# Maximum count of Strings to be selected such that there is a majority character

• Difficulty Level : Hard
• Last Updated : 31 Aug, 2021

Given an array A[]of N strings of lowercase characters, the task is to find maximum number of strings, such that one character has majority i.e. the occurrence of one character in all of the strings is greater than all the other characters combined.

Examples:

Input: A[] = {“aba”, “abcde”, “aba”}
Output: 2
Explanation: On choosing {“aba”, “aba”}, the occurrence of character ‘a’ is 4 and the occurrence of rest of the characters is 2, which is less than 4. So, a maximum of 2 strings can be chosen.

Input: A[] = {“bzc”, “zzzdz”, “e”}
Output: 3
Explanation:  All the strings can be chosen, where ‘z’ has the majority.

Approach: There is a Greedy Solution to this problem. The idea is that, consider all the lowercase characters from ‘a’ to ‘z’ and for every character,  find the maximum number of strings that can be chosen, so that the occurrence of that character is greater than all the other characters combined. Maximum of them will be the answer. Now the main problem is how to find the maximum number of strings such that one particular character has majority, to solve this use greedy algorithm. The idea is that, to find the maximum number of strings for a  particular character ‘ch’, try to choose the strings in which occurrence of ‘ch’ is greatest compared to the occurrence of all the other characters, i.e. the strings in which (occurrence of ‘ch’ – occurrence of all other character) is maximum so that it will be possible to add more strings in the future.

Follow the steps below to solve the problem:

