Prefix Sum Array in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the prefix sum array with the first element of the input array.

DSA C++

prefix[0] = [1];

Drag options to blanks, or click blank then click option'

Aarr[0]

B0

Cprefix[1]

Darr[1]

Attempts:

3 left

💡 Hint

Common Mistakes

Using prefix[1] instead of arr[0]

Initializing prefix[0] to 0 instead of arr[0]

The prefix sum array starts with the first element of the input array, so prefix[0] = arr[0].

2fill in blank

medium

Complete the code to compute the prefix sum for the current index i.

DSA C++

prefix[i] = prefix[i - 1] [1] arr[i];

Drag options to blanks, or click blank then click option'

A-

B+

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using '-' instead of '+'

Using '*' or '/' which are incorrect for sums

Each prefix sum is the sum of the previous prefix sum and the current element, so use '+'.

3fill in blank

hard

Fix the error in the loop condition to correctly iterate over the array indices.

DSA C++

for (int i = 1; i [1] n; i++) {

Drag options to blanks, or click blank then click option'

A>

B<=

C>=

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' causes out of bounds access

Using '>' or '>=' causes the loop to never run

The loop should run while i is less than n to avoid going out of bounds.

4fill in blank

hard

Fill both blanks to create a function that returns the sum of elements from index l to r using the prefix sum array.

DSA C++

int rangeSum(int l, int r, const vector<int>& prefix) {
    if (l == 0) return prefix[[1]];
    return prefix[[2]] - prefix[l - 1];
}

Drag options to blanks, or click blank then click option'

Ar

Bl

Cr + 1

Dl + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using prefix[l] instead of prefix[r]

Using prefix[r+1] which is out of bounds

To get sum from l to r, return prefix[r] - prefix[l-1], and if l is 0, return prefix[r].

5fill in blank

hard

Fill all three blanks to build a prefix sum array from input vector arr.

DSA C++

vector<int> prefixSum(const vector<int>& arr) {
    int n = arr.size();
    vector<int> prefix(n);
    prefix[0] = [1];
    for (int i = 1; i [2] n; i++) {
        prefix[i] = prefix[i - 1] [3] arr[i];
    }
    return prefix;
}

Drag options to blanks, or click blank then click option'

Aarr[0]

B<

C+

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using '-' instead of '+' in sum

Loop condition using '<=' causing out of bounds

Initializing prefix[0] incorrectly

Initialize prefix[0] with arr[0], loop while i < n, and add prefix[i-1] and arr[i].

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