将字符存储数组转换为哈希图时面临错误<int, char>然后堆。 然后打印其频率的字符和数字

Question

In this question i assume that. arr = {'a', 'a', 'b', 'c', 'c'}. then unordered_map will convert this into a-> 3, b-> 1, c-> 2. then i append this to max heap 3-> a, 2-> c, 1-> which i supposed to be this but i want to print the result as abc but i am facing error. Need help!!

code is in the link below. I dont know why my code is not accepting. https://hastebin.com/mamudogizo.cpp

#include <bits/stdc++.h>
using namespace std;
int main()
{
    unordered_map<char, int> mp;
    priority_queue<pair<int, char>> mheap;

    char arr[] = {'a','a','a','b','c','c'};

    for (int i = 0; i < 6; ++i)
    {
        mp[arr[i]]++;
    }

    for (auto i = mp.begin(); i != mp.end(); i++)
    {
        mheap.push({i->first, i->second});
    }

    pair<int, char> curr = mheap.top();
    cout<< curr.second;

    return 0;
}

Answer 1

If you want to count how many times each character occurred then you can use std::string instead of using " built in arrays and priority_queue<> " as shown below . That is there is no need for priority_queue<> . The program prints the output in the order as you desire.

Version 1 : Using std::string

#include <iostream>
#include <map>
int main() {
    std::string inputString = "anoopsinghrana";
    //std::cout<<"Enter a string: ";
    //std::cin>> inputString;
    //this map maps the char to their respective count
    std::map<char, int> charCount;
    
    for(char &c: inputString)
    {
        charCount[c]++;
    }
    
    std::size_t i = 0;
    //just go through the inputString instead of map
    for(char &c: inputString)
    {
        std::size_t index = inputString.find(c);
        if(index != inputString.npos && (index == i)){
         std::cout << c <<"-" << charCount.at(c)<<std::endl;
         
        }
        ++i;
    }
    return 0;
}

The output of the above program is as follows:

a-3
n-3
o-2
p-1
s-1
i-1
g-1
h-1
r-1

The above(version 1) program counts small and capital letters separately. So for example the character A and the character a are different.

If you still want to use built in array instead of std::string then you can use the following program. Note that still you don't need priority_queue .

Version 2 : Using built in array

#include <iostream>
#include <map>

int myFind(char arr[], int len, int seek)
{
    for (int i = 0; i < len; ++i)
    {
        if (arr[i] == seek) return i;
    }
    return -1;
}
int main() {
    char arr[] = {'a','n','o','o','p','s','i','n','g','h','r','a','n','a'};
    
    std::map<char, int> charCount;
    
    for(int i = 0; i < sizeof(arr) / sizeof(char); ++i)
    {
        charCount[arr[i]]++;
    }
    
    int k = 0;
    //just go through the inputString instead of map
    for(int i = 0; i < sizeof(arr) / sizeof(char); ++i)
    {
        //std::size_t index = inputString.find(arr[i]);
        int index = myFind(arr,sizeof(arr) / sizeof(char),arr[i]);
        if(index != -1 && (index == k)){
         std::cout << arr[i] <<"-" << charCount.at(arr[i])<<std::endl;
         
        }
        ++k;
    }
    return 0;
}

The ouptut of the above (version 2) program is:

a-3
n-3
o-2
p-1
s-1
i-1
g-1
h-1
r-1

Answer 2

Your map maps from the character to its counter, and the queue is ordered by the counter, so when you push an item from the map to the queue, you need to replace first and second :

for (auto i = mp.begin(); i != mp.end(); i++)
{
    mheap.push({ i->second, i->first });
}

Also, note that you're only printing the first element from the queue. If you want to print them all (in order), you'd need to iterate over it. Eg:

while (!mheap.empty())
{
    pair<int, char> curr = mheap.top();
    cout << curr.second << endl;
    mheap.pop();
}