prefix[r] - prefix[l-1] [OK]

Hint: Subtract prefix before l from prefix at r for range sum [OK]

Common Mistakes:

• Adding prefix[l-1] instead of subtracting
• Using multiplication or division incorrectly
• Not handling case when l = 0