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Longest suffix such that occurrence of each character is less than N after deleting atmost K characters
  • Last Updated : 29 May, 2021

Given a string S and two integer N and K, the task is to find the maximum length suffix such that occurrence of each character in the suffix string is less than N and atmost K elements can be deleted from the input string to get maximum length suffix.
Examples: 
 

Input: S = “iahagafedcba”, N = 1, K = 0 
Output: fedcba 
Explanation: 
Maximum length suffix in the given string 
Such that occurence of each character is 1 is “fedcba”, 
Because if we take the string “afedcba”, 
then the occurence of character “a” will be 2.
Input: S = “iahagafedcba”, N = 1, K = 2 
Output: hgfedcba 
Explanation: 
Maximum length suffix in the given string 
Such that occurrence of each character is 1 is “hgfedcba”, 
After deleting character “a” two times. 
 

 

Approach: The idea is to use hash-map to store the frequency of the characters of the string. 
 

  • Initialize an empty string to store the longest suffix of the string.
  • Iterate the string from last using a loop – 
    • Increment the occurrence of the character by 1 in the hash-map
    • If the occurrence of current character is less than N, then add the character into the suffix string
    • Otherwise, decrement the value of K by 1 if the value of K is greater than 0
  • Print the suffix string.

Below is the implementation of the above approach:
 



C++




// C++ implementation to find
// longest suffix of the string
// such that occurrence of each
// character is less than K
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum
// length suffix in the string
void maximumSuffix(string s,
               int n, int k){
     
    // Length of the string
    int i = s.length() - 1;
 
    // Map to store the number
    // of occurrence of character
    int arr[26] = { 0 };
    string suffix = "";
 
    // Loop to iterate string
    // from the last character
    while (i > -1) {
 
        int index = s[i] - 'a';
         
        // Condition to check if the
        // occurrence of each character
        // is less than given number
        if (arr[index] < n) {
            arr[index]++;
            suffix += s[i];
            i--;
            continue;
        }
         
        // Condition when character
        // cannot be deleted
        if (k == 0)
            break;
        k--;
        i--;
    }
    reverse(suffix.begin(), suffix.end());
 
    // Longest suffix
    cout << suffix;
}
 
// Driver Code
int main()
{
    string str = "iahagafedcba";
 
    int n = 1, k = 2;
     
    // Function call
    maximumSuffix(str, n, k);
    return 0;
}

Java




// Java implementation to find
// longest suffix of the String
// such that occurrence of each
// character is less than K
class GFG{
 
// Function to find the maximum
// length suffix in the String
static void maximumSuffix(String s,
            int n, int k){
     
    // Length of the String
    int i = s.length() - 1;
 
    // Map to store the number
    // of occurrence of character
    int arr[] = new int[26];
    String suffix = "";
     
    // Loop to iterate String
    // from the last character
    while (i > -1) {
 
        int index = s.charAt(i) - 'a';
         
        // Condition to check if the
        // occurrence of each character
        // is less than given number
        if (arr[index] < n) {
            arr[index]++;
            suffix += s.charAt(i);
            i--;
            continue;
        }
         
        // Condition when character
        // cannot be deleted
        if (k == 0)
            break;
        k--;
        i--;
    }
    suffix = reverse(suffix);
     
    // Longest suffix
    System.out.print(suffix);
}
static String reverse(String input) {
    char[] a = input.toCharArray();
    int l, r = a.length - 1;
    for (l = 0; l < r; l++, r--) {
        char temp = a[l];
        a[l] = a[r];
        a[r] = temp;
    }
    return String.valueOf(a);
}
 
// Driver Code
public static void main(String[] args)
{
    String str = "iahagafedcba";
 
    int n = 1, k = 2;
     
    // Function call
    maximumSuffix(str, n, k);
}
}
 
// This code is contributed by 29AjayKumar

Python 3




# Python 3 implementation to find
# longest suffix of the string
# such that occurrence of each
# character is less than K
 
# Function to find the maximum
# length suffix in the string
def maximumSuffix(s, n, k):
     
    # Length of the string
    i = len(s)- 1
 
    # Map to store the number
    # of occurrence of character
    arr = [0 for i in range(26)]
    suffix = ""
     
    # Loop to iterate string
    # from the last character
    while (i > -1):
        index = ord(s[i]) - ord('a');
         
        # Condition to check if the
        # occurrence of each character
        # is less than given number
        if (arr[index] < n):
            arr[index] += 1
            suffix += s[i]
            i -= 1
            continue
         
        # Condition when character
        # cannot be deleted
        if (k == 0):
            break
        k -= 1
        i -= 1
     
    suffix = suffix[::-1]
     
    # Longest suffix
    print(suffix)
 
# Driver Code
if __name__ == '__main__':
    str = "iahagafedcba"
 
    n = 1
    k = 2
     
    # Function call
    maximumSuffix(str, n, k)
 
# This code is contributed by Surendra_Gangwar

C#




// C# implementation to find
// longest suffix of the String
// such that occurrence of each
// character is less than K
using System;
 
class GFG{
  
// Function to find the maximum
// length suffix in the String
static void maximumSuffix(String s,
            int n, int k){
      
    // Length of the String
    int i = s.Length - 1;
  
    // Map to store the number
    // of occurrence of character
    int []arr = new int[26];
    String suffix = "";
      
    // Loop to iterate String
    // from the last character
    while (i > -1) {
  
        int index = s[i] - 'a';
          
        // Condition to check if the
        // occurrence of each character
        // is less than given number
        if (arr[index] < n) {
            arr[index]++;
            suffix += s[i];
            i--;
            continue;
        }
          
        // Condition when character
        // cannot be deleted
        if (k == 0)
            break;
        k--;
        i--;
    }
    suffix = reverse(suffix);
      
    // longest suffix
    Console.Write(suffix);
}
static String reverse(String input) {
    char[] a = input.ToCharArray();
    int l, r = a.Length - 1;
    for (l = 0; l < r; l++, r--) {
        char temp = a[l];
        a[l] = a[r];
        a[r] = temp;
    }
    return String.Join("",a);
}
  
// Driver Code
public static void Main(String[] args)
{
    String str = "iahagafedcba";
  
    int n = 1, k = 2;
      
    // Function call
    maximumSuffix(str, n, k);
}
}
 
// This code is contributed by Rajput-Ji
Output: 
hgfedcba

 

Performance Analysis: 
 

  • Time Complexity: In the above-given approach, there is one loop for iterating over string which takes O(L) time in worst case. Therefore, the time complexity for this approach will be O(L).
  • Auxiliary Space Complexity: In the above-given approach, there is extra space used to store the frequency of the character. Therefore, the auxiliary space complexity for the above approach will be O(L)

 

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