Continuous Subarray Sum Problem Solution

The continuous subarray sum problem requires identifying whether there exists a contiguous subarray whose sum is divisible by a given integer k. This classic algorithmic challenge tests understanding of modular arithmetic, prefix sums, and hash map optimizations for efficient runtime complexity.

1. Problem Restatement

Clarifying questions to ask the interviewer:

1. What is the maximum size of the input array? Will it fit in memory?
2. Can we modify the input array during our solution?
3. Do we need to return the actual subarray or just a boolean indicating its existence?
4. Are negative numbers allowed in the input array?
5. What is the target sum value? Is it a fixed integer or should we find any valid subarray?

Restated problem: Given an integer array and a target sum, determine whether there exists a continuous subarray of length at least 2 whose elements sum to a multiple of the target sum.

Input/output specification:

• Input: Integer array nums (length n), integer k (target divisor)
• Output: Boolean indicating whether such a subarray exists

Constraints:

• Array length: 2 ≤ n ≤ 10⁵
• Elements can be negative or positive integers
• Target k can be zero or any integer

Assumptions:

• We're looking for a continuous subarray (contiguous elements)
• Subarray must have at least 2 elements
• The sum of the subarray should be divisible by k

Data shape and hidden structure: The problem involves prefix sums with modular arithmetic properties. The key insight is that if we have two prefix sums with the same remainder when divided by k, then their difference forms a subarray sum divisible by k. This suggests using a hash map to store remainders and their earliest occurrence indices.

2. Pattern Recognition

Identified patterns:

• Prefix sum with modular arithmetic
• Subarray sum divisibility problem
• Hash map for remainder tracking

Wording that signals the pattern:

• "Continuous subarray" → prefix sum approach
• "Sum divisible by k" → modular arithmetic
• "At least 2 elements" → minimum length constraint

Why this fits the pattern: This is a classic modular arithmetic problem where we use the mathematical property that if (prefix[i] - prefix[j]) % k == 0, then prefix[i] % k == prefix[j] % k. This allows us to efficiently detect valid subarrays using a hash map.

Core insight: We can transform the problem into finding two prefix sums that have the same remainder when divided by k, which means their difference is divisible by k. We must also ensure the subarray has at least 2 elements.

3. Approach Evolution

Interview communication script:

"My first instinct is to use prefix sums with modular arithmetic because we're looking for a subarray whose sum is divisible by k. But if I naively check all pairs of prefix sums, that would be O(n²) time complexity because for each position i, I'd need to check all previous positions j where j < i.

Let me think about how to optimize this. I notice that if I maintain a hash map storing the first occurrence of each remainder when prefix sums are divided by k, I can reduce this to O(n) time.

The key insight is that if two prefix sums have the same remainder modulo k, their difference forms a subarray sum divisible by k. However, I also need to ensure that the subarray has at least 2 elements, so I should check that the difference in indices is at least 2.

I'll maintain a hash map where I store the first occurrence of each remainder. For each new prefix sum, I'll check if its remainder has been seen before. If so, I'll verify that the distance between current index and first occurrence is at least 2. I also need to handle the special case where a prefix sum itself is divisible by k (remainder 0) and we're at index 1 or later."

Brute-force approach: Check all possible subarrays of length ≥ 2 and verify if their sum is divisible by k. Time complexity O(n³) - O(n²) subarrays, each requiring O(n) sum calculation.

Improved approach: Use prefix sums with hash map to store first occurrence of each remainder. Time complexity O(n) - single pass through array.

Final approach with step-by-step reasoning:

1. Use prefix sum technique to compute cumulative sums
2. For each prefix sum, calculate its remainder modulo k
3. Store first occurrence of each remainder in a hash map
4. If same remainder appears again, check if subarray length is at least 2
5. Handle special case where prefix sum itself is divisible by k

Important invariants:

• Hash map stores only first occurrence of each remainder
• Subarray length constraint (at least 2 elements)
• Remainder modulo k property for divisibility

4. Data Structure Choice

Primary data structures:

1. HashMap<Integer, Integer> - stores remainder → first index mapping
2. int variables for current prefix sum and index tracking

Why each is needed:

• HashMap allows O(1) lookup of previously seen remainders, enabling linear time solution
• Integer variables track running sum and indices for length validation

Time and space impact:

• Time: O(n) for single pass through array
• Space: O(min(n, k)) for hash map storage (at most k distinct remainders)

Helper structures:

• No additional helper data structures needed beyond the hash map

5. Algorithm Walkthrough

Step-by-step execution on example [23, 2, 4, 6, 7], k = 6:

1. Initialize: map = {0: -1}, sum = 0
2. Index 0: sum = 23, remainder = 5, not in map → add map[5] = 0
3. Index 1: sum = 25, remainder = 1, not in map → add map[1] = 1
4. Index 2: sum = 29, remainder = 5, found in map at index 0
   • Distance = 2 - 0 = 2 ≥ 2 ✓ → return true

What happens at each major step:

• We maintain prefix sums and their remainders
• When we find a repeated remainder, we check subarray length constraint
• Special case: remainder 0 at index 1 or later means prefix itself is divisible

Edge cases handled:

• Empty array (not possible due to constraints)
• Single element arrays (not possible due to constraints)
• Zero k case (handled by special condition)
• Negative numbers (handled by modular arithmetic)

6. Java Solution

public class Solution {
    public boolean checkSubarraySum(int[] nums, int k) {
        // Handle edge cases
        if (nums == null || nums.length < 2) {
            return false;
        }
        
        // HashMap to store first occurrence of each remainder
        Map<Integer, Integer> remainderMap = new HashMap<>();
        
        // Initialize with remainder 0 at index -1 to handle cases where prefix sum is divisible by k
        remainderMap.put(0, -1);
        
        int prefixSum = 0;
        
        for (int i = 0; i < nums.length; i++) {
            prefixSum += nums[i];
            
            // Calculate remainder
            int remainder = k == 0 ? prefixSum : prefixSum % k;
            
            // If we've seen this remainder before
            if (remainderMap.containsKey(remainder)) {
                // Check if subarray length is at least 2
                if (i - remainderMap.get(remainder) >= 2) {
                    return true;
                }
            } else {
                // Store first occurrence of this remainder
                remainderMap.put(remainder, i);
            }
        }
        
        return false;
    }
}

7. Complexity Analysis

Time complexity: O(n) where n is the length of the input array. We make a single pass through the array with O(1) operations at each step.

Space complexity: O(min(n, k)) where n is the array length and k is the target divisor. In worst case, we store at most min(n, k) distinct remainders in the hash map.

Why bounds are correct:

• We process each element exactly once, making it linear time
• The hash map size is bounded by the number of possible remainders (k distinct values) or array length, whichever is smaller
• Each hash map operation (put/get) is O(1) average case

Optimality: This solution is optimal for the problem as it achieves linear time complexity, which is necessary since we must examine each element at least once to determine if any valid subarray exists.

8. Edge Cases & Pitfalls

Empty and minimal inputs:

• Array with less than 2 elements (handled by early return)
• Single element array (not possible due to constraints)

Duplicate values and repeated states:

• Multiple occurrences of same remainder handled correctly by storing only first occurrence
• Negative numbers work correctly with modular arithmetic

Boundary conditions:

• k = 0 case requires special handling (division by zero)
• Subarray length exactly 2 must be validated
• Index 0 handling with initial map entry

Common implementation mistakes:

• Forgetting to initialize the hash map with {0: -1} to handle prefix sums divisible by k
• Not handling the special case where k = 0 properly
• Incorrectly checking subarray length (should be i - prevIndex >= 2)
• Not accounting for negative numbers in modular arithmetic

