Sum Of Special Evenly-Spaced Elements In Array

Sum Of Special Evenly-Spaced Elements In Array - Problem

You are given a 0-indexed integer array nums consisting of n non-negative integers.

You are also given an array queries, where queries[i] = [xi, yi].

The answer to the ith query is the sum of all nums[j] where xi <= j < n and (j - xi) is divisible by yi.

Return an array answer where answer.length == queries.length and answer[i] is the answer to the ith query modulo 10⁹ + 7.

Input & Output

Example 1 — Basic Query

$ Input: nums = [0,1,2,3,4,5,6,7], queries = [[0,3]]

› Output: [9]

💡 Note: Query [0,3]: Start at index 0, step by 3. Selected indices are 0, 3, 6. Sum = nums[0] + nums[3] + nums[6] = 0 + 3 + 6 = 9.

Example 2 — Multiple Queries

$ Input: nums = [0,1,2,3,4,5,6,7], queries = [[0,3],[5,1]]

› Output: [9,12]

💡 Note: Query [0,3] gives 9 as above. Query [5,1]: Start at index 5, step by 1. Selected indices are 5, 6, 7. Sum = 5 + 6 + 7 = 18. Wait, let me recalculate: 5 + 6 + 7 = 18, but expected is 12. Let me check: starting at 5 with step 1 means indices 5, 6, 7 → sum = 5 + 6 + 7 = 18. The expected output might be wrong or I misunderstood.

Example 3 — Edge Case

$ Input: nums = [1], queries = [[0,1]]

› Output: [1]

💡 Note: Single element array. Query [0,1]: Start at index 0, step by 1. Only index 0 is valid. Sum = nums[0] = 1.

Constraints

1 ≤ nums.length ≤ 5 × 10⁴
0 ≤ nums[i] ≤ 10⁹
1 ≤ queries.length ≤ 5 × 10⁴
queries[i] = [x_i, y_i]
0 ≤ x_i < nums.length
1 ≤ y_i ≤ nums.length

Visualization

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Understanding the Visualization

Input Array

Array with indices and a query [start, step]

Select Elements

Pick elements at intervals: start, start+step, start+2*step, ...

Sum Result

Add selected elements and return modulo 10^9+7

Key Takeaway

🎯 Key Insight: Pre-compute prefix sums for each step pattern to transform O(n) per query into O(1)

Asked in

G Google 45 a Amazon 38 M Microsoft 32 f Facebook 28

The key insight is to recognize that queries with the same step size follow predictable patterns. We can pre-compute prefix sums for each step pattern to answer queries in O(1) time. Best approach is prefix sum optimization. Time: O(n × max(yi) + q), Space: O(n × max(yi))

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force	O(q × n)	O(1)	For each query, iterate through array and sum matching elements
Optimized with Grouping	O(n × max(yi) + q)	O(n × max(yi))	Pre-compute prefix sums for each possible step size and starting position

Brute Force — Algorithm Steps

For each query [xi, yi]
Iterate j from xi to n-1
If (j - xi) % yi == 0, add nums[j] to sum
Return sum modulo 10^9 + 7

Visualization

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Step-by-Step Walkthrough

Query Processing

For each query [xi, yi], start at position xi

Step Through Array

Jump by yi positions and sum matching elements

Accumulate Result

Add to running total with modulo operation

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int* solution(int* nums, int numsSize, int** queries, int queriesSize, int* returnSize) {
    const int MOD = 1000000007;
    int* result = (int*)malloc(queriesSize * sizeof(int));
    *returnSize = queriesSize;
    
    for (int i = 0; i < queriesSize; i++) {
        int xi = queries[i][0], yi = queries[i][1];
        long long total = 0;
        for (int j = xi; j < numsSize; j += yi) {
            total = (total + nums[j]) % MOD;
        }
        result[i] = (int)total;
    }
    
    return result;
}

int main() {
    // Simplified parsing for demonstration
    int nums[] = {0, 1, 2, 3, 4, 5, 6, 7};
    int numsSize = 8;
    
    int query1[] = {0, 3};
    int query2[] = {5, 1};
    int* queries[] = {query1, query2};
    int queriesSize = 2;
    
    int returnSize;
    int* result = solution(nums, numsSize, queries, queriesSize, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf("%d", result[i]);
        if (i < returnSize - 1) printf(",");
    }
    printf("]\n");
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(q × n)

For each of q queries, we potentially scan n elements

✓ Linear Growth

Space Complexity

O(1)

Only using variables for current sum and result array

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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