Leetcode 1208. Try to make the strings equal as much as possible (finally solved, good night)

Give you two strings of the same length ,s and t.

take s No i Four characters change to t No i Two characters need to be |s[i] - t[i]| The cost of （ The cost may be 0）, That's two characters ASCII The absolute value of the difference between code values .

The maximum budget for changing strings is maxCost. When converting strings , The total cost should be less than or equal to the budget , This also means that string conversion may be incomplete .

If you can put s The substring of is converted to it in t The substring corresponding to , Then returns the maximum length that can be converted .

If s There are no substrings that can be converted to t The substring corresponding to , Then return to 0.

Example 1：

 Input ：s = "abcd", t = "bcdf", maxCost = 3
 Output ：3
 explain ：s  Medium  "abc"  It can be  "bcd". The cost is  3, So the maximum length is  3.

Example 2：

 Input ：s = "abcd", t = "cdef", maxCost = 3
 Output ：1
 explain ：s  If you want any character in to become  t  The corresponding characters in , The cost is  2. therefore , Maximum length is  1.

Example 3：

 Input ：s = "abcd", t = "acde", maxCost = 0
 Output ：1
 explain ：a -> a, cost = 0, The string has not changed , So the maximum length is  1.
 

Tips ：

• 1 <= s.length, t.length <= 10^5
• 0 <= maxCost <= 10^6
• s and t It's all lowercase letters .

Code:

class Solution {
    
public:
    int equalSubstring(string s, string t, int maxCost) {
    
        
        vector<int>vec;
        for(int i=0;i<s.length();i++)
        {
    
            vec.push_back(abs(s[i]-t[i]));
            
        }
        
        int cnt=0;
        int sum=0;
        int res=0;
        
        
        bool flag=false;
        queue<int>q;
        for(int i=0;i<vec.size();i++)
        {
    
            
            sum+=vec[i];
            q.push(vec[i]);
            if(sum<=maxCost)
            {
    
                cnt=max((int)q.size(),cnt);
            }
            else
            {
    
                
                
                sum-=q.front();
                q.pop();
                cnt=max((int)q.size(),cnt);
                
                
            }
        }
        
        
        
        
        // cout<<cnt<<"+"<<endl;
        return cnt;
        
    }
};