9. Alternative Approaches

Alternative 1: Brute Force with Prefix Sum

• Time: O(n²) - compute all subarrays and check divisibility
• Space: O(1) - only use variables for sums
• When preferred: For small arrays or when simplicity is more important than performance

Alternative 2: Two-Pass Approach with Prefix Sum Array

• Create prefix sum array, then check all pairs of indices
• Time: O(n²) - nested loop over prefix sums
• Space: O(n) - additional array for prefix sums
• When preferred: When we need to preserve the original array and can afford O(n) space

Trade-offs:

• Brute force is simpler to understand but much slower
• Two-pass approach uses more space and is still O(n²) time
• Our chosen solution is optimal with linear time and reasonable space usage

10. Test Cases

// Sample test cases
public class TestCases {
    public void runTests() {
        // Test case 1: Basic valid case
        int[] nums1 = {23, 2, 4, 6, 7};
        int k1 = 6;
        // Expected: true (subarray [2,4] sums to 6, divisible by 6)
        
        // Test case 2: k = 0
        int[] nums2 = {5, 3, 4};
        int k2 = 0;
        // Expected: false (no valid subarray when k = 0)
        
        // Test case 3: Subarray length exactly 2
        int[] nums3 = {1, 0, 1, 0};
        int k3 = 2;
        // Expected: true (subarray [0,1] sums to 1, not divisible by 2)
        // Actually: [1,0] sums to 1, [0,1] sums to 1, [1,0] sums to 1
        // Let me correct: [0,1] at indices 1,2 sums to 1, not divisible by 2
        // But [1,0] at indices 0,1 sums to 1, not divisible by 2
        // Actually [0,1] at indices 1,2 sums to 1, [1,0] at indices 0,1 sums to 1
        // Need [0,0] or [1,1] etc. Let's say [1,0,1,0] with k=2
        // [0,1] at indices 1,2 sums to 1, not divisible by 2
        // [0,1] at indices 3,4 sums to 0, divisible by 2 - but that's indices 2,3
        // [1,0] at indices 0,1 sums to 1, not divisible by 2
        // [0,1] at indices 1,2 sums to 1, not divisible by 2
        // [1,0] at indices 2,3 sums to 1, not divisible by 2
        // [0,1] at indices 3,4 sums to 0, divisible by 2 - but that's [0] at index 3, [1] at index 4
        // Actually [1,0] at indices 0,1 sums to 1, not divisible by 2  
        // [0,1] at indices 1,2 sums to 1, not divisible by 2
        // [0,1] at indices 2,3 sums to 1, not divisible by 2
        // [1,0] at indices 3,4 sums to 1, not divisible by 2
        // [0,1] at indices 1,2 sums to 1, not divisible by 2  
        // [0,1] at indices 2,3 sums to 1, not divisible by 2
        // [0,1] at indices 3,4 sums to 1, not divisible by 2
        // [0,0] at indices 1,2 sums to 0, divisible by 2 - yes!
        // So [0,0] at indices 1,2 works! 
        int[] nums3 = {1, 0, 0, 1};
        int k3 = 2;
        // Expected: true (subarray [0,0] at indices 1,2 sums to 0, divisible by 2)
        
        // Test case 4: Empty array (not valid due to constraints)
        // Test case 5: Single element array (not valid due to constraints)
    }
}

11. Final Summary

One-sentence recap: The solution uses prefix sums with modular arithmetic to efficiently detect if any continuous subarray of at least 2 elements has a sum divisible by the target value.

Why this solution works: It leverages the mathematical property that if two prefix sums have the same remainder when divided by k, their difference represents a subarray sum divisible by k. By storing only the first occurrence of each remainder, we can quickly identify valid subarrays in linear time.

What makes it efficient: The solution achieves optimal O(n) time complexity by making a single pass through the input array and using a hash map for constant-time remainder lookups, avoiding the need to check all possible subarrays.

2. Pattern Recognition

Identifying the most likely pattern or patterns

The most likely pattern here is prefix sum with modular arithmetic. This is a classic application of the pigeonhole principle and modular arithmetic properties.

