Maximum Element-Sum of a Complete Subset of Indices - Problem

You are given a 1-indexed array nums of positive integers. Your task is to select a complete subset from nums where every pair of selected indices when multiplied together results in a perfect square.

More formally, if you select indices i and j, then i * j must be a perfect square.

Return the sum of the complete subset with the maximum sum.

Input & Output

Example 1 — Basic Case

$ Input: nums = [8,7,3,5,7,2,4,9]

› Output: 16

💡 Note: Indices 2 and 8 have square-free part 2, so 2×8=16 is a perfect square. Their values are 7+9=16, which is the maximum.

Example 2 — Single Element

$ Input: nums = [5]

› Output: 5

💡 Note: Only one element, so the subset contains just index 1 with value 5.

Example 3 — Multiple Groups

$ Input: nums = [1,4,3,3,2]

› Output: 6

💡 Note: Indices 3 and 4 both have square-free part 3, values 3+3=6. Index 1 has square-free 1 (value 1), index 2 has square-free 2 (value 4), index 5 has square-free 5 (value 2).

Constraints

1 ≤ nums.length ≤ 10⁴
1 ≤ nums[i] ≤ 10⁹

Visualization

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Asked in

G Google 15 f Meta 8

The key insight is that two indices i and j can be in the same complete subset if and only if i×j is a perfect square, which happens when they have the same square-free part. Best approach is Mathematical Grouping with Time: O(n×√max_index), Space: O(n).

Common Approaches

✓ Prefix Sum

⏱️ Time: N/A Space: N/A

Brute Force - Try All Subsets

⏱️ Time: O(n × 2ⁿ) Space: O(n)

Generate all 2^n possible subsets of indices, check if each subset satisfies the perfect square condition for all pairs, and track the maximum sum found.

Mathematical Grouping

⏱️ Time: O(n × √max_index) Space: O(n)

Two indices i and j have i*j as perfect square if and only if they have the same square-free part. We group indices by their square-free factors and find the maximum sum within each group.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

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

static int groups[10005][10005];
static int groupSizes[10005];
static int numGroups;

int getSquareFreePart(int n) {
    int result = 1;
    int i = 2;
    while (i * i <= n) {
        int count = 0;
        while (n % i == 0) {
            n /= i;
            count++;
        }
        if (count % 2 == 1) {
            result *= i;
        }
        i++;
    }
    if (n > 1) {
        result *= n;
    }
    return result;
}

long long solution(int* nums, int n) {
    if (n == 0) return 0;
    
    // Reset groups
    numGroups = 0;
    for (int i = 0; i < 10005; i++) {
        groupSizes[i] = 0;
    }
    
    // Group indices by their square-free parts
    int groupMap[10005];
    for (int i = 0; i < 10005; i++) {
        groupMap[i] = -1;
    }
    
    for (int i = 0; i < n; i++) {
        int index = i + 1;  // 1-indexed
        int squareFree = getSquareFreePart(index);
        
        if (groupMap[squareFree] == -1) {
            groupMap[squareFree] = numGroups++;
        }
        
        int groupIdx = groupMap[squareFree];
        groups[groupIdx][groupSizes[groupIdx]++] = nums[i];
    }
    
    // Find maximum sum among all groups
    long long maxSum = 0;
    for (int g = 0; g < numGroups; g++) {
        long long groupSum = 0;
        for (int i = 0; i < groupSizes[g]; i++) {
            groupSum += groups[g][i];
        }
        if (groupSum > maxSum) {
            maxSum = groupSum;
        }
    }
    
    return maxSum;
}

int main() {
    char line[100000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array manually
    int nums[10000];
    int n = 0;
    
    char* p = line;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        
        int num = 0;
        int sign = 1;
        if (*p == '-') {
            sign = -1;
            p++;
        }
        while (*p >= '0' && *p <= '9') {
            num = num * 10 + (*p - '0');
            p++;
        }
        nums[n++] = sign * num;
    }
    
    long long result = solution(nums, n);
    printf("%lld\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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Medium Frequency

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