Subarray Sum Equals K Using Hash Map in DSA Java - Time & Space Complexity

Time Complexity: Subarray Sum Equals K Using Hash Map

O(n)

Understanding Time Complexity

We want to know how fast the algorithm finds subarrays that add up to a target sum.

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

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public int subarraySum(int[] nums, int k) {
    Map map = new HashMap<>();
    map.put(0, 1);
    int count = 0, sum = 0;
    for (int num : nums) {
        sum += num;
        if (map.containsKey(sum - k)) {
            count += map.get(sum - k);
        }
        map.put(sum, map.getOrDefault(sum, 0) + 1);
    }
    return count;
}
    

This code counts how many continuous parts of the array add up exactly to k using a hash map.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: One loop through the array elements.
• How many times: Exactly once for each element in the array.

How Execution Grows With Input

As the array gets longer, the loop runs once per element, doing a few quick checks each time.

Input Size (n)	Approx. Operations
10	About 10 checks and updates
100	About 100 checks and updates
1000	About 1000 checks and updates

Pattern observation: The work grows directly with the number of elements, so doubling the input roughly doubles the work.

Final Time Complexity

Time Complexity: O(n)

This means the time to finish grows in a straight line with the input size; more elements mean proportionally more work.

Common Mistake

[X] Wrong: "Because there is a nested check inside the loop, the time is O(n²)."

[OK] Correct: The inner check uses a hash map lookup, which is very fast and does not loop over the array again, so it stays O(n).

Interview Connect

Understanding this pattern helps you solve many problems where you track sums or counts efficiently, a skill valued in interviews and real projects.

Self-Check

"What if we replaced the hash map with a list to store sums? How would the time complexity change?"

Practice

1. What is the main purpose of using a hash map in the Subarray Sum Equals K problem?

easy

A. To store cumulative sums and their frequencies for quick lookup

B. To sort the array elements before processing

C. To store the original array elements as keys

D. To keep track of the maximum subarray length

Solution

1. Step 1: Understand the problem requirement

   The problem requires finding the number of continuous subarrays whose sum equals a target value k.

2. Step 2: Role of hash map in solution

   The hash map stores cumulative sums encountered so far and their counts, allowing quick checks if a subarray sum equals k by checking if (current_sum - k) exists.

3. Final Answer:

   To store cumulative sums and their frequencies for quick lookup -> Option A

4. Quick Check:

   Hash map stores sums for fast subarray sum checks [OK]

Hint: Hash map tracks sums to find subarrays fast [OK]

Common Mistakes:

• Thinking hash map stores array elements as keys
• Assuming sorting is needed
• Using hash map to track max length instead of counts

2. Which of the following Java code snippets correctly initializes the hash map used to store cumulative sums and their counts?

easy

A. HashMap map = new HashMap();"

B. HashMap map = new HashMap<>();

C. HashMap map = new HashMap<>();

D. HashMap map = new HashMap<Integer>();

Solution

1. Step 1: Check Java generics syntax

   Java requires wrapper classes like Integer for generics, and diamond operator <> is used for type inference.

2. Step 2: Identify correct initialization

   HashMap map = new HashMap<>(); uses HashMap map = new HashMap<>(); which is correct and modern syntax.

3. Final Answer:

   HashMap map = new HashMap<>(); -> Option C

4. Quick Check:

   Use wrapper classes and diamond operator for HashMap [OK]

Hint: Use Integer, not int, and diamond operator <> [OK]

Common Mistakes:

• Using primitive types int in generics
• Missing diamond operator <>
• Incorrect constructor syntax

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

int[] nums = {1, 2, 3};
int k = 3;
int count = 0;
HashMap map = new HashMap<>();
map.put(0, 1);
int sum = 0;
for (int num : nums) {
    sum += num;
    if (map.containsKey(sum - k)) {
        count += map.get(sum - k);
    }
    map.put(sum, map.getOrDefault(sum, 0) + 1);
}
System.out.println(count);

medium

A. 3

B. 2

C. 1

D. 0

Solution

1. Step 1: Calculate cumulative sums and check subarrays

   nums = [1,2,3], k=3 - sum=1, map has 0:1, sum-k=1-3=-2 not in map - sum=3, sum-k=0 in map with count 1, count=1 - sum=6, sum-k=3 in map with count 1, count=2

2. Step 2: Final count of subarrays with sum k

   Two subarrays sum to 3: [1,2] and [3].

3. Final Answer:

   2 -> Option B

4. Quick Check:

   Two subarrays sum to k=3 [OK]

Hint: Track sums and check (sum-k) in map to count subarrays [OK]

Common Mistakes:

• Not initializing map with (0,1)
• Forgetting to update count correctly
• Misunderstanding cumulative sum logic

4. Identify the error in the following Java code snippet for counting subarrays with sum k:

int[] nums = {1, -1, 0};
int k = 0;
int count = 0;
HashMap map = new HashMap<>();
int sum = 0;
for (int num : nums) {
    sum += num;
    if (map.containsKey(sum - k)) {
        count += map.get(sum - k);
    }
    map.put(sum, 1);
}
System.out.println(count);

medium

A. Incorrect array declaration syntax

B. Using sum - k instead of sum + k in map lookup

C. Not updating sum inside the loop

D. Missing initialization of map with (0,1)

Solution

1. Step 1: Check map initialization

   The map should start with (0,1) to count subarrays starting at index 0. This is missing here.

2. Step 2: Check map update logic

   Also, map.put(sum, 1) overwrites counts instead of incrementing. But the main error is missing initial (0,1).

3. Final Answer:

   Missing initialization of map with (0,1) -> Option D

4. Quick Check:

   Initialize map with (0,1) to count empty prefix sum [OK]

Hint: Always start map with (0,1) before loop [OK]

Common Mistakes:

• Not initializing map with (0,1)
• Overwriting counts instead of incrementing
• Confusing sum-k with sum+k

5. Given an array nums = {3, 4, 7, 2, -3, 1, 4, 2} and k = 7, how many continuous subarrays sum to k? Use the hash map approach to find the answer.

hard

A. 4

B. 5

C. 3

D. 6

Solution

1. Step 1: Calculate cumulative sums and track counts

   Initialize map with (0,1), sum=0, count=0 Iterate nums: - sum=3, sum-k=-4 not in map, map[3]=1 - sum=7, sum-k=0 in map(1), count=1, map[7]=1 - sum=14, sum-k=7 in map(1), count=2, map[14]=1 - sum=16, sum-k=9 not in map, map[16]=1 - sum=13, sum-k=6 not in map, map[13]=1 - sum=14, sum-k=7 in map(1), count=3, map[14]=2 - sum=18, sum-k=11 not in map, map[18]=1 - sum=20, sum-k=13 in map(1), count=4, map[20]=1

2. Step 2: Identify subarrays

   Four subarrays sum to 7: [3,4], [7], [7,2,-3,1], [1,4,2].

3. Final Answer:

   4 -> Option A

4. Quick Check:

   Count subarrays with sum k using cumulative sums [OK]

Hint: Count subarrays by checking (sum-k) in map during iteration [OK]

Common Mistakes:

• Missing subarrays with negative numbers
• Not updating counts correctly when sum repeats
• Forgetting to initialize map with (0,1)