Why this problem fits that pattern

The problem asks us to find a continuous subarray whose sum is divisible by k. This immediately signals that we're dealing with modular arithmetic properties. When we compute prefix sums and take them modulo k, we're looking for two prefix sums that have the same remainder when divided by k. If two prefix sums have the same remainder, their difference (which represents a subarray sum) is divisible by k.

Core insight needed to solve it

The key insight is that if we have prefix sums at indices i and j where prefixSum[i] % k == prefixSum[j] % k, then the subarray from index i+1 to j has a sum divisible by k. This is because:

(prefixSum[j] - prefixSum[i]) % k == 0

This pattern appears in problems involving subarrays with specific sum properties, particularly when dealing with divisibility constraints.

Wording that signals the pattern

• "continuous subarray" - suggests prefix sum approach
• "sum divisible by k" - signals modular arithmetic and the pigeonhole principle
• "at least length 2" - adds constraint that we need to track indices properly

This is a well-known pattern in competitive programming where we use the mathematical property that if two prefix sums have the same remainder modulo k, their difference is divisible by k.

3. Approach Evolution

My first instinct is to use a brute force approach where I check every possible subarray to see if its sum is divisible by k. But that's O(n²) because for each starting position, I'd need to calculate sums for all ending positions. Let me think about how to optimize this.

I notice that if I maintain a running sum and use modular arithmetic, I can reduce redundant calculations. The key insight is that if two prefix sums have the same remainder when divided by k, then their difference (the subarray sum) is divisible by k.

Let me think about this more carefully. If prefixSum[i] % k == prefixSum[j] % k, then (prefixSum[j] - prefixSum[i]) % k == 0, which means the subarray sum from i+1 to j is divisible by k.

This suggests I can use a HashMap to store the first occurrence of each remainder. If I encounter the same remainder again, and the distance between positions is at least 2, then I've found a valid subarray.

The key invariant that must hold is: we need at least one element between the two indices where we see the same remainder, so that our subarray has at least length 2.

I'll also need to handle the edge case where a prefix sum itself is divisible by k, which means we found a valid subarray starting from index 0.

The optimization works because instead of recalculating all subarray sums, I only need to track remainders and their first occurrence indices. When I see a repeated remainder, I can immediately determine if the subarray between those positions has sum divisible by k.

This reduces the time complexity from O(n²) to O(n) while maintaining correctness through the mathematical property of modular arithmetic.

// Brute force approach - O(n²) time, O(1) space
// For each starting position i, calculate sum for all ending positions j >= i
// Check if any sum is divisible by k

// Improved approach - O(n) time, O(min(n,k)) space
// Use prefix sums and modular arithmetic
// Store first occurrence of each remainder in HashMap
// When same remainder appears again, check if subarray length >= 2

4. Data Structure Choice

For this problem, I'll use a HashMap to store prefix sums and their indices. This is the key data structure that enables the O(n) solution.

The HashMap stores <prefix_sum % k, earliest_index> pairs. This allows us to quickly check if we've seen a prefix sum that differs by a multiple of k from our current prefix sum.

Why HashMap is the right choice:

• Fast lookup and insertion: O(1) average time complexity for both operations
• Space efficiency: We only store what we need to detect the target subarray
• Simplifies logic: Eliminates the need for nested loops or additional data structures

The key invariant maintained is that we only store the earliest occurrence of each modulo value to ensure we find the longest valid subarray.

HashMap is preferred over other structures like ArrayList or TreeSet because:

• ArrayList would require O(n) search time for each lookup
• TreeSet would add unnecessary complexity for this specific use case
• HashMap provides the optimal balance of time and space efficiency

The HashMap is our primary optimization lever that transforms this from O(n²) to O(n) time complexity.

// HashMap to store <remainder, first occurrence index>
Map<Integer, Integer> remainderMap = new HashMap<>();

5. Algorithm Walkthrough

Let me trace through the algorithm with a concrete example to demonstrate how it works.

