###### tags: `Weekly Contest`
# Weekly Contest 412

## [3264. Final Array State After K Multiplication Operations I](https://leetcode.com/problems/final-array-state-after-k-multiplication-operations-i/) (<font color=#00B8A3>Easy</font>)

限制 :

<ul>
<li><code>1 <= nums.length <= 100</code></li>
<li><code>1 <= nums[i] <= 100</code></li>
<li><code>1 <= k <= 10</code></li>
<li><code>1 <= multiplier <= 5</code></li>
</ul>
### Solution

#### 時間複雜度: $O(N^2)$

#### 空間複雜度: $O(1)$

程式碼:
這題要在 list 當中找最小的值，之後對這個值 * multiplier。重複 k 次。
暴力法就是直接重複 k 次就對了。

class Solution {
public:
    vector<int> getFinalState(vector<int>& nums, int k, int multiplier) {
        for(int i=0;i<k;i++) {
            int min_value = INT_MAX, index = 0;
            for(int i=0;i<nums.size();i++) {
                if(min_value > nums[i]) {
                    index = i;
                    min_value = nums[i];
                }
            }
            nums[index] = multiplier* nums[index];
        }
        return nums;
    }
};

## [3265. Count Almost Equal Pairs I](https://leetcode.com/problems/count-almost-equal-pairs-i/description/) (<font color=#FFC011>Medium</font>)

限制 :

<ul>
<li><code>2 <= nums.length <= 100</code></li>
<li><code>1 <= nums[i] <= 10⁶</code></li>
</ul>
### Solution

#### 時間複雜度: $O(N^2)$

#### 空間複雜度: $O(1)$

程式碼:

class Solution {
public:
    bool isEqual(int num1, int num2) {
        string str1, str2;
        stringstream ch1, ch2;
        ch1 << num1;
        ch1 >> str1;
        ch2 << num2;
        ch2 >> str2;
        
        str1 = string(max(str1.size(), str2.size()) - str1.size(), '0') + str1;
        str2 = string(max(str1.size(), str2.size()) - str2.size(), '0') + str2;

        int diff = 0;
        char diff_record[4] = {};
        for (int i = 0; i < str2.size(); i++) {
            if (str1[i] != str2[i]) {
                if (diff >= 2)
                    return false;
                diff_record[diff] = str1[i];
                diff_record[diff + 2] = str2[i];
                diff++;
            }
        }
        if (diff_record[0] != diff_record[3] || diff_record[1] != diff_record[2])
            return false;
        return true;
    }

    int countPairs(vector<int>& nums) {
        int result = 0;
        for (int i = 0; i < nums.size(); i++) {
            for (int j = i + 1; j < nums.size(); j++) {
                result += isEqual(nums[i], nums[j]);
            }
        }
        return result;
    }
};

## [3]()(<font color=#FFC011>Medium</font>)

限制 :

<ul>
<li><code>10⁴</code></li>
</ul>
### Solution

#### 時間複雜度: $O()$

#### 空間複雜度: $O()$

程式碼:

## [4]()(<font color=#FF375F>Hard</font>)

限制 :

<ul>
<li><code>10⁴</code></li>
</ul>
### Solution

#### 時間複雜度: $O()$

#### 空間複雜度: $O()$

程式碼: