We use the concept of Prefix Sums. If the sum from index 0 to i is S1 and the sum from index 0 to j is S2, then the sum of the subarray from i+1 to j is S2 - S1. We want S2 - S1 = k, which means S1 = S2 - k. We use a hash map to store the frequencies of prefix sums.

1. Initialize count = 0, curr_sum = 0.
2. prefix_sums = {0: 1} (Base case: a sum of 0 has appeared 1 time).
3. Iterate through nums:
   • curr_sum += num.
   • If curr_sum - k is in prefix_sums:
     • count += prefix_sums[curr_sum - k].
   • prefix_sums[curr_sum] = prefix_sums.get(curr_sum, 0) + 1.
4. Return count.

• Time Complexity: O(N).
• Space Complexity: O(N).

def subarray_sum(nums, k):
    count = 0
    curr_sum = 0
    prefix_sums = {0: 1}
    
    for n in nums:
        curr_sum += n
        if curr_sum - k in prefix_sums:
            count += prefix_sums[curr_sum - k]
        prefix_sums[curr_sum] = prefix_sums.get(curr_sum, 0) + 1
        
    return count