Given a string S of length N and an array Q[][] of dimension M × 3 consisting of queries of type {L, R, C}, the task is to print the first index of the character C in the range [L, R], if found. Otherwise, print -1.
Examples:
Input: S= “abcabcabc”, Q[][] = { { 0, 3, ‘a’ }, { 0, 2, ‘b’ }, { 2, 4, ‘z’ } }
Output: 0 1 -1
Explanation:
- First query: Print 0 which is the first index of character ‘a’ in the range [0, 3].
- Second query: Print 1, which is the first index of character ‘b’ in the range [0, 2].
- Third query: Print -1 as the character ‘z’ does not occur in the range [2, 4].
Input: S= “neveropen”, Q[][] = { { 0, 12, ‘f’ }, { 0, 2, ‘g’ }, { 6, 12, ‘f’ } }
Output: 5 0 -1
Explanation:
- First query: Print 5, which is the first index of character ‘f’ in the range [0, 12].
- Second query: Print 0 which is
the first index of character ‘g’ in the range [0, 2].- Third query: Print -1 as the character ‘f’ does not occur in the range [6, 12].
Naive Approach: The simplest approach is to traverse the string over the range of indices [L, R] for each query and print the first occurrence of character C if found. Otherwise, print -1.
Time Complexity: O(M * Q)
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized by pre-storing the indices of characters in the map of vectors and using binary search to find the index in the range [L, R] in the vector of characters C. Follow the steps below to solve the problem:
- Initialize a Map < char, vector < int > >, say V, to store indices of all occurrences of a character.
- Traverse the string and update V.
- Traverse the array Q:
- If the size of V[C] is zero then print -1.
- Otherwise, find the index by using binary search in vector V[C] i.e lower_bound(V[C].begin(), V[C].end(), L) – V[C].begin() and store it in a variable, say idx.
- If idx = N or idx > R, then print -1.
- Otherwise, print the index idx.
Below is the implementation of the above approach:
C++
// C++ implementation// for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the first occurrence// for a given rangeint firstOccurrence(string s, vector<pair<pair<int, int>, char> > Q){ // N = length of string int N = s.size(); // M = length of queries int M = Q.size(); // Stores the indices of a character map<char, vector<int> > v; // Traverse the string for (int i = 0; i < N; i++) { // Push the index i // into the vector[s[i]] v[s[i]].push_back(i); } // Traverse the query for (int i = 0; i < M; i++) { // Stores the value L int left = Q[i].first.first; // Stores the value R int right = Q[i].first.second; // Stores the character C char c = Q[i].second; if (v.size() == 0) { cout << "-1 "; continue; } // Find index >= L in // the vector v int idx = lower_bound(v.begin(), v.end(), left) - v.begin(); // If there is no index of C >= L if (idx == (int)v.size()) { cout << "-1 "; continue; } // Stores the value at idx idx = v[idx]; // If idx > R if (idx > right) { cout << "-1 "; } // Otherwise else { cout << idx << " "; } }}// Driver Codeint main(){ string S = "abcabcabc"; vector<pair<pair<int, int>, char> > Q = { { { 0, 3 }, 'a' }, { { 0, 2 }, 'b' }, { { 2, 4 }, 'z' } }; firstOccurrence(S, Q); return 0;} |
Java
// Java program to implement the above approachimport java.util.*;public class Main { // Function to find the first occurrence // for a given range public static void firstOccurrence(String s, ArrayList<Pair<Pair<Integer, Integer>, Character>> Q) { // N = length of string int N = s.length(); // M = length of queries int M = Q.size(); // Stores the indices of a character HashMap<Character, ArrayList<Integer>> v = new HashMap<Character, ArrayList<Integer>>(); // Traverse the string for (int i = 0; i < N; i++) { // Push the index i // into the vector[s.charAt(i)] char c = s.charAt(i); if (!v.containsKey(c)) { v.put(c, new ArrayList<Integer>()); } v.get(c).add(i); } // Traverse the query for (int i = 0; i < M; i++) { // Stores the value L int left = Q.get(i).first.first; // Stores the value R int right = Q.get(i).first.second; // Stores the character C char c = Q.get(i).second; if (!v.containsKey(c) || v.get(c).size() == 0) { System.out.print("-1 "); continue; } // Find index >= L in // the vector v.get(c) ArrayList<Integer> charIndices = v.get(c); int idx = Collections.binarySearch(charIndices, left); // If there is no index of C >= L if (idx < 0) { idx = -(idx + 1); if (idx == charIndices.size() || charIndices.get(idx) > right) { System.out.print("-1 "); continue; } } else if (charIndices.get(idx) > right) { System.out.print("-1 "); continue; } // Stores the value at idx idx = charIndices.get(idx); // If idx > R if (idx > right) { System.out.print("-1 "); } // Otherwise else { System.out.print(idx + " "); } } } // Driver Code public static void main(String[] args) { String S = "abcabcabc"; ArrayList<Pair<Pair<Integer, Integer>, Character>> Q = new ArrayList<Pair<Pair<Integer, Integer>, Character>>(); Q.add(new Pair<Pair<Integer, Integer>, Character>(new Pair<Integer, Integer>(0, 3), 'a')); Q.add(new Pair<Pair<Integer, Integer>, Character>(new Pair<Integer, Integer>(0, 2), 'b')); Q.add(new Pair<Pair<Integer, Integer>, Character>(new Pair<Integer, Integer>(2, 4), 'z')); firstOccurrence(S, Q); }}class Pair<U, V> { public U first; public V second; public Pair(U first, V second) { this.first = first; this.second = second; }}// Contributed by adityashae15 |
