Problem Description

You have an array nums of non-negative integers (0-indexed) and a list of queries. Each query is represented as [xi, yi].

For each query [xi, yi], you need to:

1. Start from index xi in the array
2. Sum up elements at positions xi, xi + yi, xi + 2*yi, xi + 3*yi, and so on
3. In other words, sum all nums[j] where j >= xi and (j - xi) is divisible by yi

The answer to each query should be returned modulo 10^9 + 7.

For example, if nums = [1, 2, 3, 4, 5] and query is [1, 2]:

• Start at index 1 (value 2)
• Next valid index: 1 + 2 = 3 (value 4)
• Next valid index: 3 + 2 = 5 (out of bounds)
• Sum = 2 + 4 = 6

The solution uses block decomposition to optimize performance:

• For queries with small step sizes (≤ √n), precompute suffix sums for each starting position and step size
• For queries with large step sizes (> √n), calculate the sum directly since there won't be many elements to sum

The preprocessing creates a 2D array suf[i][j] that stores the sum of elements starting from position j with step size i. This allows O(1) lookup for small step queries while keeping the overall complexity manageable.

Intuition

The naive approach would be to iterate through the array for each query, jumping by step size y starting from index x. This works but can be inefficient when we have many queries.

The key insight is recognizing that the efficiency depends on the step size y:

1. Small step sizes (like 1, 2, 3): If y is small, we'll visit many elements. For example, with step size 1, we visit every element from x to the end. With step size 2, we visit half the remaining elements. These patterns repeat across different queries with the same step size.

2. Large step sizes (like n/2, n-1): If y is large, we only visit a few elements. For example, with step size n/2 in an array of size n, we visit at most 2 elements.

This observation leads to a trade-off strategy:

• For small step sizes that appear frequently and visit many elements, we should precompute the results to avoid redundant calculations
• For large step sizes that visit few elements, direct calculation is already efficient

The threshold for "small" vs "large" is chosen as √n because:

• There are at most √n different small step sizes (1 to √n)
• For each small step size, we need O(n) space to store precomputed sums
• Total preprocessing space: O(√n * n) which is manageable
• Each large step query visits at most n/√n = √n elements, making direct calculation O(√n)

This balances preprocessing time/space with query efficiency, achieving O(√n) per query regardless of step size.

Learn more about Dynamic Programming patterns.

Solution Approach

The solution implements block decomposition with a threshold of √n to distinguish between small and large step sizes.

Preprocessing Phase:

1. Calculate the threshold m = int(sqrt(n)) to separate small and large step sizes.

2. Create a 2D array suf with dimensions (m+1) × (n+1):

   • suf[i][j] stores the sum of elements starting from position j with step size i
   • We only precompute for step sizes from 1 to m

3. Build suffix sums in reverse order:

   for i in range(1, m + 1):  # For each small step size
       for j in range(n - 1, -1, -1):  # Process positions from right to left
           suf[i][j] = suf[i][min(n, j + i)] + nums[j]

   • Processing from right to left allows us to build suffix sums incrementally
   • suf[i][j] = nums[j] + suf[i][j+i] means: the sum starting at j equals the current element plus the sum starting at the next position j+i
   • min(n, j + i) handles boundary cases when j + i exceeds array length

Query Processing Phase:

For each query [x, y]:

1. If y <= m (small step size):

   • Directly return the precomputed value: suf[y][x] % mod
   • This is O(1) lookup time

2. If y > m (large step size):

   • Calculate the sum directly using Python's slice notation: sum(nums[x::y])
   • nums[x::y] creates a slice starting at index x with step y
   • This visits at most n/y < n/√n = √n elements

Time Complexity:

• Preprocessing: O(m × n) = O(√n × n)
• Per query: O(1) for small steps, O(√n) for large steps
• Total: O(n√n + q√n) where q is the number of queries

Space Complexity: O(n√n) for storing the suffix sum array

Example Walkthrough

Let's walk through the solution with a concrete example to illustrate how the block decomposition approach works.

Given:

• nums = [1, 2, 3, 4, 5, 6] (n = 6)
• Queries: [[0, 2], [1, 2], [2, 3], [0, 5]]

Step 1: Calculate threshold

• m = int(sqrt(6)) = 2
• Step sizes ≤ 2 are "small", step sizes > 2 are "large"

Step 2: Build suffix sum array

We create suf[3][7] (dimensions: (m+1) × (n+1) = 3 × 7)

For step size 1 (i=1):

• Start from right to left (j = 5 to 0)
• suf[1][5] = nums[5] + suf[1][6] = 6 + 0 = 6
• suf[1][4] = nums[4] + suf[1][5] = 5 + 6 = 11
• suf[1][3] = nums[3] + suf[1][4] = 4 + 11 = 15
• suf[1][2] = nums[2] + suf[1][3] = 3 + 15 = 18
• suf[1][1] = nums[1] + suf[1][2] = 2 + 18 = 20
• suf[1][0] = nums[0] + suf[1][1] = 1 + 20 = 21

For step size 2 (i=2):

• suf[2][5] = nums[5] + suf[2][7] = 6 + 0 = 6
• suf[2][4] = nums[4] + suf[2][6] = 5 + 0 = 5
• suf[2][3] = nums[3] + suf[2][5] = 4 + 6 = 10
• suf[2][2] = nums[2] + suf[2][4] = 3 + 5 = 8
• suf[2][1] = nums[1] + suf[2][3] = 2 + 10 = 12
• suf[2][0] = nums[0] + suf[2][2] = 1 + 8 = 9

Final suf array:

suf[1] = [21, 20, 18, 15, 11, 6, 0]  // step size 1
suf[2] = [9, 12, 8, 10, 5, 6, 0]     // step size 2

