Problem Description

You are given an integer array nums and two integers l and r. Your task is to find a subarray that satisfies the following conditions:

1. The subarray must have a length between l and r (inclusive)
2. The sum of the subarray must be greater than 0
3. Among all such valid subarrays, you need to find the one with the minimum sum

A subarray is defined as a contiguous non-empty sequence of elements from the original array. For example, if nums = [1, 2, 3], then [1], [2, 3], and [1, 2, 3] are all subarrays, but [1, 3] is not because it's not contiguous.

The function should return the minimum sum of such a subarray. If no subarray exists that meets all the criteria, return -1.

Example scenarios:

• If nums = [3, -2, 1, 4], l = 2, r = 3, you need to find subarrays of length 2 or 3 whose sum is positive, then return the smallest such sum.
• The subarray length is calculated as right_index - left_index + 1.
• The sum must be strictly greater than 0 (not equal to 0).

Intuition

Since we need to find the minimum sum among all valid subarrays, and the constraints involve both the length of the subarray and its sum being positive, we need to examine all possible subarrays that could potentially be our answer.

The key insight is that we cannot use typical optimization techniques like sliding window or dynamic programming here because:

1. We're looking for the minimum positive sum, not maximum
2. The array can contain negative numbers, which means subarrays don't have a monotonic property

This leads us to a brute force approach: we need to check every possible subarray whose length falls within [l, r].

For each starting position i, we can build subarrays incrementally by extending the right endpoint j. As we extend, we maintain a running sum s by adding nums[j] to it. This is more efficient than recalculating the sum from scratch for each subarray.

Once we have a subarray of length at least l (when j - i + 1 >= l), we start checking if it meets our criteria:

• Is the length within bounds? (l <= j - i + 1 <= r)
• Is the sum positive? (s > 0)

If both conditions are met, we update our answer with the minimum between the current answer and this sum.

The reason this works is that we're guaranteed to explore all possible subarrays of valid lengths, and by keeping track of the minimum positive sum we encounter, we'll find the optimal answer. If no valid subarray exists after checking all possibilities, we return -1.

Learn more about Prefix Sum and Sliding Window patterns.

Solution Approach

The solution uses a nested loop approach to enumerate all possible subarrays:

1. Outer Loop - Starting Position: We iterate through each index i from 0 to n-1 as the starting position of our subarray.

2. Inner Loop - Ending Position: For each starting position i, we iterate through each index j from i to n-1 as the ending position. This ensures we examine all subarrays starting at position i.

3. Running Sum Calculation: We maintain a running sum s that starts at 0 for each new starting position. As we extend the subarray by moving j, we add nums[j] to s. This gives us the sum of the subarray [i, j] in constant time.

4. Validation Check: For each subarray [i, j], we check two conditions:

   • Length constraint: l <= j - i + 1 <= r (the subarray length must be between l and r)
   • Positive sum: s > 0 (the sum must be strictly positive)

5. Answer Update: If both conditions are satisfied, we update our answer with min(ans, s), keeping track of the minimum positive sum found so far.

6. Initialization and Return:

   • We initialize ans with infinity (inf) to ensure any valid sum will be smaller
   • After examining all subarrays, if ans is still infinity, it means no valid subarray was found, so we return -1
   • Otherwise, we return the minimum sum found

The time complexity is O(n²) where n is the length of the array, as we examine all possible subarrays. The space complexity is O(1) as we only use a few variables to track the current state.

Example Walkthrough

Let's walk through the solution with a concrete example:

• nums = [3, -2, 1, 4]
• l = 2, r = 3
• We need subarrays of length 2 or 3 with positive sum, and want the minimum such sum.

