Prefix Sum Array in DSA C++ - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Prefix Sum Array Construction

What is the output of the following C++ code that constructs a prefix sum array from the given input array?

DSA C++

std::vector<int> arr = {3, 1, 4, 1, 5};
std::vector<int> prefixSum(arr.size());
prefixSum[0] = arr[0];
for (int i = 1; i < arr.size(); i++) {
    prefixSum[i] = prefixSum[i - 1] + arr[i];
}
for (int val : prefixSum) {
    std::cout << val << " ";
}

A 3 4 8 9 14

B 3 1 4 1 5

C 3 7 11 12 17

D 3 4 5 6 7

💡 Hint

Remember that each prefix sum element is the sum of all elements up to that index.

The prefix sum array is built by adding the current element to the sum of all previous elements. So prefixSum[0] = 3, prefixSum[1] = 3+1=4, prefixSum[2] = 4+4=8, prefixSum[3] = 8+1=9, prefixSum[4] = 9+5=14.

❓ Predict Output

intermediate

Prefix Sum Query Result

Given the prefix sum array prefixSum = {2, 5, 9, 14, 20}, what is the sum of elements from index 1 to 3 (inclusive) in the original array?

A 7

B 12

C 9

D 14

💡 Hint

Use prefixSum[j] - prefixSum[i-1] to find sum from i to j.

Sum from index 1 to 3 = prefixSum[3] - prefixSum[0] = 14 - 2 = 12.

🔧 Debug

advanced

Identify the Bug in Prefix Sum Calculation

What error does the following C++ code produce when trying to compute a prefix sum array?

DSA C++

std::vector<int> arr = {1, 2, 3, 4};
std::vector<int> prefixSum(arr.size());
for (int i = 1; i <= arr.size(); i++) {
    prefixSum[i] = prefixSum[i - 1] + arr[i];
}

A Syntax error: missing semicolon

B No error, outputs correct prefix sum

C Runtime error: out_of_range

D Logical error: sums are incorrect

💡 Hint

Check the loop boundary and indexing carefully.

The loop runs with i <= arr.size(), which tries to access prefixSum[arr.size()] and arr[arr.size()], both out of bounds causing runtime error.

🧠 Conceptual

advanced

Time Complexity of Prefix Sum Queries

If you have an array of size n and you precompute its prefix sum array, what is the time complexity to answer a sum query for any subarray [i, j]?

A O(1)

B O(n)

C O(log n)

D O(n^2)

💡 Hint

Think about how prefix sums allow direct calculation without looping.

Once prefix sums are computed, sum queries can be answered in constant time by subtracting prefixSum[i-1] from prefixSum[j].

🚀 Application

expert

Using Prefix Sum to Count Subarrays with Sum K

Given an integer array and an integer K, which approach correctly uses prefix sums to count the number of continuous subarrays whose sum equals K?

A Use prefix sums but only check sums starting from index 0 to j.

B Compute prefix sums and for each pair (i, j), check if prefixSum[j] - prefixSum[i] == K using nested loops.

C Sort the array and use two pointers to find subarrays summing to K.

D Use a hash map to store prefix sums and count how many times prefixSum[j] - K appears while iterating.

💡 Hint

Think about how to find subarrays efficiently without checking all pairs.

Option D uses a hash map to store counts of prefix sums seen so far, allowing O(n) time counting of subarrays summing to K.

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