Given a string S, containing lowercase English alphabets, and an integer K, the task is to find any index of the string which consists of more than K active characters. If found, print Yes. Otherwise, print No.
Count of active characters for any index is the number of characters having previous occurrences before or at the current index and last occurrence at or after the current index.
Examples:
Input: S = “aabbcd”, K = 1
Output: No
Explanation:
Index 1: Active character: a
Index 2: Active character: a
Index 3: Active character: b
Index 4: Active character: b
Index 5: Active character: c
Index 6: Active character: d
There are no more than one active character at any index in the string.Input: S = “aabbcdca”, K = 2
Output: Yes
Explanation:
Index 1: Active character : a
Index 2: Active character : a
Index 3: Active characters : a, b
Index 4: Active characters : a, b
Index 5: Active characters : a, c
Index 6: Active characters : a, c, d
Therefore, there exists an index with more than 2 active characters.
Approach:
Follow the steps below to solve the problem:
- The idea is to store the last occurrence of each character present in the string in a Map.
- Traverse the string and keep storing active letters int a Set.
- If at any index, the size of the Set exceeds K, print “Yes”.
- Otherwise, check if the current index is the last occurrence of the current character. If so, remove the character from the Set.
- Finally, if no index is found with more than K active characters, print “No”.
Below is the implementation of the above approach:
C++
// C++ Program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to check if any index// contains more than K active charactersstring checkString(string s, int K){ int n = s.length(); // Store the last occurrence of // each character in the map. unordered_map<char, int> mp; for (int i = 0; i < n; i++) { mp[s[i]] = i; } int cnt = 0, f = 0; // Stores the active // characters unordered_set<int> st; for (int i = 0; i < n; i++) { // Insert the character st.insert(s[i]); // If the size of set // exceeds K if (st.size() > K) { f = 1; break; } // Remove the character from // set if i is the last index // of the current character if (mp[s[i]] == i) st.erase(s[i]); } return (f == 1 ? "Yes" : "No");}// Driver Codeint main(){ string s = "aabbcdca"; int k = 2; cout << checkString(s, k); return 0;} |
Java
// Java program to implement the// above approachimport java.util.*;class GFG{// Function to check if any index// contains more than K active charactersstatic String checkString(String s, int K){ int n = s.length(); // Store the last occurrence of // each character in the map. Map<Character, Integer> mp = new HashMap<>(); for(int i = 0; i < n; i++) { mp.put(s.charAt(i), i); } int cnt = 0, f = 0; // Stores the active // characters Set<Character> st = new HashSet<>(); for(int i = 0; i < n; i++) { // Insert the character st.add(s.charAt(i)); // If the size of set // exceeds K if (st.size() > K) { f = 1; break; } // Remove the character from // set if i is the last index // of the current character if (mp.get(s.charAt(i)) == i) st.remove(s.charAt(i)); } return (f == 1 ? "Yes" : "No");}// Driver codepublic static void main(String[] args){ String s = "aabbcdca"; int k = 2; System.out.println(checkString(s, k));}}// This code is contributed by offbeat |
Python3
# Python3 program to implement # the above approach # Function to check if any index # contains more than K active characters def checkString(s, K): n = len(s) # Store the last occurrence of # each character dict = {} for i in range(n): dict[s[i]] = i; # Stores the active # characters st = set() for i in range(n): # Insert the character st.add(s[i]) # If the size of set # exceeds K if len(st) > K: print("Yes") return # Remove the character from # set if i is the last index # of the current character if dict[s[i]] == i: st.remove(s[i]) print("No")# Driver codes = "aabbcdca"K = 2checkString(s, K)# This code is contributed by vashisthamadhur2 |
C#
// C# program to implement the// above approachusing System;using System.Collections.Generic;class GFG{// Function to check if any index// contains more than K active charactersstatic String checkString(String s, int K){ int n = s.Length; // Store the last occurrence of // each character in the map. Dictionary<char, int> mp = new Dictionary<char, int>(); for(int i = 0; i < n; i++) { if(mp.ContainsKey(s[i])) mp[s[i]] = i; else mp.Add(s[i], i); } int f = 0; // Stores the active // characters HashSet<char> st = new HashSet<char>(); for(int i = 0; i < n; i++) { // Insert the character st.Add(s[i]); // If the size of set // exceeds K if (st.Count > K) { f = 1; break; } // Remove the character from // set if i is the last index // of the current character if (mp[s[i]] == i) st.Remove(s[i]); } return (f == 1 ? "Yes" : "No");}// Driver codepublic static void Main(String[] args){ String s = "aabbcdca"; int k = 2; Console.WriteLine(checkString(s, k));}}// This code is contributed by shikhasingrajput |
Javascript
<script>// Javascript Program to implement// the above approach// Function to check if any index// contains more than K active charactersfunction checkString(s, K){ var n = s.length; // Store the last occurrence of // each character in the map. var mp = new Map(); for (var i = 0; i < n; i++) { if(mp.has(s[i])) { mp.set(s[i], mp.get(s[i])+1); } else mp.set(s[i], 1); } var cnt = 0, f = 0; // Stores the active // characters var st = new Set(); for (var i = 0; i < n; i++) { // Insert the character st.add(s[i]); // If the size of set // exceeds K if (st.size > K) { f = 1; break; } // Remove the character from // set if i is the last index // of the current character if (mp.get(s[i]) == i) st.delete(s[i]); } return (f == 1 ? "Yes" : "No");}// Driver Codevar s = "aabbcdca";var k = 2;document.write( checkString(s, k));// This code is contributed by importantly.</script> |
Output:
Yes
Time Complexity: O(N)
Auxiliary Space: O(1) because size of map and set will be constant
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