LeetCode #1862 — HARD

Sum of Floored Pairs

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

Given an integer array nums, return the sum of floor(nums[i] / nums[j]) for all pairs of indices 0 <= i, j < nums.length in the array. Since the answer may be too large, return it modulo 10⁹ + 7.

The floor() function returns the integer part of the division.

Example 1:

Input: nums = [2,5,9]
Output: 10
Explanation:
floor(2 / 5) = floor(2 / 9) = floor(5 / 9) = 0
floor(2 / 2) = floor(5 / 5) = floor(9 / 9) = 1
floor(5 / 2) = 2
floor(9 / 2) = 4
floor(9 / 5) = 1
We calculate the floor of the division for every pair of indices in the array then sum them up.

Example 2:

Input: nums = [7,7,7,7,7,7,7]
Output: 49

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁵

Step 01

Brute Force Baseline

Problem summary: Given an integer array nums, return the sum of floor(nums[i] / nums[j]) for all pairs of indices 0 <= i, j < nums.length in the array. Since the answer may be too large, return it modulo 109 + 7. The floor() function returns the integer part of the division.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math · Binary Search

Example 1

[2,5,9]

Example 2

[7,7,7,7,7,7,7]

Step 02

Core Insight

What unlocks the optimal approach

Find the frequency (number of occurrences) of all elements in the array.
For each element, iterate through its multiples and multiply frequencies to find the answer.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1862: Sum of Floored Pairs
class Solution {
    public int sumOfFlooredPairs(int[] nums) {
        final int mod = (int) 1e9 + 7;
        int mx = 0;
        for (int x : nums) {
            mx = Math.max(mx, x);
        }
        int[] cnt = new int[mx + 1];
        int[] s = new int[mx + 1];
        for (int x : nums) {
            ++cnt[x];
        }
        for (int i = 1; i <= mx; ++i) {
            s[i] = s[i - 1] + cnt[i];
        }
        long ans = 0;
        for (int y = 1; y <= mx; ++y) {
            if (cnt[y] > 0) {
                for (int d = 1; d * y <= mx; ++d) {
                    ans += 1L * cnt[y] * d * (s[Math.min(mx, d * y + y - 1)] - s[d * y - 1]);
                    ans %= mod;
                }
            }
        }
        return (int) ans;
    }
}

// Accepted solution for LeetCode #1862: Sum of Floored Pairs
func sumOfFlooredPairs(nums []int) (ans int) {
	mx := slices.Max(nums)
	cnt := make([]int, mx+1)
	s := make([]int, mx+1)
	for _, x := range nums {
		cnt[x]++
	}
	for i := 1; i <= mx; i++ {
		s[i] = s[i-1] + cnt[i]
	}
	const mod int = 1e9 + 7
	for y := 1; y <= mx; y++ {
		if cnt[y] > 0 {
			for d := 1; d*y <= mx; d++ {
				ans += d * cnt[y] * (s[min((d+1)*y-1, mx)] - s[d*y-1])
				ans %= mod
			}
		}
	}
	return
}

# Accepted solution for LeetCode #1862: Sum of Floored Pairs
class Solution:
    def sumOfFlooredPairs(self, nums: List[int]) -> int:
        mod = 10**9 + 7
        cnt = Counter(nums)
        mx = max(nums)
        s = [0] * (mx + 1)
        for i in range(1, mx + 1):
            s[i] = s[i - 1] + cnt[i]
        ans = 0
        for y in range(1, mx + 1):
            if cnt[y]:
                d = 1
                while d * y <= mx:
                    ans += cnt[y] * d * (s[min(mx, d * y + y - 1)] - s[d * y - 1])
                    ans %= mod
                    d += 1
        return ans

// Accepted solution for LeetCode #1862: Sum of Floored Pairs
impl Solution {
    pub fn sum_of_floored_pairs(nums: Vec<i32>) -> i32 {
        const MOD: i32 = 1_000_000_007;
        let mut mx = 0;
        for &x in nums.iter() {
            mx = mx.max(x);
        }

        let mut cnt = vec![0; (mx + 1) as usize];
        let mut s = vec![0; (mx + 1) as usize];

        for &x in nums.iter() {
            cnt[x as usize] += 1;
        }

        for i in 1..=mx as usize {
            s[i] = s[i - 1] + cnt[i];
        }

        let mut ans = 0;
        for y in 1..=mx as usize {
            if cnt[y] > 0 {
                let mut d = 1;
                while d * y <= (mx as usize) {
                    ans += ((cnt[y] as i64)
                        * (d as i64)
                        * (s[std::cmp::min(mx as usize, d * y + y - 1)] - s[d * y - 1]))
                        % (MOD as i64);
                    ans %= MOD as i64;
                    d += 1;
                }
            }
        }

        ans as i32
    }
}

// Accepted solution for LeetCode #1862: Sum of Floored Pairs
function sumOfFlooredPairs(nums: number[]): number {
    const mx = Math.max(...nums);
    const cnt: number[] = Array(mx + 1).fill(0);
    const s: number[] = Array(mx + 1).fill(0);
    for (const x of nums) {
        ++cnt[x];
    }
    for (let i = 1; i <= mx; ++i) {
        s[i] = s[i - 1] + cnt[i];
    }
    let ans = 0;
    const mod = 1e9 + 7;
    for (let y = 1; y <= mx; ++y) {
        if (cnt[y]) {
            for (let d = 1; d * y <= mx; ++d) {
                ans += cnt[y] * d * (s[Math.min((d + 1) * y - 1, mx)] - s[d * y - 1]);
                ans %= mod;
            }
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(M × log M)

Space

O(M)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.