Given a string S of size N consisting of lower case alphabets and an integer Q which represents the number of queries for S. Our task is to print the number of duplicate characters in the substring L to R for all the queries Q.
Note: 1 ?N ? 106 and 1 ? Q? 106
Examples:
Input :
S = “neveropen”, Q = 2
L = 1 R = 5
L = 4 R = 8
Output :
1
0
Explanation:
For the first query ‘e’ is the only duplicate character in S from range 1 to 5.
For the second query there is no duplicate character in S.Input :
S = “Geekyy”, Q = 1
L = 1 R = 6
Output :
2
Explanation:
For the first query ‘e’ and ‘y’ are duplicate characters in S from range 1 to 6.
Naive Approach:
The naive approach would be to maintain a frequency array of size 26, to store the count of each character. For each query, given a range [L, R] we will traverse substring S[L] to S[R] and keep counting the occurrence of each character. Now, if the frequency of any character is greater than 1 then we would add 1 to answer.
Efficient Approach:
To solve the above problem in an efficient way we will store the position of each character as it appears in the string in a dynamic array. For each given query we will iterate over all the 26 lower case alphabets. If the current letter is in the substring S[L: R] then the next element of the first element which is greater than or equal L to in the corresponding vector should exist and be less than or equal to R.
Diagram below shows how we store characters in the dynamic array:
Below is the implementation of the above approach:
CPP
// CPP implementation to Find the total// number of duplicate character in a// range L to R for Q number of queries in a string S#include <bits/stdc++.h>using namespace std;// Vector of vector to store// position of all characters// as they appear in stringvector<vector<int> > v(26);// Function to store position of each charactervoid calculate(string s){ for (int i = 0; i < s.size(); i++) { // Inserting position of each // character as they appear v[s[i] - 'a'].push_back(i); }}// Function to calculate duplicate// characters for Q queriesvoid query(int L, int R){ // Variable to count duplicates int duplicates = 0; // Iterate over all 26 characters for (int i = 0; i < 26; i++) { // Finding the first element which // is less than or equal to L auto first = lower_bound(v[i].begin(), v[i].end(), L - 1); // Check if first pointer exists // and is less than R if (first != v[i].end() && *first < R) { // Incrementing first pointer to check // if the next duplicate element exists first++; // Check if the next element exists // and is less than R if (first != v[i].end() && *first < R) duplicates++; } } cout << duplicates << endl;}// Driver Codeint main(){ string s = "neveropen"; int Q = 2; int l1 = 1, r1 = 5; int l2 = 4, r2 = 8; calculate(s); query(l1, r1); query(l2, r2); return 0;} |
Python3
# Python implementation to Find the total# number of duplicate character in a# range L to R for Q number of queries in a string Simport bisect# Vector of vector to store# position of all characters# as they appear in stringv = [[] for _ in range(26)]# Function to store position of each characterdef calculate(s: str) -> None: for i in range(len(s)): # Inserting position of each # character as they appear v[ord(s[i]) - ord('a')].append(i)# Function to calculate duplicate# characters for Q queriesdef query(L: int, R: int) -> None: # Variable to count duplicates duplicates = 0 # Iterate over all 26 characters for i in range(26): # Finding the first element which # is less than or equal to L first = bisect.bisect_left(v[i], L - 1) # Check if first pointer exists # and is less than R if (first < len(v[i]) and v[i][first] < R): # Incrementing first pointer to check # if the next duplicate element exists first += 1 # Check if the next element exists # and is less than R if (first < len(v[i]) and v[i][first] < R): duplicates += 1 print(duplicates)# Driver Codeif __name__ == "__main__": s = "neveropen" Q = 2 l1 = 1 r1 = 5 l2 = 4 r2 = 8 calculate(s) query(l1, r1) query(l2, r2)# This code is contributed by sanjeev2552 |
Java
