Can someone help me why my code is giving me wrong output ?

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Автор vishudon, история, 6 лет назад, По-английски

I was trying to solve a problem, but I encountered a weird error. I don't know why my code is giving me wrong output. Can someone help me what is wrong in my code ? Thanks in advance.

Code

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

int main()
{
   int n,m;
   cin>>n>>m;
   
   map<string,int> mp1,mp2;
   
   for(int i=0;i<n;i++)
   {
      string s;
      cin>>s;
      mp1[s]++;
   }
   
   vector<string> common;
   for(int i=0;i<m;i++)
   {
      string s;
      cin>>s;
      if(mp1[s]!=0)                // line 1
         common.emplace_back(s);      // line 2
      mp2[s]++;
   }
   
   int len1 = (int)mp1.size();
   int len2 = (int)mp2.size();
   
   cout<<"len1 = "<<len1<<" , len2 = "<<len2<<"\n";
   
}

If I am giving this input :

1 2
abc
cdf
edf

then my code is giving me this output : len1 = 2 , len2 = 2 ( while len1 must be 1 )

Note : If I am removing that common string checking part — these two lines ( line1 and line2 ) from 2nd for loop, then it is giving me correct output as len1 = 1 , len2 = 2

Please help me.

c++

vishudon
6 лет назад
6

Комментарии (5)

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scopeInfinity

6 лет назад, скрыть # |

Seems like mp1[s]!=0 is creating new keys in the map.

https://stackoverflow.com/questions/10124679/what-happens-if-i-read-a-maps-value-where-the-key-does-not-exist

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vishudon

6 лет назад, скрыть # ^ |

Do you want to say something about that note which i have specified here ?

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Ryoumen_Sukuna

6 лет назад, скрыть # ^ |

Yeah as scopeInfinity said mp1[s]!=0 adds another entry in the mp1 if s is not present already. Hence your len1 is 2 (For the given case).

If you want to check if a string s is present map s or not then use mp1.find(s)!=mp1.end() Because map::find returns an iterator to the element if present otherwise its equal to mp1.end()

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