Prefix Sum Array in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Prefix Sum Array Construction

What is the output of the prefix sum array after processing the input array?

DSA Java

int[] arr = {2, 4, 6, 8};
int[] prefixSum = new int[arr.length];
prefixSum[0] = arr[0];
for (int i = 1; i < arr.length; i++) {
    prefixSum[i] = prefixSum[i - 1] + arr[i];
}
System.out.println(java.util.Arrays.toString(prefixSum));

A [0, 2, 6, 12]

B [2, 4, 6, 8]

C [2, 8, 14, 20]

D [2, 6, 12, 20]

💡 Hint

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

The prefix sum array stores cumulative sums. Starting with 2, then 2+4=6, then 6+6=12, then 12+8=20.

❓ Predict Output

intermediate

Prefix Sum Array Query Result

Given the prefix sum array, what is the sum of elements from index 1 to 3 (inclusive) in the original array?

DSA Java

int[] prefixSum = {3, 7, 12, 18, 25};
int left = 1, right = 3;
int result = prefixSum[right] - prefixSum[left - 1];
System.out.println(result);

A 15

B 12

C 18

D 22

💡 Hint

Use prefixSum[right] - prefixSum[left - 1] to get sum between left and right.

Sum from index 1 to 3 is prefixSum[3] - prefixSum[0] = 18 - 3 = 15.

🧠 Conceptual

advanced

Understanding Prefix Sum Array Initialization

Why do we initialize the first element of the prefix sum array with the first element of the input array?

A To avoid null pointer exceptions during summation.

B Because prefix sum at index 0 is just the first element itself, no previous sums exist.

C To make the prefix sum array the same size as the input array.

D Because the first element is always zero in prefix sums.

💡 Hint

Think about what prefix sum means at the start of the array.

Prefix sum at index 0 equals the first element because it sums all elements up to index 0, which is just that element.

❓ Predict Output

advanced

Output of Prefix Sum with Negative Numbers

What is the prefix sum array output for the input array containing negative numbers?

DSA Java

int[] arr = {5, -3, 7, -2, 4};
int[] prefixSum = new int[arr.length];
prefixSum[0] = arr[0];
for (int i = 1; i < arr.length; i++) {
    prefixSum[i] = prefixSum[i - 1] + arr[i];
}
System.out.println(java.util.Arrays.toString(prefixSum));

A [5, 2, 9, 7, 11]

B [5, -3, 7, -2, 4]

C [5, 8, 15, 13, 17]

D [0, 5, 2, 9, 7]

💡 Hint

Add each element to the sum of all previous elements.

Prefix sums: 5, 5+(-3)=2, 2+7=9, 9+(-2)=7, 7+4=11.

🚀 Application

expert

Using Prefix Sum to Count Subarrays with Given Sum

Given an integer array and a target sum, which approach correctly uses prefix sums to count the number of subarrays that sum to the target?

DSA Java

int countSubarraysWithSum(int[] arr, int target) {
    int count = 0;
    int[] prefixSum = new int[arr.length + 1];
    prefixSum[0] = 0;
    for (int i = 1; i <= arr.length; i++) {
        prefixSum[i] = prefixSum[i - 1] + arr[i - 1];
    }
    java.util.Map<Integer, Integer> map = new java.util.HashMap<>();
    for (int sum : prefixSum) {
        if (map.containsKey(sum - target)) {
            count += map.get(sum - target);
        }
        map.put(sum, map.getOrDefault(sum, 0) + 1);
    }
    return count;
}

A Sort the array and use two pointers to find subarrays with target sum.

B Iterate all subarrays and sum elements each time without prefix sums.

C Use a hashmap to store frequencies of prefix sums and count pairs where difference equals target.

D Use prefix sums but only count sums equal to target without hashmap.

💡 Hint

Think about how prefix sums and hashmap can find subarrays efficiently.

The hashmap stores counts of prefix sums. For each prefix sum, checking if (current sum - target) exists counts subarrays summing to target.

Practice

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

easy

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

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

C. An array where each element is the difference between consecutive elements in arr

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

Solution

1. Step 1: Understand prefix sum definition

   A prefix sum array stores sums of elements from the start of the array up to each index.

2. Step 2: Compare options with definition

   Only An array where each element is the sum of all elements up to that index in arr matches this definition exactly.

