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

Time Complexity: Subarray Sum Equals K Using Hash Map

O(n)

Understanding Time Complexity

We want to understand how fast the algorithm finds subarrays that sum to a target value using a hash map.

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

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func subarraySum(nums []int, k int) int {
    count, sum := 0, 0
    prefixSums := map[int]int{0: 1}
    for _, num := range nums {
        sum += num
        if val, ok := prefixSums[sum-k]; ok {
            count += val
        }
        prefixSums[sum]++
    }
    return count
}
    

This code counts how many continuous parts of the array add up to k using a hash map to store sums seen so far.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Single loop over the array elements.
• How many times: Exactly once for each element in the input array.

How Execution Grows With Input

Each element is processed once, and hash map lookups and inserts happen in constant time on average.

Input Size (n)	Approx. Operations
10	About 10 steps
100	About 100 steps
1000	About 1000 steps

Pattern observation: The number of steps grows roughly in direct proportion to the input size.

Final Time Complexity

Time Complexity: O(n)

This means the time needed grows linearly with the size of the input array.

Common Mistake

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

[OK] Correct: The inner check is a hash map lookup, which is on average constant time, so it does not multiply the loop's cost.

Interview Connect

Understanding this pattern helps you solve many problems involving sums and counts efficiently, a valuable skill in interviews and real projects.

Self-Check

"What if we replaced the hash map with a list to store prefix 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 the original array elements as keys

B. To sort the array elements before processing

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

D. To keep track of the maximum subarray length

Solution

1. Step 1: Understand the problem goal

   The problem requires finding subarrays whose sums equal a target value k.

2. Step 2: Role of hash map in solution

   A hash map stores cumulative sums seen so far and their counts, enabling quick checks if a previous sum exists that forms a subarray summing to k.

3. Final Answer:

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

4. Quick Check:

   Hash map = cumulative sums storage [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
• Confusing hash map with array for sums

2. Which of the following Go code snippets correctly initializes a hash map to store integer keys and values for the subarray sum problem?

easy

A. m := make([]int, 0)

B. m := make(map[int]int)

C. m := map[int]string{}

D. m := map[string]int{}

Solution

1. Step 1: Identify correct map type for keys and values

   We need a map with integer keys and integer values to store sums and counts.

2. Step 2: Check Go syntax for map initialization

   make(map[int]int) correctly creates a map with int keys and int values.

3. Final Answer:

   m := make(map[int]int) -> Option B

4. Quick Check:

   map[int]int = correct map type [OK]

Hint: Use make(map[int]int) for int keys and values [OK]

Common Mistakes:

• Using slice instead of map
• Wrong key or value types in map
• Using map with string keys incorrectly

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

nums := []int{1, 2, 3}
k := 3
count := 0
sum := 0
m := make(map[int]int)
m[0] = 1
for _, v := range nums {
    sum += v
    if val, ok := m[sum-k]; ok {
        count += val
    }
    m[sum]++
}
fmt.Println(count)

medium

A. 1

B. 0

C. 3

D. 2

Solution

1. Step 1: Trace cumulative sums and map updates

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

2. Step 2: Final count after loop

   Count is 2 because two subarrays sum to k=3: [1,2] and [3].

3. Final Answer:

   2 -> Option D

4. Quick Check:

   Subarrays summing to 3 = 2 [OK]

Hint: Count increments when (sum-k) found in map [OK]

Common Mistakes:

• Not initializing m[0] = 1
• Forgetting to increment map count
• Miscounting subarrays

4. Identify the bug in the following Go code snippet for counting subarrays summing to k:

nums := []int{1, 1, 1}
k := 2
count := 0
sum := 0
m := make(map[int]int)
for _, v := range nums {
    sum += v
    if val, ok := m[sum-k]; ok {
        count += val
    }
    m[sum]++
}
fmt.Println(count)

medium

A. Missing initialization m[0] = 1 before loop

B. Incorrect increment of count variable

C. Wrong loop variable used

D. Using map with wrong key type

Solution

1. Step 1: Check map initialization

   The map m is used to store cumulative sums, but m[0] is not initialized to 1 before the loop.

2. Step 2: Effect of missing m[0] = 1

   Without m[0] = 1, subarrays starting from index 0 that sum to k are not counted, causing incorrect count.

3. Final Answer:

   Missing initialization m[0] = 1 before loop -> Option A

4. Quick Check:

   Initialize m[0]=1 to count subarrays from start [OK]

Hint: Always set m[0] = 1 before loop [OK]

Common Mistakes:

• Not initializing m[0]
• Incrementing count incorrectly
• Using wrong map key type

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

hard

A. 4

B. 3

C. 6

D. 5

Solution

1. Step 1: Calculate cumulative sums and track counts

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

2. Step 2: Verify subarrays

   The subarrays summing to 7 are: [3,4], [7], [7,2,-3,1], [1,4,2]. Total count = 4.

3. Final Answer:

   4 -> Option A

4. Quick Check:

   Count subarrays with sum k = 4 [OK]

Hint: Track cumulative sums and counts carefully [OK]

Common Mistakes:

• Missing m[0] initialization
• Miscounting overlapping subarrays
• Not updating map counts correctly