时间复杂度较低的办法

#include<bits/stdc++.h>
using namespace std;
int main(){
    int n,tmp,count=114514;
    cin >> n;
    string s;
    int nums[10][10]={0};
    for(int i = 0;i<n;i++){
        cin >> s;
        for(int j = 0;j<10;j++){
            nums[s[j]-48][j] ++;
        }
    }
    for(int i = 0;i<10;i++){
        tmp = 0;
        for(int j = 0;j<10;j++){
            if(tmp<(nums[i][j]-1)*10+j){
                tmp =  (nums[i][j]-1)*10+j;
            }
        }
        if(tmp<count){
            count = tmp;
        }
        //cout<<tmp<<endl;
    }
    cout<<count;
}