• Define a function CalDif(string s, char ch) and perform the following tasks:
• Initialize the variables chcount as 0 to store the count of that character and othercount as 0 to store the count of other characters.
• Traverse the string s using the variable x and perform the following tasks:
• If character at current position is ch, then increase the value of chcount by 1 else increase the value of othercount by 1.
• Return the difference between chcount and othercount.
• Initialize the variable ans and assign 0, it will store the final answer.
• Iterate over the lower-case character range [a, z] using the variable i and perform the following steps:
• Initialize a vector arr[] of length N, where arr[i] will store (occurrence of ch – occurrence of all other character) for i-th string.
• Run a loop from i=0 to N-1
• For the ith string compute (occurrence of ch – occurrence of all other character) using the function CalDiff(A[i], ch) and assign it to arr[i].
• Sort arr[] in decreasing order.
• Initialize two variables temp and count and assign 0 to them.
• Traverse the array arr[] and do temp += arr[i] and count++, where i starts from 0, until temp <= 0.
• Set the value of ans max of ans and count.
• After performing the above steps, print the value of ans as the answer.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to calculate (occurrence of ch -``// occurrence of all other characters combined)``int` `CalDiff(string s, ``char` `ch)``{``    ``// Initialize ch count as 0``    ``int` `chcount = 0;` `    ``// Initialize all other char count as 0``    ``int` `othercount = 0;` `    ``// Traversing the string``    ``for` `(``auto` `x : s) {``        ``// Current character is ch``        ``if` `(x == ch)``            ``chcount++;``        ``// Current character is not ch``        ``else``            ``othercount++;``    ``}` `    ``// Return the final result``    ``return` `(chcount - othercount);``}` `// Function to calculate maximum number of string``// with one character as majority``int` `MaximumNumberOfString(string A[], ``int` `N)``{``    ``// Initializing ans with 0, to store final answer``    ``int` `ans = 0;` `    ``// For every character from 'a' to 'z' run loop``    ``for` `(``char` `ch = ``'a'``; ch <= ``'z'``; ch++) {` `        ``// Initialize arr to store character count``        ``// difference``        ``vector<``int``> arr(N);` `        ``// Traverse every string``        ``for` `(``int` `i = 0; i < N; i++) {` `            ``// Calculate the required value by``            ``// function call and assign it to arr[i]``            ``arr[i] = CalDiff(A[i], ch);``        ``}` `        ``// Sort arr[] in decreasing order``        ``sort(arr.begin(), arr.end(), greater<``int``>());` `        ``// Initialize temp and count as 0``        ``int` `temp = 0, count = 0;` `        ``// Adding the first arr[] element to temp``        ``temp += arr[0];` `        ``// Maintaining j as index``        ``int` `j = 1;` `        ``// Run loop until temp <= 0``        ``while` `(temp > 0) {``            ``// Increasing count``            ``count++;` `            ``// Adding temp with next arr[] element``            ``if` `(j != N)``                ``temp += arr[j++];``            ``else``                ``break``;``        ``}` `        ``// Set ans as max of ans and count``        ``ans = max(ans, count);``    ``}` `    ``// Returning the final result``    ``return` `ans;``}` `// Driver Code``int` `main()``{``    ``// Input``    ``string A[] = { ``"aba"``, ``"abcde"``, ``"aba"` `};``    ``int` `N = ``sizeof``(A) / ``sizeof``(A[0]);` `    ``// Function call``    ``cout << MaximumNumberOfString(A, N);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;` `class` `GFG``{` `// Function to calculate (occurrence of ch -``// occurrence of all other characters combined)``static` `int` `CalDiff(String s, ``char` `ch)``{``  ` `    ``// Initialize ch count as 0``    ``int` `chcount = ``0``;` `    ``// Initialize all other char count as 0``    ``int` `othercount = ``0``;` `    ``// Traversing the String``    ``for` `(``int` `x : s.toCharArray())``    ``{``      ` `        ``// Current character is ch``        ``if` `(x == ch)``            ``chcount++;``      ` `        ``// Current character is not ch``        ``else``            ``othercount++;``    ``}` `    ``// Return the final result``    ``return` `(chcount - othercount);``}` `// Function to calculate maximum number of String``// with one character as majority``static` `int` `MaximumNumberOfString(String A[], ``int` `N)``{``  ` `    ``// Initializing ans with 0, to store final answer``    ``int` `ans = ``0``;` `    ``// For every character from 'a' to 'z' run loop``    ``for` `(``char` `ch = ``'a'``; ch <= ``'z'``; ch++) {` `        ``// Initialize arr to store character count``        ``// difference``       ``int` `[]arr = ``new` `int``[N];` `        ``// Traverse every String``        ``for` `(``int` `i = ``0``; i < N; i++) {` `            ``// Calculate the required value by``            ``// function call and assign it to arr[i]``            ``arr[i] = CalDiff(A[i], ch);``        ``}` `        ``// Sort arr[] in decreasing order``        ``Arrays.sort(arr);``        ``arr = reverse(arr);` `        ``// Initialize temp and count as 0``        ``int` `temp = ``0``, count = ``0``;` `        ``// Adding the first arr[] element to temp``        ``temp += arr[``0``];` `        ``// Maintaining j as index``        ``int` `j = ``1``;` `        ``// Run loop until temp <= 0``        ``while` `(temp > ``0``)``        ``{``          ` `            ``// Increasing count``            ``count++;` `            ``// Adding temp with next arr[] element``            ``if` `(j != N)``                ``temp += arr[j++];``            ``else``                ``break``;``        ``}` `        ``// Set ans as max of ans and count``        ``ans = Math.max(ans, count);``    ``}` `    ``// Returning the final result``    ``return` `ans;``}``static` `int``[] reverse(``int` `a[]) {``    ``int` `i, n = a.length, t;``    ``for` `(i = ``0``; i < n / ``2``; i++) {``        ``t = a[i];``        ``a[i] = a[n - i - ``1``];``        ``a[n - i - ``1``] = t;``    ``}``    ``return` `a;``}``  ` `// Driver Code``public` `static` `void` `main(String[] args)``{``  ` `    ``// Input``    ``String A[] = { ``"aba"``, ``"abcde"``, ``"aba"` `};``    ``int` `N = A.length;` `    ``// Function call``    ``System.out.print(MaximumNumberOfString(A, N));` `}``}` `// This code is contributed by Amit Katiyar`

