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],[5,2]]

› Output: [9,12]

💡 Note: Query [0,3] gives 9 as above. Query [5,2]: Start at index 5, step by 2. Selected indices are 5, 7. Sum = 5 + 7 = 12.

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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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

✓ Brute Force

⏱️ Time: O(q × n) Space: O(1)

For each query [xi, yi], iterate from position xi to end of array, checking if (j - xi) % yi == 0 and summing those elements.

Optimized with Grouping

⏱️ Time: O(n × max(yi) + q) Space: O(n × max(yi))

Group indices by their remainder when divided by step size, then pre-compute prefix sums for each group to answer queries efficiently.

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>

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

void parseQueries(const char* str, int queries[][2], int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',' || *p == '[') p++;
        if (*p == ']' || *p == '\0') break;
        
        queries[*size][0] = (int)strtol(p, (char**)&p, 10);
        while (*p == ' ' || *p == ',') p++;
        queries[*size][1] = (int)strtol(p, (char**)&p, 10);
        (*size)++;
        
        while (*p && *p != ']' && *p != '[') p++;
        if (*p == ']') p++;
    }
}

int* solution(int* nums, int numsSize, int queries[][2], 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() {
    char line[10000];
    int nums[10000];
    int queries[1000][2];
    int numsSize, queriesSize;
    
    // Read nums array
    fgets(line, sizeof(line), stdin);
    parseArray(line, nums, &numsSize);
    
    // Read queries array
    fgets(line, sizeof(line), stdin);
    parseQueries(line, queries, &queriesSize);
    
    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

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

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Code -

Time & Space Complexity

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