3. Final Answer:

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

4. Quick Check:

   Prefix sum = cumulative sum [OK]

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

Common Mistakes:

• Confusing prefix sum with product or max operations
• Thinking prefix sum stores differences
• Mixing prefix sum with suffix sum

2. Which of the following Java code snippets correctly initializes a prefix sum array prefix for an integer array arr?

easy

A. int[] prefix = new int[arr.length]; for(int i=0; i

B. int[] prefix = new int[arr.length]; prefix[0] = 0; for(int i=1; i

C. int[] prefix = new int[arr.length]; prefix[0] = arr[0]; for(int i=1; i

D. int[] prefix = new int[arr.length]; for(int i=0; i

Solution

1. Step 1: Check initialization of prefix array

   Prefix sum array must start with prefix[0] = arr[0] to include first element sum.

2. Step 2: Verify loop calculation

   Loop from i=1 to end adds previous prefix sum plus current arr element, which is correct.

3. Final Answer:

   int[] prefix = new int[arr.length]; prefix[0] = arr[0]; for(int i=1; i -> Option C

4. Quick Check:

   Initialize first element, then add previous sums [OK]

Hint: Start prefix[0] with arr[0], then add previous sums in loop [OK]

Common Mistakes:

• Starting prefix[0] with 0 instead of arr[0]
• Multiplying instead of adding elements
• Reversing array elements instead of summing

3. What is the output of the following Java code?

int[] arr = {2, 4, 6, 8};
int[] prefix = new int[arr.length];
prefix[0] = arr[0];
for(int i = 1; i < arr.length; i++) {
    prefix[i] = prefix[i-1] + arr[i];
}
System.out.println(java.util.Arrays.toString(prefix));

medium

A. [2, 6, 12, 20]

B. [2, 4, 6, 8]

C. [0, 2, 6, 12]

D. [2, 8, 14, 22]

Solution

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

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

2. Step 2: Verify final prefix array

   The prefix array 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 for each index [OK]

Common Mistakes:

• Printing original array instead of prefix sum
• Starting prefix sum with 0 incorrectly
• Adding wrong elements in loop

4. Identify the error in the following Java code that attempts to create a prefix sum array:

int[] arr = {1, 3, 5};
int[] prefix = new int[arr.length];
prefix[0] = 0;
for(int i = 1; i <= arr.length; i++) {
    prefix[i] = prefix[i-1] + arr[i];
}

medium

A. Loop condition should be i < arr.length instead of i <= arr.length

B. prefix[0] should be initialized to arr[0] instead of 0

C. Array index arr[i] will cause an error when i equals arr.length

D. All of the above

Solution

1. Step 1: Check loop boundary

   The loop uses i <= arr.length, which causes index out of bounds because valid indices are 0 to arr.length-1.

2. Step 2: Check prefix[0] initialization and array access

   prefix[0] should be arr[0] to start sums correctly; also arr[i] access fails at i = arr.length.

3. Step 3: Combine all errors

   All issues cause runtime errors or incorrect prefix sums.

4. Final Answer:

   All of the above -> Option D

5. Quick Check:

   Loop bounds, initialization, and indexing errors combined [OK]

Hint: Check loop limits and array indices carefully [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Initializing prefix[0] incorrectly
• Accessing array out of bounds

5. Given an integer array arr of length 5: {3, 1, 4, 1, 5}, you want to find the sum of elements from index 1 to 3 (inclusive) efficiently using a prefix sum array. Which expression correctly calculates this sum using the prefix sum array prefix where prefix[i] is the sum of elements from index 0 to i?

hard

A. prefix[3] - prefix[2]

B. prefix[3] - prefix[0]

C. prefix[3] - prefix[1]

D. prefix[3] - prefix[0] + arr[1]

Solution

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

   Sum from index l to r = prefix[r] - prefix[l-1]. Here l=1, r=3.

2. Step 2: Apply formula with given indices

   Sum = prefix[3] - prefix[0]. prefix[0] is sum up to index 0, so subtracting it leaves sum from 1 to 3.

3. Final Answer:

   prefix[3] - prefix[0] -> Option B

4. Quick Check:

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

Hint: Range sum = prefix[end] - prefix[start - 1] [OK]

Common Mistakes:

• Subtracting prefix at wrong index
• Adding extra elements unnecessarily
• Confusing inclusive indices