Queries to find first occurrence of a character in a given range
Last Updated :
16 Mar, 2023
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= "geeksforgeeks", 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 range
int 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[c].size() == 0) {
cout << "-1 ";
continue;
}
// Find index >= L in
// the vector v[c]
int idx
= lower_bound(v[c].begin(),
v[c].end(), left)
- v[c].begin();
// If there is no index of C >= L
if (idx == (int)v[c].size()) {
cout << "-1 ";
continue;
}
// Stores the value at idx
idx = v[c][idx];
// If idx > R
if (idx > right) {
cout << "-1 ";
}
// Otherwise
else {
cout << idx << " ";
}
}
}
// Driver Code
int 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 program to implement the above approach
import 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 implementation
# for the above approach
from bisect import bisect_left
# Function to find the first occurrence
# for a given range
def 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[c]
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 Code
if __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# program to implement the above approach
using 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[c] = new List<int>();
}
v[c].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[c].Count == 0)
{
Console.Write("-1 ");
continue;
}
// Find index >= L in
// the vector v.get(c)
List<int> charIndices = v[c];
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 implementation
// for the above approach
// lower_bound function
function 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 range
function 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[c]
let idx = lower_bound(v[c], left);
// If there is no index of C >= L
if (idx == v[c].length) {
process.stdout.write("-1 ");
continue;
}
// Stores the value at idx
idx = v[c][idx];
// If idx > R
if (idx > right) {
process.stdout.write("-1 ");
}
// Otherwise
else {
process.stdout.write(idx + " ");
}
}
}
// Driver Code
let S = "abcabcabc";
let Q
= [ [ [ 0, 3 ], 'a' ],
[ [ 0, 2 ], 'b' ],
[ [ 2, 4 ], 'z' ] ];
firstOccurrence(S, Q);
// The code is contributed by Arushi Jindal.
Time Complexity: O(M * log(N))
Auxiliary Space: O(N)
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