1283. Find the Smallest Divisor Given a Threshold

LeetCode medium original: C# #array #csharp #leetcode #medium #search

题目文本会按所选界面语言从俄语翻译；代码保持不变。

LeetCode Static

Дан 数组 целых чисел nums и 整数 threshold. Мы выберем положительный целый делитель, разделим все elementы 数组а на него и суммируем результат деления. find наименьший делитель, такой что результат, упомянутый выше, меньше или равен threshold.

Каждый результат деления округляется до ближайшего большего целого числа. (На示例: 7/3 = 3 и 10/2 = 5).

C# 解法

匹配/原始

public class Solution {
    public int SmallestDivisor(int[] nums, int threshold) {
        int maxElement = nums.Max();
        
        for (int divisor = 1; divisor <= maxElement; divisor++) {
            int sumOfDivisionResults = 0;
            bool thresholdExceeded = true;
            
            foreach (int num in nums) {
                sumOfDivisionResults += (num + divisor - 1) / divisor;
                if (sumOfDivisionResults > threshold) {
                    thresholdExceeded = false;
                    break;
                }
            }
            
            if (thresholdExceeded) {
                return divisor;
            }
        }
        
        return -1;
    }
}

C++ 解法

自动草稿，提交前请检查

#include <bits/stdc++.h>
using namespace std;

// Auto-generated C++ draft from the C# solution. Review containers, LINQ and helper types before submit.
class Solution {
public:
    public int SmallestDivisor(vector<int>& nums, int threshold) {
        int maxElement = nums.Max();
        
        for (int divisor = 1; divisor <= maxElement; divisor++) {
            int sumOfDivisionResults = 0;
            bool thresholdExceeded = true;
            
            foreach (int num in nums) {
                sumOfDivisionResults += (num + divisor - 1) / divisor;
                if (sumOfDivisionResults > threshold) {
                    thresholdExceeded = false;
                    break;
                }
            }
            
            if (thresholdExceeded) {
                return divisor;
            }
        }
        
        return -1;
    }
}

Java 解法

匹配/原始

class Solution {
    public int smallestDivisor(int[] nums, int threshold) {
        int maxElement = Arrays.stream(nums).max().getAsInt();
        
        for (int divisor = 1; divisor <= maxElement; divisor++) {
            int sumOfDivisionResults = 0;
            boolean thresholdExceeded = true;
            
            for (int num : nums) {
                sumOfDivisionResults += (int) Math.ceil((double) num / divisor);
                if (sumOfDivisionResults > threshold) {
                    thresholdExceeded = false;
                    break;
                }
            }
            
            if (thresholdExceeded) {
                return divisor;
            }
        }
        
        return -1;
    }
}

JavaScript 解法

匹配/原始

var smallestDivisor = function(nums, threshold) {
    let maxElement = Math.max(...nums);
    
    for (let divisor = 1; divisor <= maxElement; divisor++) {
        let sumOfDivisionResults = 0;
        let thresholdExceeded = true;
        
        for (let num of nums) {
            sumOfDivisionResults += Math.ceil(num / divisor);
            if (sumOfDivisionResults > threshold) {
                thresholdExceeded = false;
                break;
            }
        }
        
        if (thresholdExceeded) {
            return divisor;
        }
    }
    
    return -1;
};

Go 解法

匹配/原始

import "math"

func smallestDivisor(nums []int, threshold int) int {
    maxElement := max(nums)
    
    for divisor := 1; divisor <= maxElement; divisor++ {
        sumOfDivisionResults := 0
        thresholdExceeded := true
        
        for _, num := range nums {
            sumOfDivisionResults += int(math.Ceil(float64(num) / float64(divisor)))
            if sumOfDivisionResults > threshold {
                thresholdExceeded = false
                break
            }
        }
        
        if thresholdExceeded {
            return divisor
        }
    }
    
    return -1
}

func max(nums []int) int {
    maxNum := nums[0]
    for _, num := range nums {
        if num > maxNum {
            maxNum = num
        }
    }
    return maxNum
}

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