Given a lower alphabetic string “S” of size N, and an integer K; the task is to find the count of characters that will remain ungrouped, after dividing the given string into K groups of distinct characters.
Examples:
Input: S = “stayinghomesaveslife”, K = 1
Output: 6
Explanation:
In the string S the elements whose occurrence is more than one time are ‘e’ -> 3 times, ‘s’ -> 3 times, ‘a’ -> 2 times, ‘i’ -> 2 times and rest all elements occurs 1 time each.
There is only one group to be formed as K = 1 so only one copy of these elements can be present in the group and rest all elements cannot be in the group,
So the elements which are left out of the group are 2-times ‘e’, 2-times ‘s’, 1-time ‘a’ and 1-time ‘i’.
Total elements left out are 6.
Input: S = “stayinghomesaveslife”, K = 2
Output: 2
Explanation:
In the string S the elements whose occurrence is more than one time are ‘e’ -> 3 times, ‘s’ -> 3 times, ‘a’ -> 2 times, ‘i’ -> 2 times and rest all elements occurs 1 time each.
Since 2 groups can be formed, one group contains 1 copy of all the elements. The second group will contain 1 copy of the extra elements i.e. ‘e’, ‘s’, ‘a’ and ‘i’. The elements that will be left out are 1-time ‘e’ and 1-time ‘s’.
Total elements that will be left out are 2.
Approach: The idea is to use frequency-counting.
- Create a Hashing data structure, to store the frequency of characters ‘a’-‘z’.
- Find the initial frequency of each character in the given string and store it in the hashing data structure.
- Since a group can contain only 1 occurrence of a character. Therefore, decrement K from each character’s occurrence in the hashing data structure.
- Now add the remaining frequencies of the characters in the hashing data structure. This will be the required number of characters that will remain ungrouped.
Below is the implementation of the above approach:
C++
// C++ code to implement the above approach#include <bits/stdc++.h>using namespace std;void findUngroupedElement(string s, int k){ int n = s.length(); // create array where // index represents alphabets int b[26]; for (int i = 0; i < 26; i++) b[i] = 0; // fill count of every // alphabet to corresponding // array index for (int i = 0; i < n; i++) { char p = s.at(i); b[p - 'a'] += 1; } int sum = 0; // count for every element // how much is exceeding // from no. of groups then // sum them for (int i = 0; i < 26; i++) { if (b[i] > k) sum += b[i] - k; } // print answer cout << sum << endl;}// Driver codeint main(){ string s = "stayinghomesaveslife"; int k = 1; findUngroupedElement(s, k); return 0;} |
Java
// Java code to implement the above approachimport java.util.*;class GFG{static void findUngroupedElement(String s, int k){ int n = s.length(); // Create array where // index represents alphabets int []b = new int[26]; for(int i = 0; i < 26; i++) b[i] = 0; // Fill count of every // alphabet to corresponding // array index for(int i = 0; i < n; i++) { char p = s.charAt(i); b[p - 'a'] += 1; } int sum = 0; // Count for every element // how much is exceeding // from no. of groups then // sum them for(int i = 0; i < 26; i++) { if (b[i] > k) sum += b[i] - k; } // Print answer System.out.println(sum);}// Driver codepublic static void main(String srgs[]){ String s = "stayinghomesaveslife"; int k = 1; findUngroupedElement(s, k);}}// This code is contributed by ANKITKUMAR34 |
Python3
# Python3 code to implement the above approachdef findUngroupedElement(s, k): n = len(s); # Create array where # index represents alphabets b = [0] * 26 # Fill count of every # alphabet to corresponding # array index for i in range(n): p = s[i] b[ord(p) - ord('a')] += 1 sum = 0; # Count for every element # how much is exceeding # from no. of groups then # sum them for i in range(26): if (b[i] > k): sum += b[i] - k # Print answer print(sum)# Driver codes = "stayinghomesaveslife"k = 1findUngroupedElement(s, k)# This code is contributed by ANKITKUMAR34 |
C#
// C# code to implement the above approachusing System;class GFG{static void findUngroupedElement(String s, int k){ int n = s.Length; // Create array where // index represents alphabets int []b = new int[26]; for(int i = 0; i < 26; i++) b[i] = 0; // Fill count of every // alphabet to corresponding // array index for(int i = 0; i < n; i++) { char p = s[i]; b[p - 'a'] += 1; } int sum = 0; // Count for every element // how much is exceeding // from no. of groups then // sum them for(int i = 0; i < 26; i++) { if (b[i] > k) sum += b[i] - k; } // Print answer Console.WriteLine(sum);}// Driver codepublic static void Main(String []srgs){ String s = "stayinghomesaveslife"; int k = 1; findUngroupedElement(s, k);}}// This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript implementation of the above approachfunction findUngroupedElement(s, k){ let n = s.length; // Create array where // index represents alphabets let b = Array.from({length: 26}, (_, i) => 0); for(let i = 0; i < 26; i++) b[i] = 0; // Fill count of every // alphabet to corresponding // array index for(let i = 0; i < n; i++) { let p = s[i]; b[p.charCodeAt() - 'a'.charCodeAt()] += 1; } let sum = 0; // Count for every element // how much is exceeding // from no. of groups then // sum them for(let i = 0; i < 26; i++) { if (b[i] > k) sum += b[i] - k; } // Print answer document.write(sum);}// Driver code let s = "stayinghomesaveslife"; let k = 1; findUngroupedElement(s, k); // This code is contributed by code_hunt.</script> |
6
Time Complexity: O(N)
Auxiliary Space complexity: O(26) which is equivalent to O(1)
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
