如何针对输入的大量字符串优化我的代码

Question

I passed 4 tests on my university compiler and the problem is the 5th one. Time limit is 1second for each test. How can I optimize this code, maybe there is a better option for sorting if I compare strings? My code:

#include <iostream>
#include <string.h>

using namespace std;

void qsort(string &tab, int min, int max)
{
    if(min<max)
    {
        int min_min = min;
        for(int i=min+1;i<=max;i++)
        {
            if(tab[i]<tab[min])
            {
                swap(tab[++min_min],tab[i]);
            }
        }
        swap(tab[min],tab[min_min]);
        qsort(tab,min,min_min-1);
        qsort(tab,min_min+1,max);

    }
}
bool sprawdz(string tab,string tab2)
{
    for(int i=0;i<tab.length();i++)
    {
        if(tab[i]!=tab2[i])
        {
            return false;
            break;
        }
    }
    return true;
}

int main()
{
    string tablica1, tablica2;
    int ile;

    scanf("%d",&ile);
    for(int i=0;i<ile;i++)
    {
        cin>>tablica1>>tablica2;
        qsort(tablica1,0,tablica1.length()-1);
        qsort(tablica2,0,tablica2.length()-1);

        if(tablica1==tablica2)
        {
            printf("TAK\n");
        }
        else
        {
            printf("NIE\n");
        }


    }

    return 0;
}

Only information it thrown is min = 25177 max = 25978 so these numbers are quite large. Any ideas? Task is to check if words are anagrams.

Answer 1

maybe there is a better option for sorting if I compare strings?

My favourite tip: The optimal way to do something is to not do it ;)

You do not have to sort the strings to check if they are anagrams. Order does not matter for anagrams, so why sort them?

Sorting is typically O(n log n) , while simply counting the frequency of characters is O(n) . To count characters you can use a std::unordered_map , or if that is not allowed use an array of counters. Traverse the strings to count occurences of each character, then compare the arrays of counters.

PS: You should also check if the size of the strings is the same before applying any further logic.