Solving Leetcode Interviews in Seconds with AI: Sum of Floored Pairs
Introduction
In this blog post, we will explore how to solve the LeetCode problem "1862" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.
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
Explanation
Here's the solution to the problem, incorporating efficiency and addressing the modulo constraint:
High-Level Approach:
- Frequency Counting: Create a frequency array to count the occurrences of each number in the input. This allows for efficient calculation of divisions.
- Iterative Division: Iterate through all possible numerators (i) and denominators (j) up to the maximum value in the input array. Calculate
floor(i / j)and multiply by the number of timesiandjappear in the input to accumulate the result. - Modulo Arithmetic: Apply modulo operation after each addition to prevent integer overflow.
Complexity:
- Runtime Complexity: O(N + M*logM), where N is the number of elements in
nums, and M is the maximum value innums. - Storage Complexity: O(M), where M is the maximum value in
nums(for the frequency array).
- Runtime Complexity: O(N + M*logM), where N is the number of elements in
Code
def sum_of_floored_pairs(nums):
"""
Calculates the sum of floor(nums[i] / nums[j]) for all pairs of indices in the array, modulo 10^9 + 7.
Args:
nums: An integer array.
Returns:
The sum of floored pairs modulo 10^9 + 7.
"""
MOD = 10**9 + 7
max_val = max(nums)
counts = [0] * (max_val + 1)
for num in nums:
counts[num] += 1
total_sum = 0
for i in range(1, max_val + 1):
count_i = counts[i]
if count_i == 0:
continue
for j in range(1, max_val + 1):
count_j = counts[j]
if count_j == 0:
continue
total_sum = (total_sum + (i // j) * count_i * count_j) % MOD
return total_sum