LeetCode 2709 - Greatest Common Divisor Traversal

Understood. I will provide the complete, detailed reference guide for LeetCode 2702 - Minimum Operations to Make Numbers Non-positive, following your formatting rules exactly. The problem provides a 0-indexed integer array nums and two integers x and y.

leetcodehardarraymathunion-findnumber-theory

LeetCode Problem 2709

Difficulty: 🔴 Hard
Topics: Array, Math, Union-Find, Number Theory

Solution

Understood. I will provide the complete, detailed reference guide for LeetCode 2702 - Minimum Operations to Make Numbers Non-positive, following your formatting rules exactly.

Problem Understanding

The problem provides a 0-indexed integer array nums and two integers x and y. You are allowed to perform the following operation any number of times: choose an index i in the array, decrement nums[i] by x, and decrement all other elements by y. The goal is to determine the minimum number of such operations needed to make every element in nums less than or equal to zero.

In simpler terms, each operation strongly reduces one chosen element and slightly reduces all others. The challenge is to strategically select which index to operate on at each step to minimize the total number of operations.

Key points about the input: nums.length can be up to 10^5, individual numbers can be up to 10^9, and 1 <= y < x <= 10^9. The constraints imply that a brute-force simulation of every possible sequence of operations is infeasible. The algorithm must be efficient in both time and space.

Important edge cases include arrays with a single element, arrays where all numbers are equal, arrays with extremely large values, and cases where x is only slightly larger than y.

Approaches

Brute-force Approach:

The brute-force approach would involve simulating every possible sequence of operations. At each step, we could try choosing every index, update the array, and recursively continue until all elements are non-positive. This method guarantees correctness but is exponential in time complexity (O(n^k), where k is the number of operations needed) and is infeasible for large arrays or large numbers.

Optimal Approach Insight:

The key observation is that after m operations, if you chose index i exactly k_i times, its final value is:

nums[i] - k_i * x - (m - k_i) * y = nums[i] - m*y - k_i*(x - y)

We need each element to be non-positive:

nums[i] - m*y - k_i*(x - y) <= 0  => k_i >= max(0, ceil((nums[i] - m*y) / (x - y)))

The total number of chosen operations must satisfy:

sum(k_i for all i) <= m

Thus, we can perform binary search on the minimum number of total operations m. For each candidate m, calculate the minimum k_i needed for each element. If the sum of k_i ≤ m, it is feasible; otherwise, increase m.

This approach reduces the problem to O(n log(max(nums)/y)) complexity, which is efficient.

Approach Time Complexity Space Complexity Notes
Brute Force O(?^n) O(n) Recursively simulate all sequences of operations
Optimal O(n log(max(nums)/y)) O(n) Binary search on number of operations, compute minimum k_i for each element

Algorithm Walkthrough

  1. Define search boundaries: Initialize low = 0 and high = max element in nums divided by y, plus some buffer. This represents the possible range of total operations.
  2. Binary search: While low < high, compute mid = (low + high) // 2. Check if it is possible to achieve non-positive numbers with mid total operations.
  3. Feasibility check: For each number in nums, compute the minimum times it needs to be chosen:
required_i = max(0, ceil((nums[i] - mid*y) / (x - y)))

Sum all required_i. If sum(required_i) <= mid, mid operations are feasible; otherwise, they are not. 4. Update boundaries: If feasible, update high = mid; else, update low = mid + 1. 5. Result: When binary search completes, low contains the minimum number of operations.

Why it works:

By using binary search on the total number of operations, we exploit the monotonicity property: if m operations are feasible, any m' > m is also feasible. Calculating the minimum required picks k_i ensures that the chosen strategy minimizes operations. This guarantees correctness and optimality.

Python Solution

from typing import List
import math

class Solution:
    def minOperations(self, nums: List[int], x: int, y: int) -> int:
        def is_feasible(m: int) -> bool:
            total_required = 0
            for num in nums:
                required = max(0, math.ceil((num - m * y) / (x - y)))
                total_required += required
                if total_required > m:
                    return False
            return True

        low, high = 0, max(nums)  # Upper bound can be max(nums)
        while low < high:
            mid = (low + high) // 2
            if is_feasible(mid):
                high = mid
            else:
                low = mid + 1
        return low

Implementation walkthrough:

The helper is_feasible computes, for each element, how many times it must be picked to be non-positive. If the total required picks exceed m, the current m is too small. Binary search efficiently narrows down the minimum total operations. This avoids simulating each operation individually.

Go Solution

package main

import (
    "math"
)

func minOperations(nums []int, x int, y int) int {
    isFeasible := func(m int) bool {
        totalRequired := 0
        for _, num := range nums {
            required := int(math.Max(0, math.Ceil(float64(num - m*y)/float64(x-y))))
            totalRequired += required
            if totalRequired > m {
                return false
            }
        }
        return true
    }

    low, high := 0, 0
    for _, num := range nums {
        if num > high {
            high = num
        }
    }

    for low < high {
        mid := (low + high) / 2
        if isFeasible(mid) {
            high = mid
        } else {
            low = mid + 1
        }
    }
    return low
}

Go-specific notes:

Go lacks ceil for integers, so we convert to float64 temporarily. We also manually compute max using math.Max. Slices are handled directly, avoiding nil checks because nums is guaranteed to have at least one element.

Worked Examples

Example 1: nums = [3,4,1,7,6], x = 4, y = 2

Operation nums state Notes
0 [3,4,1,7,6] initial
1 [1,2,-1,3,4] pick index 3
2 [-1,0,-3,-1,2] pick index 3
3 [-3,-2,-5,-3,-2] pick index 4
Result all ≤ 0 total operations = 3

Example 2: nums = [1,2,1], x = 2, y = 1

Operation nums state Notes
0 [1,2,1] initial
1 [0,0,0] pick index 1
Result all ≤ 0 total operations = 1

Complexity Analysis

Measure Complexity Explanation
Time O(n log(max(nums)/y)) Binary search on total operations, each check is O(n)
Space O(1) Only constant extra variables needed

The reasoning relies on the fact that binary search narrows the feasible operations, and the feasibility check iterates through the array once.

Test Cases

# Basic examples
assert Solution().minOperations([3,4,1,7,6], 4, 2) == 3  # Example 1
assert Solution().minOperations([1,2,1], 2, 1) == 1     # Example 2

# Single element
assert Solution().minOperations([10], 5, 3) == 2        # Only one element

# Equal elements
assert Solution().minOperations([5,5,5], 4, 2) == 2     # Symmetry

# Large numbers
assert Solution().minOperations([10**9, 10**9], 10**9, 10**8) == 1

# Small x-y difference
assert Solution().minOperations([10,15,20], 5, 4) == 12
Test Why
[3,4,1,7,6], x=4,y=2 Example from problem
[1,2,1], x=2,y=1 Example from problem