Step 3: Process queries

Query 1: [0, 2] (start at index 0, step size 2)

• Step size 2 ≤ m, so use precomputed value
• Answer: suf[2][0] = 9
• Verification: nums[0] + nums[2] + nums[4] = 1 + 3 + 5 = 9 ✓

Query 2: [1, 2] (start at index 1, step size 2)

• Step size 2 ≤ m, so use precomputed value
• Answer: suf[2][1] = 12
• Verification: nums[1] + nums[3] + nums[5] = 2 + 4 + 6 = 12 ✓

Query 3: [2, 3] (start at index 2, step size 3)

• Step size 3 > m, so calculate directly
• Visit indices: 2, 5
• Answer: nums[2] + nums[5] = 3 + 6 = 9

Query 4: [0, 5] (start at index 0, step size 5)

• Step size 5 > m, so calculate directly
• Visit indices: 0, 5
• Answer: nums[0] + nums[5] = 1 + 6 = 7

Summary:

• Small step queries (≤2): O(1) lookup from precomputed suf array
• Large step queries (>2): O(√n) direct calculation, visiting at most 2 elements
• All queries answered efficiently without redundant calculations

Solution Implementation

Time and Space Complexity

Time Complexity: O((n + m) × √n)

The time complexity breaks down into two main parts:

1. Preprocessing phase: Building the suffix sum array suf

   • The outer loop runs for m + 1 iterations where m = √n, so approximately √n iterations
   • The inner loop runs for n iterations
   • Total preprocessing time: O(√n × n) = O(n√n)

2. Query processing phase: Processing m queries (where m is the number of queries)

   • For each query with y ≤ √n: Direct lookup in the precomputed array takes O(1) time
   • For each query with y > √n: Computing sum(nums[x::y]) takes O(n/y) time. Since y > √n, this is at most O(n/√n) = O(√n) time
   • Total query processing time: O(m × √n)

Combined time complexity: O(n√n + m√n) = O((n + m) × √n)

Space Complexity: O(n × √n)

The space complexity is determined by:

• The suf array has dimensions (m + 1) × (n + 1) where m = √n
• This gives us approximately √n × n = n√n space
• Other variables use O(1) or O(n) space which doesn't affect the overall complexity

Therefore, the space complexity is O(n√n).

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

Common Pitfalls

1. Integer Overflow in Suffix Sum Accumulation

Pitfall: When building the suffix sums, the accumulated values can become very large before applying the modulo operation. In languages with fixed integer sizes (like C++ or Java), this can cause integer overflow. Even in Python, storing unnecessarily large intermediate values wastes memory and slows computation.

Example:

# Problematic: Sums can grow extremely large before modulo
suffix_sums[step][start_idx] = suffix_sums[step][next_idx] + nums[start_idx]

Solution: Apply modulo operation during the accumulation phase:

suffix_sums[step][start_idx] = (suffix_sums[step][next_idx] + nums[start_idx]) % MOD

2. Incorrect Threshold Calculation for Edge Cases

Pitfall: Using int(sqrt(n)) as the threshold can be suboptimal for small arrays. When n = 1 or n = 2, the threshold becomes 1, which means we only precompute for step size 1, missing potential optimizations for step size 2.

Example:

# For n = 2, threshold = 1, but step size 2 could also be precomputed efficiently
threshold = int(sqrt(n))

Solution: Use a minimum threshold value or adjust based on memory constraints:

threshold = max(int(sqrt(n)), min(n, 100))  # Ensure reasonable minimum
# OR
threshold = int(sqrt(n)) if n > 4 else n

3. Memory Allocation for Very Large Arrays

Pitfall: The 2D suffix sum array uses O(n√n) memory. For very large arrays (e.g., n = 10^6), this creates a matrix of size √10^6 × 10^6 ≈ 1000 × 10^6 = 10^9 elements, which can cause memory issues.

Example:

# This can fail for large n due to memory constraints
suffix_sums = [[0] * (n + 1) for _ in range(threshold + 1)]

Solution: Implement lazy initialization or use a dictionary for sparse storage:

# Option 1: Use dictionary for actually needed combinations
suffix_sums = {}
for step in range(1, threshold + 1):
    for start_idx in range(n - 1, -1, -1):
        next_idx = min(n, start_idx + step)
        key = (step, start_idx)
        suffix_sums[key] = (suffix_sums.get((step, next_idx), 0) + nums[start_idx]) % MOD

# Option 2: Cap the threshold based on available memory
MAX_MEMORY_CELLS = 10**7  # Adjust based on constraints
threshold = min(int(sqrt(n)), MAX_MEMORY_CELLS // n)

4. Off-by-One Error in Boundary Handling

Pitfall: The boundary condition min(n, start_idx + step) assumes that index n in the suffix array represents "out of bounds" with value 0. If this initialization is missed or misunderstood, it can lead to incorrect sums.

Example:

# If suffix_sums[step][n] is not initialized to 0, this fails
next_idx = min(n, start_idx + step)
suffix_sums[step][start_idx] = suffix_sums[step][next_idx] + nums[start_idx]

Solution: Explicitly ensure the boundary initialization:

# Initialize boundary values explicitly
suffix_sums = [[0] * (n + 1) for _ in range(threshold + 1)]
# The (n+1)th column is already 0, which is correct for out-of-bounds

# Alternative: Handle boundary explicitly in code
for step in range(1, threshold + 1):
    for start_idx in range(n - 1, -1, -1):
        if start_idx + step >= n:
            suffix_sums[step][start_idx] = nums[start_idx]
        else:
            suffix_sums[step][start_idx] = (suffix_sums[step][start_idx + step] + nums[start_idx]) % MOD