Consider the input array nums = [23, 2, 4, 6, 7] with k = 6.

Step 1: Initialize variables

• remainderMap = {} (empty HashMap)
• runningSum = 0
• remainder = 0

Step 2: Process index 0 (nums[0] = 23)

• runningSum = 0 + 23 = 23
• remainder = 23 % 6 = 5
• Map does not contain remainder 5, so add remainderMap[5] = 0

Step 3: Process index 1 (nums[1] = 2)

• runningSum = 23 + 2 = 25
• remainder = 25 % 6 = 1
• Map does not contain remainder 1, so add remainderMap[1] = 1

Step 4: Process index 2 (nums[2] = 4)

• runningSum = 25 + 4 = 29
• remainder = 29 % 6 = 5
• Map contains remainder 5 with value 0
• Since currentIdx - storedIdx = 2 - 0 = 2 ≥ 2, we found a valid subarray
• Return true

The algorithm correctly identifies that the subarray [2, 4] (indices 1-2) sums to 6, which is divisible by k=6.

Edge case handling:

• When we have a prefix sum that is divisible by k, we check if it's at index 0 (which would mean the subarray is from index 1 to current)
• The map is initialized with remainderMap.put(0, -1) to handle this case properly

The algorithm terminates correctly because:

1. We process each element once
2. For any valid subarray, there exists a previous prefix sum with the same remainder
3. The difference in indices gives us the subarray length ≥ 2
4. We return immediately when we find such a case, ensuring optimal termination

6. Java Solution

public class Solution {
    public boolean checkSubarraySum(int[] nums, int k) {
        // Handle edge cases
        if (nums == null || nums.length < 2) {
            return false;
        }
        
        // Map to store remainder and its first occurrence index
        Map<Integer, Integer> remainderMap = new HashMap<>();
        // Initialize with remainder 0 at index -1 to handle subarrays starting from index 0
        remainderMap.put(0, -1);
        
        int prefixSum = 0;
        
        for (int i = 0; i < nums.length; i++) {
            prefixSum += nums[i];
            int remainder = prefixSum % k;
            
            // If we've seen this remainder before, check if the subarray length is at least 2
            if (remainderMap.containsKey(remainder)) {
                int prevIndex = remainderMap.get(remainder);
                // Subarray length must be at least 2
                if (i - prevIndex > 1) {
                    return true;
                }
            } else {
                // Store the first occurrence of this remainder
                remainderMap.put(remainder, i);
            }
        }
        
        return false;
    }
}

This solution uses the mathematical property that if two prefix sums have the same remainder when divided by k, then their difference (the subarray sum) is divisible by k. We maintain a map of remainders to their first occurrence indices to efficiently detect valid subarrays while ensuring the length constraint is met.

7. Complexity Analysis

Time Complexity: O(n)

The algorithm makes a single pass through the input array with a constant-time operation at each step. For each element, we perform:

• Hash map lookup and insertion (O(1) average case)
• Modular arithmetic operations
• Constant-time conditional checks

Since we process each element exactly once, the overall time complexity is linear.

Space Complexity: O(min(n, k))

The space usage comes entirely from the HashMap storing prefix sums modulo k. In the worst case, we might store up to n different remainders if all prefix sums have distinct remainders. However, since there are only k possible remainders (0 through k-1), the HashMap will contain at most k entries before we find a duplicate.

The actual space complexity is O(min(n, k)) because:

• If n ≤ k, we might store up to n entries (though this is rare in practice)
• If n > k, we can't have more than k distinct remainders, so the HashMap size is bounded by k

The solution is optimal for this problem because:

1. We must examine each element at least once to determine if a valid subarray exists
2. The modular arithmetic approach is the most efficient way to detect subarrays with sum divisible by k
3. No better algorithm exists that can solve this problem in less than O(n) time

This complexity analysis assumes the hash map operations are O(1) amortized, which is valid for standard Java HashMap implementations with good hash functions.

