algoadvance

You are given an array of integers nums and an integer k. Your task is to find the total number of continuous subarrays whose sum equals to k.

Example 1:

Input: nums = [1,1,1], k = 2
Output: 2

Example 2:

Input: nums = [1,2,3], k = 3
Output: 2

Constraints:

• 1 <= nums.length <= 2 * 10^4
• -1000 <= nums[i] <= 1000
• -10^7 <= k <= 10^7

Clarifying Questions

1. Can the input array contain negative numbers? Yes, the array can contain negative numbers.

2. Can the input array be empty? No, the length of nums is at least 1 according to the constraints.

3. Is there a maximum length to the input array? Yes, the maximum length is 2 * 10^4.

4. Should the subarrays be contiguous? Yes, the subarrays must be continuous.

Strategy

To efficiently solve this problem, we can use a hashmap (dictionary) to store the cumulative sums (prefix sums) and their frequencies as we iterate through the array. The key idea is:

1. current_sum is the cumulative sum up to the current index.
2. If current_sum - k exists in the hashmap, it means there is a subarray ending at the current index which sums to k.

Here is the step-by-step approach:

1. Initialize current_sum to 0.
2. Initialize a hashmap prefix_sums to store the frequencies of cumulative sums. Initially, set prefix_sums[0] = 1 because a sum of 0 means there’s a subarray from the start that sums to k.
3. Iterate through the array, updating the current_sum.
4. For each element, check if current_sum - k is in prefix_sums. If it is, it means there are prefix_sums[current_sum - k] subarrays ending at this index which sum to k.
5. Update the hashmap with the new current_sum.

Code

def subarraySum(nums, k):
    count = 0
    current_sum = 0
    prefix_sums = {0: 1}
    
    for num in nums:
        current_sum += num
        
        # Check if there is a prefix sum that, when removed from current_sum, equals k
        if current_sum - k in prefix_sums:
            count += prefix_sums[current_sum - k]
        
        # Update the hashmap with the current sum
        if current_sum in prefix_sums:
            prefix_sums[current_sum] += 1
        else:
            prefix_sums[current_sum] = 1
    
    return count

Time Complexity

• Time Complexity: O(n)
  • We use a single pass to iterate through the nums array, and each operation (updating and checking in the hashmap) is O(1).
• Space Complexity: O(n)
  • In the worst case, we store a cumulative sum for each element in the hashmap.

The above solution efficiently computes the total number of continuous subarrays whose sum equals k within optimal time and space constraints.

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