Given a string str of lowercase characters, the task is to find the minimum number of characters that need to add to the string in order to make it balanced. A string is said to be balanced if and only if the number of occurrences of each of the characters is equal.
Examples:
Input: str = “neveropen”
Output: 15
Add 2 ‘g’, 2 ‘k’, 2 ‘s’, 3 ‘f’, 3 ‘o’ and 3 ‘r’.Input: str = “abcd”
Output: 0
The string is already balanced.
Approach: In order to minimize the additions required, every character’s frequency must be made equal to the frequency of the element occurring most frequently. So first, create a frequency array and find the frequency of all the characters of the given string. Now, the required answer will be the sum of the absolute differences in the frequency of every character with the maximum frequency from the frequency array.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;#define MAX 26// Function to return the minimum additions// required to balance the given stringint minimumAddition(string str, int len){ // To store the frequency of // the characters of str int freq[MAX] = { 0 }; // Update the frequency of the characters for (int i = 0; i < len; i++) { freq[str[i] - 'a']++; } // To store the maximum frequency from the array int maxFreq = *max_element(freq, freq + MAX); // To store the minimum additions required int minAddition = 0; for (int i = 0; i < MAX; i++) { // Every character's frequency must be // equal to the frequency of the most // frequently occurring character if (freq[i] > 0) { minAddition += abs(maxFreq - freq[i]); } } return minAddition;}// Driver codeint main(){ string str = "neveropen"; int len = str.length(); cout << minimumAddition(str, len); return 0;} |
Java
// Java implementation of the approach class GFG { final static int MAX = 26; static int max_element(int freq[]) { int max_ele = freq[0]; for(int i = 0; i < MAX; i++) { if(max_ele < freq[i]) max_ele = freq[i]; } return max_ele; } // Function to return the minimum additions // required to balance the given string static int minimumAddition(String str, int len) { // To store the frequency of // the characters of str int freq[] = new int[MAX]; // Update the frequency of the characters for (int i = 0; i < len; i++) { freq[str.charAt(i) - 'a']++; } // To store the maximum frequency from the array int maxFreq = max_element(freq); // To store the minimum additions required int minAddition = 0; for (int i = 0; i < MAX; i++) { // Every character's frequency must be // equal to the frequency of the most // frequently occurring character if (freq[i] > 0) { minAddition += Math.abs(maxFreq - freq[i]); } } return minAddition; } // Driver code public static void main (String[] args) { String str = "neveropen"; int len = str.length(); System.out.println(minimumAddition(str, len)); } }// This code is contributed by AnkitRai01 |
Python3
# Python3 implementation of the approachMAX = 26# Function to return the minimum additions# required to balance the given stringdef minimumAddition(str1, Len): # To store the frequency of # the characters of str1 freq = [0 for i in range(MAX)] # Update the frequency of the characters for i in range(Len): freq[ord(str1[i]) - ord('a')] += 1 # To store the maximum frequency from the array maxFreq = max(freq) # To store the minimum additions required minAddition = 0 for i in range(MAX): # Every character's frequency must be # equal to the frequency of the most # frequently occurring character if (freq[i] > 0): minAddition += abs(maxFreq - freq[i]) return minAddition# Driver codestr1 = "neveropen"Len = len(str1)print(minimumAddition(str1, Len))# This code is contributed Mohit Kumar |
C#
// C# implementation of the approachusing System; class GFG { static int MAX = 26; static int max_element(int []freq) { int max_ele = freq[0]; for(int i = 0; i < MAX; i++) { if(max_ele < freq[i]) max_ele = freq[i]; } return max_ele; } // Function to return the minimum additions // required to balance the given string static int minimumAddition(String str, int len) { // To store the frequency of // the characters of str int []freq = new int[MAX]; // Update the frequency of the characters for (int i = 0; i < len; i++) { freq[str[i] - 'a']++; } // To store the maximum frequency from the array int maxFreq = max_element(freq); // To store the minimum additions required int minAddition = 0; for (int i = 0; i < MAX; i++) { // Every character's frequency must be // equal to the frequency of the most // frequently occurring character if (freq[i] > 0) { minAddition += Math.Abs(maxFreq - freq[i]); } } return minAddition; } // Driver code public static void Main (String[] args) { String str = "neveropen"; int len = str.Length; Console.WriteLine(minimumAddition(str, len)); } }// This code is contributed by 29AjayKumar |
Javascript
<script> // Javascript implementation of the approach let MAX = 26; function max_element(freq) { let max_ele = freq[0]; for(let i = 0; i < MAX; i++) { if(max_ele < freq[i]) max_ele = freq[i]; } return max_ele; } // Function to return the minimum additions // required to balance the given string function minimumAddition(str, len) { // To store the frequency of // the characters of str let freq = new Array(MAX); freq.fill(0); // Update the frequency of the characters for (let i = 0; i < len; i++) { freq[str[i].charCodeAt() - 'a'.charCodeAt()]++; } // To store the maximum frequency from the array let maxFreq = max_element(freq); // To store the minimum additions required let minAddition = 0; for (let i = 0; i < MAX; i++) { // Every character's frequency must be // equal to the frequency of the most // frequently occurring character if (freq[i] > 0) { minAddition += Math.abs(maxFreq - freq[i]); } } return minAddition; } let str = "neveropen"; let len = str.length; document.write(minimumAddition(str, len)); </script> |
Output
15
Time Complexity: O(n), where n is the length of the given string.
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