8. Edge Cases & Pitfalls

Empty and Minimal Inputs

• Empty array: Should return false since no subarray exists
• Single element array: Should return false since we need at least 2 elements for a valid subarray
• Two element array: Valid case that should be handled properly

Duplicate Values and Repeated States

• Arrays with many duplicate values can cause issues with modular arithmetic
• The same remainder can appear multiple times, but we must ensure the subarray length is at least 2
• Example: [23, 2, 4, 6, 7] with k = 6 - remainder 2 appears at indices 0 and 2, but we need to ensure the distance is at least 2

Boundary Conditions

• k = 0: Special case requiring careful handling - when k=0, we're looking for subarrays that sum to 0
• Array length < 2: Immediate return false
• All zeros: Should handle properly (e.g., [0, 0] with k = 0 should return true)
• Negative numbers: Must handle negative prefix sums correctly in modular arithmetic

Common Implementation Mistakes

1. Incorrect modular arithmetic with negatives: In Java, -1 % 6 = -1, but we want 5. Must handle negative remainders properly.
2. Index tracking errors: Not maintaining proper distance between indices when same remainder appears
3. Not checking minimum length requirement: Failing to verify subarray has at least 2 elements
4. HashMap key collision: Using integer remainder as HashMap key without considering hash collisions (not an issue in this case since we're using direct integer keys)
5. Off-by-one errors: Miscounting array indices when calculating subarray length

Implementation-specific Issues

• HashMap initialization: Should handle the case where prefix sum mod k = 0 at index 0 (this is a valid subarray of length > 1)
• Zero handling: When k = 0, we must check if any subarray sums to zero rather than checking modular arithmetic
• Integer overflow: Large arrays with large numbers could cause overflow in prefix sum calculations, though this is less critical for the algorithm's correctness

9. Alternative Approaches

There are several alternative approaches to solving the Continuous Subarray Sum problem, each with different trade-offs:

Brute Force Approach: Check all possible subarrays using nested loops. For each starting index, iterate through all ending indices and calculate the sum. This approach has O(n³) time complexity due to recalculating sums repeatedly, but can be optimized to O(n²) by maintaining running sums. While simple to implement and understand, it's inefficient for large inputs.

Prefix Sum with Hash Map: Instead of checking all subarrays, we can use the mathematical property that (sum[j] - sum[i]) % k = 0 implies sum[j] % k = sum[i] % k. We maintain a hash map storing the first occurrence of each remainder when prefix sums are divided by k. This allows us to detect if there exists a subarray with sum divisible by k in O(n) time. This is essentially the same as our chosen approach but with a different perspective on the mathematical relationship.

Sliding Window Technique: This approach works well when we're looking for subarrays with specific sum values, but requires the array to have certain properties (like all positive numbers). For this problem with modular arithmetic, sliding window isn't directly applicable since we're not looking for a fixed sum but rather one that's divisible by k.

Two Pointers with Modular Arithmetic: We could attempt to use two pointers where we maintain a window and track remainders, but this becomes complex due to the modular arithmetic constraint and doesn't naturally lead to an efficient solution like our chosen approach.

The hash map-based prefix sum approach is preferred because it's both time and space optimal, handles all edge cases gracefully, and leverages mathematical properties of modular arithmetic to achieve linear time complexity. The brute force approach is only suitable for very small inputs due to its quadratic time complexity, while the sliding window approach doesn't apply well to this modular sum problem.

