Why Subarray Sum Equals K Using Hash Map in DSA Java?

The Big Idea

What if you could find matching sums instantly without checking every possibility?

The Scenario

Imagine you have a long list of daily expenses and want to find if any continuous days add up exactly to a certain budget.

Doing this by checking every possible group of days one by one is tiring and slow.

The Problem

Manually checking all possible continuous groups means trying many combinations.

This takes a lot of time and can easily cause mistakes, especially with long lists.

The Solution

Using a hash map, we keep track of sums we've seen so far.

This helps us quickly find if a previous sum can pair with the current sum to reach the target.

It makes the search fast and simple.

Before vs After

✗ Before

for (int start = 0; start < nums.length; start++) {
  int sum = 0;
  for (int end = start; end < nums.length; end++) {
    sum += nums[end];
    if (sum == k) return true;
  }
}

✓ After

Map<Integer, Integer> sumCounts = new HashMap<>();
sumCounts.put(0, 1);
int sum = 0;
for (int num : nums) {
  sum += num;
  if (sumCounts.containsKey(sum - k)) return true;
  sumCounts.put(sum, sumCounts.getOrDefault(sum, 0) + 1);
}

What It Enables

This method lets you find continuous sums matching a target in just one pass through the list.

Real Life Example

Checking if a certain amount of calories was consumed over any continuous days in a diet plan.

Key Takeaways

Manual checking is slow and error-prone.

Hash map tracks sums to find matches quickly.

Enables fast detection of continuous sums equal to target.

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)