Minimize the length of string by removing occurrence of only one character

Given a string str containing only lowercase alphabets, the task is to minimize the length of the string after performing the following operation:
- Delete all occurrences of any one character from this string.
Examples:
Input: str = “abccderccwq”
Output: 7
character ‘c’ will be deleted since it has maximum occurrence.Input: str = “dddded”
Output: 1
Explanation: character ‘d’ will be deleted
Approach:
- Maintain the frequency of each and every character in an array.
- Find the character with the maximum frequency.
- Finally, subtract the length of the original string with the frequency of that character to get the desired answer.
Implementation:
C++
// C++ program to minimize the length of string by // removing occurrence of only one character#include <bits/stdc++.h>using namespace std;// Function to find the minimum lengthint minimumLength(string s){ int maxOcc = 0, n = s.length(); int arr[26] = {0}; // Count the frequency of each alphabet for (int i = 0; i < n; i++) arr[s[i] - 'a']++; // Find the alphabets with maximum frequency for (int i = 0; i < 26; i++) if (arr[i] > maxOcc) maxOcc = arr[i]; // Subtract the frequency of character // from length of string return (n - maxOcc);}// Driver Codeint main(){ string str = "afddewqd"; cout << minimumLength(str); return 0;} |
Java
// Java program to minimize the length of string // by removing occurrence of only one characterimport java.io.*;class GFG { // Function to find the minimum length static int minimumLength(String s) { int maxOcc = 0, n = s.length(); int arr[] = new int[26]; // Count the frequency of each alphabet for (int i = 0; i < n; i++) arr[s.charAt(i) - 'a']++; // Find the alphabets with maximum frequency for (int i = 0; i < 26; i++) if (arr[i] > maxOcc) maxOcc = arr[i]; // Subtract the frequency of character // from length of string return (n - maxOcc); } // Driver Code public static void main (String[] args) { String str = "afddewqd"; System.out.println( minimumLength(str)); }}//This code is contributed by chandan_jnu... |
Python 3
# Python 3 program to minimize the length of string # by removing occurrence of only one character# Function to find the minimum lengthdef minimumLength(s) : maxOcc = 0 n = len(s) arr = [0]*26 # Count the frequency of each alphabet for i in range(n) : arr[ord(s[i]) -ord('a')] += 1 # Find the alphabets with maximum frequency for i in range(26) : if arr[i] > maxOcc : maxOcc = arr[i] # Subtract the frequency of character # from length of string return n - maxOcc# Driver Codeif __name__ == "__main__" : str = "afddewqd" print(minimumLength(str))# This code is contributed by ANKITRAI1 |
C#
// C# program to minimize the// length of string by removing// occurrence of only one characterusing System;class GFG { // Function to find the minimum length static int minimumLength(String s) { int maxOcc = 0, n = s.Length; int []arr = new int[26]; // Count the frequency of each alphabet for (int i = 0; i < n; i++) arr[s[i] - 'a']++; // Find the alphabets with // maximum frequency for (int i = 0; i < 26; i++) if (arr[i] > maxOcc) maxOcc = arr[i]; // Subtract the frequency of character // from length of string return (n - maxOcc); } // Driver Code public static void Main (String[] args) { String str = "afddewqd"; Console.WriteLine( minimumLength(str)); }}// This code is contributed by// PrinciRaj1992 |
Javascript
<script>// Javascript program to minimize the length of string by // removing occurrence of only one character// Function to find the minimum lengthfunction minimumLength(s){ var maxOcc = 0, n = s.length; var arr = Array(26).fill(0); // Count the frequency of each alphabet for (var i = 0; i < n; i++) arr[s[i].charCodeAt(0) - 'a'.charCodeAt(0)]++; // Find the alphabets with maximum frequency for (var i = 0; i < 26; i++) if (arr[i] > maxOcc) maxOcc = arr[i]; // Subtract the frequency of character // from length of string return (n - maxOcc);}// Driver Codevar str = "afddewqd";document.write( minimumLength(str));</script> |
Output
5
Time complexity: O(n) where n is the length of string
Auxiliary space: O(1) since constant space is being used
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