# Frequency of smallest character in first sentence less than that of second sentence

Given two array of strings, arr1[] and arr2[], the task is to count the number of string in arr2[] whose frequency of smallest characters is less than frequency of smallest character for each string in arr1[].

Examples:

Input: arr1[] = {“cbd”}, arr2[] = {“zaaaz”}
Output: 1
Explanation:
Frequency of smallest characters in “cbd” is 1 which is less than the frequency of smallest characters in “zaaaz” which is 2.
Therefore the total count is 1 for string “cbd”.

Input: arr1[] = {“yyy”,”zz”}, arr2[] = {“x”,”xx”,”xxx”,”xxxx”}
Output: 1 2
Explanation:
1. frequency of smallest characters in “yyy” is 3 which is less than the frequency of smallest characters in “xxxx” which is 4.
Therefore the total count is 1 for string “yyy”.
2. frequency of smallest characters in “zz” is 2 which is less than the frequency of smallest characters in “xxx” and “xxxx” which is 3 and 4 respectively.
Therefore the total count is 2 for string “zz”.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: This problem can be solved using Greedy Approach. Below are the steps:

1. For each string in the array arr2[] count the frequency of smallest characters and store it in the array (say freq[]).
2. Sort the frequency array freq[].
3. Now for each string in the array arr1[] count the frequency of smallest characters in the string (say X).
4. For each X, find the number of elements in greater than X in freq[] using Binary Search by using the approach discussed in this article.

Below is the implementation of the above approach:

 `// C++ program for the above approach ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to count the frequency of ` `// minimum character ` `int` `countMinFreq(string s) ` `{ ` ` `  `    ``// Sort the string s ` `    ``sort(s.begin(), s.end()); ` ` `  `    ``// Return the count with smallest ` `    ``// character ` `    ``return` `count(s.begin(), s.end(), s); ` `} ` ` `  `// Function to count number of frequency ` `// of smallest character of string arr1[] ` `// is less than the string in arr2[] ` `void` `countLessThan(vector& arr1, ` `                   ``vector& arr2) ` `{ ` `    ``// To store the frequency of smallest ` `    ``// character in each string of arr2 ` `    ``vector<``int``> freq; ` ` `  `    ``// Traverse the arr2[] ` `    ``for` `(string s : arr2) { ` ` `  `        ``// Count the frequency of smallest ` `        ``// character in string s ` `        ``int` `f = countMinFreq(s); ` ` `  `        ``// Append the frequency to freq[] ` `        ``freq.push_back(f); ` `    ``} ` ` `  `    ``// Sort the frequency array ` `    ``sort(freq.begin(), freq.end()); ` ` `  `    ``// Traverse the array arr1[] ` `    ``for` `(string s : arr1) { ` ` `  `        ``// Count the frequency of smallest ` `        ``// character in string s ` `        ``int` `f = countMinFreq(s); ` ` `  `        ``// find the element greater than f ` `        ``auto` `it = upper_bound(freq.begin(), ` `                              ``freq.end(), f); ` ` `  `        ``// Find the count such that ` `        ``// arr1[i] < arr2[j] ` `        ``int` `cnt = freq.size() ` `                  ``- (it - freq.begin()); ` ` `  `        ``// Print the count ` `        ``cout << cnt << ``' '``; ` `    ``} ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` ` `  `    ``vector arr1, arr2; ` `    ``arr1 = { ``"yyy"``, ``"zz"` `}; ` `    ``arr2 = { ``"x"``, ``"xx"``, ``"xxx"``, ``"xxxx"` `}; ` ` `  `    ``// Function Call ` `    ``countLessThan(arr1, arr2); ` `    ``return` `0; ` `} `

 `# Python3 program for the above approach ` `from` `bisect ``import` `bisect_right as upper_bound ` ` `  `# Function to count the frequency  ` `# of minimum character ` `def` `countMinFreq(s): ` ` `  `    ``# Sort the string s ` `    ``s ``=` `sorted``(s) ` ` `  `    ``# Return the count with smallest ` `    ``# character ` `    ``x ``=` `0` `    ``for` `i ``in` `s: ` `        ``if` `i ``=``=` `s[``0``]: ` `            ``x ``+``=` `1` `    ``return` `x ` ` `  `# Function to count number of frequency ` `# of smallest character of string arr1[] ` `# is less than the string in arr2[] ` `def` `countLessThan(arr1, arr2): ` `     `  `    ``# To store the frequency of smallest ` `    ``# character in each string of arr2 ` `    ``freq ``=` `[] ` ` `  `    ``# Traverse the arr2[] ` `    ``for` `s ``in` `arr2: ` ` `  `        ``# Count the frequency of smallest ` `        ``# character in string s ` `        ``f ``=` `countMinFreq(s) ` ` `  `        ``# Append the frequency to freq[] ` `        ``freq.append(f) ` ` `  `    ``# Sort the frequency array ` `    ``feq ``=` `sorted``(freq) ` ` `  `    ``# Traverse the array arr1[] ` `    ``for` `s ``in` `arr1: ` ` `  `        ``# Count the frequency of smallest ` `        ``# character in string s ` `        ``f ``=` `countMinFreq(s); ` ` `  `        ``# find the element greater than f ` `        ``it ``=` `upper_bound(freq,f) ` ` `  `        ``# Find the count such that ` `        ``# arr1[i] < arr2[j] ` `        ``cnt ``=` `len``(freq)``-``it ` ` `  `        ``# Print the count ` `        ``print``(cnt, end ``=` `" "``) ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` ` `  `    ``arr1 ``=` `[``"yyy"``, ``"zz"``] ` `    ``arr2 ``=` `[ ``"x"``, ``"xx"``, ``"xxx"``, ``"xxxx"``] ` ` `  `    ``# Function Call ` `    ``countLessThan(arr1, arr2); ` ` `  `# This code is contributed by Mohit Kumar  `

Output:
```1 2
```

Time Complexity: O(N + M*log M), where N and M is the length of given arrays respectively.

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Improved By : mohit kumar 29

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