Step-by-step execution:

Starting at i = 0 (nums[0] = 3):

• j = 0: subarray [3], length = 1, sum = 3
  • Length 1 < l (2), so skip
• j = 1: subarray [3, -2], length = 2, sum = 3 + (-2) = 1
  • Length 2 ∈ [2, 3] ✓, sum 1 > 0 ✓
  • Update ans = min(∞, 1) = 1
• j = 2: subarray [3, -2, 1], length = 3, sum = 1 + 1 = 2
  • Length 3 ∈ [2, 3] ✓, sum 2 > 0 ✓
  • Update ans = min(1, 2) = 1
• j = 3: subarray [3, -2, 1, 4], length = 4, sum = 2 + 4 = 6
  • Length 4 > r (3), so skip

Starting at i = 1 (nums[1] = -2):

• j = 1: subarray [-2], length = 1, sum = -2
  • Length 1 < l (2), so skip
• j = 2: subarray [-2, 1], length = 2, sum = -2 + 1 = -1
  • Length 2 ∈ [2, 3] ✓, but sum -1 ≤ 0 ✗, so skip
• j = 3: subarray [-2, 1, 4], length = 3, sum = -1 + 4 = 3
  • Length 3 ∈ [2, 3] ✓, sum 3 > 0 ✓
  • Update ans = min(1, 3) = 1

Starting at i = 2 (nums[2] = 1):

• j = 2: subarray [1], length = 1, sum = 1
  • Length 1 < l (2), so skip
• j = 3: subarray [1, 4], length = 2, sum = 1 + 4 = 5
  • Length 2 ∈ [2, 3] ✓, sum 5 > 0 ✓
  • Update ans = min(1, 5) = 1

Starting at i = 3 (nums[3] = 4):

• j = 3: subarray [4], length = 1, sum = 4
  • Length 1 < l (2), so skip

Result: The minimum positive sum among all valid subarrays is 1, achieved by the subarray [3, -2] at indices [0, 1].

Solution Implementation

Time and Space Complexity

The time complexity is O(n^2), where n is the length of the array nums. This is because we have two nested loops: the outer loop iterates through all possible starting positions i from 0 to n-1, and for each starting position, the inner loop iterates through all possible ending positions j from i to n-1. In the worst case, this results in approximately n * (n+1) / 2 iterations, which simplifies to O(n^2).

The space complexity is O(1). The algorithm only uses a constant amount of extra space for variables n, ans, i, j, and s, regardless of the input size. No additional data structures that grow with the input size are created.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Confusing Minimum Positive Sum with Overall Minimum

A common mistake is trying to find the overall minimum sum first and then checking if it's positive. This approach fails because the minimum sum subarray might be negative, while there could be other positive-sum subarrays that are valid.

Incorrect approach:

# Wrong: Finding minimum sum first, then checking if positive
min_sum = float('inf')
for start in range(n):
    current_sum = 0
    for end in range(start, n):
        current_sum += nums[end]
        if l <= end - start + 1 <= r:
            min_sum = min(min_sum, current_sum)
return min_sum if min_sum > 0 else -1  # This misses valid positive subarrays!

Correct approach: Only consider subarrays with positive sums when updating the minimum.

2. Early Termination When Sum Becomes Negative

Another pitfall is breaking the inner loop when the current sum becomes negative, thinking no further extensions will be valid. This is wrong because adding more elements might make the sum positive again.

Incorrect approach:

for start in range(n):
    current_sum = 0
    for end in range(start, n):
        current_sum += nums[end]
        if current_sum <= 0:
            break  # Wrong! Future elements might make it positive

Example: nums = [-1, -2, 10], l = 3, r = 3

• If we break when sum becomes negative, we miss the valid subarray [-1, -2, 10] with sum = 7

3. Not Handling Edge Cases Properly

Failing to initialize the minimum sum correctly or not checking if any valid subarray was found can lead to incorrect outputs.

Common mistakes:

• Initializing min_sum to 0 instead of infinity
• Returning 0 or the unmodified infinity value instead of -1 when no valid subarray exists
• Not considering that all subarrays within the length range might have non-positive sums

Solution: Always initialize with float('inf') and explicitly check if it remains unchanged before returning the result.