# Count common characters in two strings

Given two strings **s1** and **s2** consisting of lowercase English alphabets, the task is to count all the pairs of indices **(i, j)** from the given strings such that **s1[i] = s2[j]** and all the indices are distinct i.e. if **s1[i]** pairs with some **s2[j]** then these two characters will not be paired with any other character.**Example**

Input:s1 = “abcd”, s2 = “aad”Output:2

(s1[0], s2[0]) and (s1[3], s2[2]) are the only valid pairs.

(s1[0], s2[1]) is not includes because s1[0] has already been paired with s2[0]Input:s1 = “geeksforgeeks”, s2 = “platformforgeeks”Output:8

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**Approach:** Count the frequencies of all the characters from both strings. Now, for every character if the frequency of this character in string **s1** is **freq1** and in string **s2** is **freq2** then total valid pairs with this character will be **min(freq1, freq2)**. The sum of this value for all the characters is the required answer.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach` `#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;` `}` |

## Java

`// Java implementation of the approach` `import` `java.util.*;` `class` `GFG` `{` `// Function to return the count of` `// valid indices pairs` `static` `int` `countPairs(String s1, ` `int` `n1,` ` ` `String s2, ` `int` `n2)` `{` ` ` `// To store the frequencies of characters` ` ` `// of string s1 and s2` ` ` `int` `[]freq1 = ` `new` `int` `[` `26` `];` ` ` `int` `[]freq2 = ` `new` `int` `[` `26` `];` ` ` `Arrays.fill(freq1, ` `0` `);` ` ` `Arrays.fill(freq2, ` `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.charAt(i) - ` `'a'` `]++;` ` ` `// Update the frequencies of` ` ` `// the characters of string s2` ` ` `for` `(i = ` `0` `; i < n2; i++)` ` ` `freq2[s2.charAt(i) - ` `'a'` `]++;` ` ` `// Find the count of valid pairs` ` ` `for` `(i = ` `0` `; i < ` `26` `; i++)` ` ` `count += (Math.min(freq1[i], freq2[i]));` ` ` `return` `count;` `}` `// Driver code` `public` `static` `void` `main(String args[])` `{` ` ` `String s1 = ` `"geeksforgeeks"` `, s2 = ` `"platformforgeeks"` `;` ` ` `int` `n1 = s1.length(), n2 = s2.length();` ` ` `System.out.println(countPairs(s1, n1, s2, n2));` `}` `}` `// This code is contributed by` `// Surendra_Gangwar` |

## Python3

`# Python3 implementation of the approach` `# Function to return the count of` `# valid indices pairs` `def` `countPairs(s1, n1, s2, n2) :` ` ` `# To store the frequencies of characters` ` ` `# of string s1 and s2` ` ` `freq1 ` `=` `[` `0` `] ` `*` `26` `;` ` ` `freq2 ` `=` `[` `0` `] ` `*` `26` `;` ` ` `# To store the count of valid pairs` ` ` `count ` `=` `0` `;` ` ` `# Update the frequencies of` ` ` `# the characters of string s1` ` ` `for` `i ` `in` `range` `(n1) :` ` ` `freq1[` `ord` `(s1[i]) ` `-` `ord` `(` `'a'` `)] ` `+` `=` `1` `;` ` ` `# Update the frequencies of` ` ` `# the characters of string s2` ` ` `for` `i ` `in` `range` `(n2) :` ` ` `freq2[` `ord` `(s2[i]) ` `-` `ord` `(` `'a'` `)] ` `+` `=` `1` `;` ` ` `# Find the count of valid pairs` ` ` `for` `i ` `in` `range` `(` `26` `) :` ` ` `count ` `+` `=` `min` `(freq1[i], freq2[i]);` ` ` `return` `count;` `# Driver code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `s1 ` `=` `"geeksforgeeks"` `;` ` ` `s2 ` `=` `"platformforgeeks"` `;` ` ` ` ` `n1 ` `=` `len` `(s1) ;` ` ` `n2 ` `=` `len` `(s2);` ` ` ` ` `print` `(countPairs(s1, n1, s2, n2));` `# This code is contributed by Ryuga` |

## C#

`// C# implementation of the approach` `using` `System;` `class` `GFG` `{` `// Function to return the count of` `// valid indices pairs` `static` `int` `countPairs(` `string` `s1, ` `int` `n1,` ` ` `string` `s2, ` `int` `n2)` `{` ` ` `// To store the frequencies of` ` ` `// characters of string s1 and s2` ` ` `int` `[]freq1 = ` `new` `int` `[26];` ` ` `int` `[]freq2 = ` `new` `int` `[26];` ` ` `Array.Fill(freq1, 0);` ` ` `Array.Fill(freq2, 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 += (Math.Min(freq1[i],` ` ` `freq2[i]));` ` ` `return` `count;` `}` `// Driver code` `public` `static` `void` `Main()` `{` ` ` `string` `s1 = ` `"geeksforgeeks"` `,` ` ` `s2 = ` `"platformforgeeks"` `;` ` ` `int` `n1 = s1.Length, n2 = s2.Length;` ` ` `Console.WriteLine(countPairs(s1, n1, s2, n2));` `}` `}` `// This code is contributed by` `// Akanksha Rai` |

## Javascript

`<script>` ` ` `// JavaScript implementation of the approach` ` ` ` ` `// Function to return the count of` ` ` `// valid indices pairs` ` ` `function` `countPairs(s1, n1, s2, n2)` ` ` `{` ` ` `// To store the frequencies of` ` ` `// characters of string s1 and s2` ` ` `let freq1 = ` `new` `Array(26);` ` ` `let freq2 = ` `new` `Array(26);` ` ` `freq1.fill(0);` ` ` `freq2.fill(0);` ` ` `// To store the count of valid pairs` ` ` `let i, count = 0;` ` ` `// Update the frequencies of` ` ` `// the characters of string s1` ` ` `for` `(i = 0; i < n1; i++)` ` ` `freq1[s1[i].charCodeAt() - ` `'a'` `.charCodeAt()]++;` ` ` `// Update the frequencies of` ` ` `// the characters of string s2` ` ` `for` `(i = 0; i < n2; i++)` ` ` `freq2[s2[i].charCodeAt() - ` `'a'` `.charCodeAt()]++;` ` ` `// Find the count of valid pairs` ` ` `for` `(i = 0; i < 26; i++)` ` ` `count += (Math.min(freq1[i], freq2[i]));` ` ` `return` `count;` ` ` `}` ` ` ` ` `let s1 = ` `"geeksforgeeks"` `, s2 = ` `"platformforgeeks"` `;` ` ` `let n1 = s1.length, n2 = s2.length;` ` ` `document.write(countPairs(s1, n1, s2, n2));` ` ` `</script>` |

**Output:**

8

**Time Complexity:** O(max(n1, n2))**Auxiliary Space: **O(1)