10. Test Cases

public class Solution {
    public boolean checkSubarraySum(int[] nums, int k) {
        // Implementation here
    }
}

The test cases below demonstrate various scenarios including edge cases, boundary conditions, and tricky inputs:

// Test case 1: Basic valid case
int[] nums1 = {23, 2, 4, 6, 7};
int k1 = 6;
// Expected: true (subarray [2, 4] sums to 6, which is divisible by 6)

// Test case 2: Valid case with negative numbers
int[] nums2 = {23, 2, -4, 6, 7};
int k2 = 6;
// Expected: true (subarray [2, -4, 6] sums to 4, and 4 % 6 = 4, but [23, 2, -4] sums to 21, and 21 % 6 = 3)

// Test case 3: Empty array
int[] nums3 = {};
int k3 = 1;
// Expected: false

// Test case 4: Single element array
int[] nums4 = {1};
int k4 = 1;
// Expected: false

// Test case 5: Two elements that sum to k
int[] nums5 = {1, 2};
int k5 = 3;
// Expected: true (entire array sums to 3, divisible by 3)

// Test case 6: Subarray with sum zero
int[] nums6 = {1, 2, 3, 4, 5};
int k6 = 5;
// Expected: true (subarray [2,3] sums to 5, divisible by 5)

// Test case 7: Large numbers and k = 1
int[] nums7 = {1000000, 1000000, 1000000};
int k7 = 1;
// Expected: true (any number divisible by 1)

// Test case 8: All zeros
int[] nums8 = {0, 0};
int k8 = 0;
// Expected: true (subarray [0,0] sums to 0, and 0 % 0 is undefined, but we handle this case)

// Test case 9: Large subarray sum
int[] nums9 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
int k9 = 11;
// Expected: true (subarray [2,3,4,5] sums to 14, and 14 % 11 = 3, but [1,2,3,4,5] sums to 15, and 15 % 11 = 4)

// Test case 10: Multiple valid subarrays
int[] nums10 = {2, 4, 3, 6, 8, 9};
int k10 = 5;
// Expected: true (subarray [4,3] sums to 7, and 7 % 5 = 2, but [2,4,3] sums to 9, and 9 % 5 = 4)

// Test case 11: k = 0 with valid subarray
int[] nums11 = {2, 4, 3, 6};
int k11 = 0;
// Expected: false (division by zero case, but we should return false)

// Test case 12: Large numbers with k = 7
int[] nums12 = {5, 1, 2, 3, 4, 5, 6};
int k12 = 7;
// Expected: true (subarray [5,1,2,3,4] sums to 15, and 15 % 7 = 1, but [1,2,3,4,5] sums to 15, and 15 % 7 = 1)

// Test case 13: Valid subarray of length exactly 2
int[] nums13 = {5, 1, 2};
int k13 = 6;
// Expected: true (subarray [5,1] sums to 6, divisible by 6)

// Test case 14: Valid subarray with negative numbers
int[] nums14 = {-1, -2, 3, 4};
int k14 = 6;
// Expected: true (subarray [-2,3,4] sums to 5, and 5 % 6 = 5, but [-1,-2,3] sums to 0, and 0 % 6 = 0)

// Test case 15: Large array with large k
int[] nums15 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
int k15 = 1000000;
// Expected: false (no subarray sum divisible by large k)

The key test cases cover:

1. Basic functionality with valid subarrays
2. Edge cases like empty arrays and single elements
3. Negative numbers and zero handling
4. Boundary conditions with k = 0 and k = 1
5. Large numbers and large k values
6. Subarray length exactly 2 (minimum valid length)
7. Multiple valid subarrays in one array

These cases ensure robustness across various input scenarios and help validate the correctness of the implementation.

11. Final Summary

The core insight is to leverage the mathematical property that if two prefix sums have the same remainder when divided by k, then the subarray between them has a sum divisible by k. This transforms the problem from checking all possible subarrays into tracking remainders with a hash map.

The solution works by maintaining a running prefix sum and storing the first occurrence of each remainder modulo k. When we encounter the same remainder again, we know there's a valid subarray between the two positions. The key optimization is that we only need to check if the distance between positions is at least 2 to ensure we have a valid subarray.

This approach is optimal with O(n) time and space complexity, making it efficient for large inputs while maintaining correctness through careful handling of edge cases like zero remainders and minimum subarray length requirements.