tgoop.com/tphunterscpp/198
Create:
Last Update:
Last Update:
🧠 #کد_چالش
حل توسط : چند نفر از دوستان, اما ساده و بهترین کد کد زیر میباشد.
// C++ implementation of the approach🟢لطفا سوالات, نظرات و کد های خود را برای ما کامنت کنید.
#include <bits/stdc++.h>
using namespace std;
// Function to return the count of
// valid indices pairs
int countPairs(string s1, int n1, string s2, int n2)
{
// To store the frequencies of characters
// of string s1 and s2
int freq1[26] = { 0 };
int freq2[26] = { 0 };
// To store the count of valid pairs
int i, count = 0;
// Update the frequencies of
// the characters of string s1
for (i = 0; i < n1; i++)
freq1[s1[i] - 'a']++;
// Update the frequencies of
// the characters of string s2
for (i = 0; i < n2; i++)
freq2[s2[i] - 'a']++;
// Find the count of valid pairs
for (i = 0; i < 26; i++)
count += (min(freq1[i], freq2[i]));
return count;
}
// Driver code
int main()
{
string s1 = "geeksforgeeks", s2 = "platformforgeeks";
int n1 = s1.length(), n2 = s2.length();
cout << countPairs(s1, n1, s2, n2);
return 0;
}
BY Tp hunters cpp

Share with your friend now:
tgoop.com/tphunterscpp/198