// java implementation to Find the total// number of duplicate character in a// range L to R for Q number of queries in a string Simport java.util.ArrayList;import java.util.List;public class DuplicateCharacter { // List of List to store // position of all characters // as they appear in string static List<List<Integer> > v = new ArrayList<>(); // Function to store position of each character static void calculate(String s) { for (int i = 0; i < 26; i++) { v.add(new ArrayList<>()); } // Inserting position of each // character as they appear for (int i = 0; i < s.length(); i++) { v.get(s.charAt(i) - 'a').add(i); } } // Function to calculate duplicate // characters for Q queries static void query(int L, int R) { int duplicates = 0; for (int i = 0; i < 26; i++) { // Finding the first element which // is less than or equal to L int j = 0; while (j < v.get(i).size() && v.get(i).get(j) < L) { j++; } // Check if first pointer exists // and is less than R if (j < v.get(i).size() && v.get(i).get(j) < R) { // Incrementing first pointer to check // if the next duplicate element exists j++; // Check if the next element exists // and is less than R if (j < v.get(i).size() && v.get(i).get(j) < R) { duplicates++; } } } System.out.println(duplicates); } public static void main(String[] args) { String s = "neveropen"; int Q = 2; int l1 = 1, r1 = 5; int l2 = 4, r2 = 8; calculate(s); query(l1, r1); query(l2, r2); }} |
Javascript
// javascript code for the above approach// Function to find the index of // the first element in a sorted // array which is greater than or // equal to a given valuefunction bisect_left(arr, x) { let lo = 0, hi = arr.length; while (lo < hi) { let mid = Math.floor((lo + hi) / 2); if (arr[mid] < x) { lo = mid + 1; } else { hi = mid; } } return lo;}// Vector of array to store// position of all characters// as they appear in stringlet v = [...Array(26)].map(() => []);// Function to store position of each characterfunction calculate(s) { for (let i = 0; i < s.length; i++) { // Inserting position of each // character as they appear v[s.charCodeAt(i) - 'a'.charCodeAt(0)].push(i); }}// Function to calculate duplicate// characters for Q queriesfunction query(L, R) { // Variable to count duplicates let duplicates = 0; // Iterate over all 26 characters for (let i = 0; i < 26; i++) { // Finding the first element which // is less than or equal to L let first = bisect_left(v[i], L - 1); // Check if first pointer exists // and is less than R if (first < v[i].length && v[i][first] < R) { // Incrementing first pointer to check // if the next duplicate element exists first += 1; // Check if the next element exists // and is less than R if (first < v[i].length && v[i][first] < R) { duplicates += 1; } } } console.log(duplicates);}// Driver Codelet s = "neveropen";let Q = 2;let l1 = 1;let r1 = 5;let l2 = 4;let r2 = 8;calculate(s);query(l1, r1);query(l2, r2);// This code is contributed by princekumaras |
C#
// C# program for the above approachusing System;using System.Collections.Generic;public class DuplicateCharacter{ // List of List to store // position of all characters // as they appear in string static List<List<int>> v = new List<List<int>>(); // Function to store position of each character static void calculate(string s) { for (int i = 0; i < 26; i++) { v.Add(new List<int>()); } // Inserting position of each // character as they appear for (int i = 0; i < s.Length; i++) { v[s[i] - 'a'].Add(i); } } // Function to calculate duplicate // characters for Q queries static void query(int L, int R) { int duplicates = 0; for (int i = 0; i < 26; i++) { // Finding the first element which // is less than or equal to L int j = 0; while (j < v[i].Count && v[i][j] < L) { j++; } // Check if first pointer exists // and is less than R if (j < v[i].Count && v[i][j] < R) { // Incrementing first pointer to check // if the next duplicate element exists j++; // Check if the next element exists // and is less than R if (j < v[i].Count && v[i][j] < R) { duplicates++; } } } Console.WriteLine(duplicates); } public static void Main(string[] args) { string s = "neveropen"; int Q = 2; int l1 = 1, r1 = 5; int l2 = 4, r2 = 8; calculate(s); query(l1, r1); query(l2, r2); }}// This code is contributed by adityashatmfh |
Output:
1 0
Time Complexity: O( Q * 26 * log N)
Auxiliary Space: O(N)
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