## Python3

 `# Python 3 program for the above approach` `# Function to calculate (occurrence of ch -``# occurrence of all other characters combined)``def` `CalDiff(s, ch):``    ``# Initialize ch count as 0``    ``chcount ``=` `0` `    ``# Initialize all other char count as 0``    ``othercount ``=` `0` `    ``# Traversing the string``    ``for` `x ``in` `s:``        ``# Current character is ch``        ``if` `(x ``=``=` `ch):``            ``chcount ``+``=` `1``        ``# Current character is not ch``        ``else``:``            ``othercount ``+``=` `1` `    ``# Return the final result``    ``return` `(chcount ``-` `othercount)` `# Function to calculate maximum number of string``# with one character as majority``def` `MaximumNumberOfString(A, N):``    ``# Initializing ans with 0, to store final answer``    ``ans ``=` `0` `    ``# For every character from 'a' to 'z' run loop``    ``for` `ch ``in` `range``(``97``,``123``,``1``):``        ``# Initialize arr to store character count``        ``# difference``        ``arr ``=` `[``0` `for` `i ``in` `range``(N)]` `        ``# Traverse every string``        ``for` `i ``in` `range``(N):``            ``# Calculate the required value by``            ``# function call and assign it to arr[i]``            ``arr[i] ``=` `CalDiff(A[i], ``chr``(ch))` `        ``# Sort arr[] in decreasing order``        ``arr.sort(reverse ``=` `True``)` `        ``# Initialize temp and count as 0``        ``temp ``=` `0``        ``count ``=` `0` `        ``# Adding the first arr[] element to temp``        ``temp ``+``=` `arr[``0``]` `        ``# Maintaining j as index``        ``j ``=` `1` `        ``# Run loop until temp <= 0``        ``while` `(temp > ``0``):``            ``# Increasing count``            ``count ``+``=` `1` `            ``# Adding temp with next arr[] element``            ``if` `(j !``=` `N):``                ``temp ``+``=` `arr[j]``                ``j ``+``=` `1``            ``else``:``                ``break` `        ``# Set ans as max of ans and count``        ``ans ``=` `max``(ans, count)` `    ``# Returning the final result``    ``return` `ans` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``# Input``    ``A ``=` `[``"aba"``, ``"abcde"``, ``"aba"``]``    ``N ``=` `len``(A)` `    ``# Function call``    ``print``(MaximumNumberOfString(A, N))``    ` `    ``# This code is contributed by SURENDRA_GANGWAR.`

## C#

 `// C# program for the above approach``using` `System;` `class` `GFG{``    ` `// Function to calculate (occurrence of ch -``// occurrence of all other characters combined)``static` `int` `CalDiff(``string` `s, ``char` `ch)``{``    ` `    ``// Initialize ch count as 0``    ``int` `chcount = 0;` `    ``// Initialize all other char count as 0``    ``int` `othercount = 0;` `    ``// Traversing the String``    ``foreach``(``int` `x ``in` `s.ToCharArray())``    ``{``        ` `        ``// Current character is ch``        ``if` `(x == ch)``            ``chcount++;``      ` `        ``// Current character is not ch``        ``else``            ``othercount++;``    ``}` `    ``// Return the final result``    ``return``(chcount - othercount);``}` `// Function to calculate maximum number of String``// with one character as majority``static` `int` `MaximumNumberOfString(``string``[] A, ``int` `N)``{``    ` `    ``// Initializing ans with 0, to store final answer``    ``int` `ans = 0;` `    ``// For every character from 'a' to 'z' run loop``    ``for``(``char` `ch = ``'a'``; ch <= ``'z'``; ch++)``    ``{``        ` `        ``// Initialize arr to store character count``        ``// difference``        ``int` `[]arr = ``new` `int``[N];``        ` `        ``// Traverse every String``        ``for``(``int` `i = 0; i < N; i++)``        ``{``            ` `            ``// Calculate the required value by``            ``// function call and assign it to arr[i]``            ``arr[i] = CalDiff(A[i], ch);``        ``}` `        ``// Sort arr[] in decreasing order``        ``Array.Sort(arr);``        ``arr = reverse(arr);` `        ``// Initialize temp and count as 0``        ``int` `temp = 0, count = 0;` `        ``// Adding the first arr[] element to temp``        ``temp += arr[0];` `        ``// Maintaining j as index``        ``int` `j = 1;` `        ``// Run loop until temp <= 0``        ``while` `(temp > 0)``        ``{``            ` `            ``// Increasing count``            ``count++;` `            ``// Adding temp with next arr[] element``            ``if` `(j != N)``                ``temp += arr[j++];``            ``else``                ``break``;``        ``}` `        ``// Set ans as max of ans and count``        ``ans = Math.Max(ans, count);``    ``}` `    ``// Returning the final result``    ``return` `ans;``}` `static` `int``[] reverse(``int``[] a)``{``    ``int` `i, n = a.Length, t;``    ``for``(i = 0; i < n / 2; i++)``    ``{``        ``t = a[i];``        ``a[i] = a[n - i - 1];``        ``a[n - i - 1] = t;``    ``}``    ``return` `a;``}` `// Driver Code``public` `static` `void` `Main(String []args)``{``    ` `    ``// Input``    ``string``[] A = { ``"aba"``, ``"abcde"``, ``"aba"` `};``    ``int` `N = A.Length;``    ` `    ``// Function call``    ``Console.WriteLine(MaximumNumberOfString(A, N));``}``}` `// This code is contributed by sanjoy_62`

## Javascript

 ``
Output
`2`

Time Complexity: O(N * |s|), where |s| is maximum string length.
Auxiliary Space: O(N)

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