Python3
# Python3 implementation# for the above approachfrom bisect import bisect_left# Function to find the first occurrence# for a given rangedef firstOccurrence(s, Q): # N = length of string N = len(s) # M = length of queries M = len(Q) # Stores the indices of a character v = [[] for i in range(26)] # Traverse the string for i in range(N): # Push the index i # into the vector[s[i]] v[ord(s[i]) - ord('a')].append(i) # Traverse the query for i in range(M): # Stores the value L left = Q[i][0] # Stores the value R right = Q[i][1] # Stores the character C c = Q[i][2] if (len(v[ord(c) - ord('a')]) == 0): print ("-1") continue # Find index >= L in # the vector v idx = bisect_left(v[ord(c) - ord('a')], left) # If there is no index of C >= L if (idx == len(v[ord(c) - ord('a')])): print("-1 ") continue # Stores the value at idx idx = v[ord(c) - ord('a')][idx] # If idx > R if (idx > right): print ("-1 ") # Otherwise else: print(idx, end=" ")# Driver Codeif __name__ == '__main__': S = "abcabcabc"; Q = [ [ 0, 3 , 'a'],[ 0, 2 , 'b' ],[ 2, 4, 'z']] firstOccurrence(S, Q) # This code is contributed by mohit kumar 29. |
C#
// C# program to implement the above approachusing System;using System.Collections.Generic;public class Program{ // Function to find the first occurrence // for a given range public static void FirstOccurrence(string s, List<Tuple<Tuple<int, int>, char>> Q) { // N = length of string int N = s.Length; // M = length of queries int M = Q.Count; // Stores the indices of a character Dictionary<char, List<int>> v = new Dictionary<char, List<int>>(); // Traverse the string for (int i = 0; i < N; i++) { // Push the index i // into the vector[s.charAt(i)] char c = s[i]; if (!v.ContainsKey(c)) { v = new List<int>(); } v.Add(i); } // Traverse the query for (int i = 0; i < M; i++) { // Stores the value L int left = Q[i].Item1.Item1; // Stores the value R int right = Q[i].Item1.Item2; // Stores the character C char c = Q[i].Item2; if (!v.ContainsKey(c) || v.Count == 0) { Console.Write("-1 "); continue; } // Find index >= L in // the vector v.get(c) List<int> charIndices = v; int idx = charIndices.BinarySearch(left); // If there is no index of C >= L if (idx < 0) { idx = -(idx + 1); if (idx == charIndices.Count || charIndices[idx] > right) { Console.Write("-1 "); continue; } } else if (charIndices[idx] > right) { Console.Write("-1 "); continue; } // Stores the value at idx idx = charIndices[idx]; // If idx > R if (idx > right) { Console.Write("-1 "); } // Otherwise else { Console.Write(idx + " "); } } } // Driver Code public static void Main() { string S = "abcabcabc"; List<Tuple<Tuple<int, int>, char>> Q = new List<Tuple<Tuple<int, int>, char>>(); Q.Add(Tuple.Create(Tuple.Create(0, 3), 'a')); Q.Add(Tuple.Create(Tuple.Create(0, 2), 'b')); Q.Add(Tuple.Create(Tuple.Create(2, 4), 'z')); FirstOccurrence(S, Q); }}// Contributed by adityasharmadev01 |
Javascript
// javascript implementation// for the above approach// lower_bound functionfunction lower_bound(a, x){ let l = 0; let h = a.length - 1; while(l <= h){ let m = Math.floor((l + h)/2); if(a[m] < x){ l = m + 1; } else if(a[m] > x){ h = m - 1; } else return l; } return l;}// Function to find the first occurrence// for a given rangefunction firstOccurrence(s, Q){ // N = length of string let N = s.length; // M = length of queries let M = Q.length; // Stores the indices of a character let v = {}; // map<char, vector<int> > v; // Traverse the string for (let i = 0; i < N; i++) { // Push the index i // into the vector[s[i]] if(s[i] in v){ v[s[i]].push(i); } else{ v[s[i]] = []; v[s[i]].push(i); } } // Traverse the query for (let i = 0; i < M; i++) { // Stores the value L let left = Q[i][0][0]; // Stores the value R let right = Q[i][0][1]; // Stores the character C let c = Q[i][1]; if (c in v){ } else{ console.log("-1 "); continue; } // Find index >= L in // the vector v let idx = lower_bound(v, left); // If there is no index of C >= L if (idx == v.length) { process.stdout.write("-1 "); continue; } // Stores the value at idx idx = v[idx]; // If idx > R if (idx > right) { process.stdout.write("-1 "); } // Otherwise else { process.stdout.write(idx + " "); } }}// Driver Codelet S = "abcabcabc";let Q = [ [ [ 0, 3 ], 'a' ], [ [ 0, 2 ], 'b' ], [ [ 2, 4 ], 'z' ] ];firstOccurrence(S, Q); // The code is contributed by Arushi Jindal. |
Output:
0 1 -1
Time Complexity: O(M * log(N))
Auxiliary Space: O(N)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
