Given a string str and another string patt. Find the character in patt that is present at the minimum index in str. If no character of patt is present in str then print ‘No character present’.
Examples:
Input: str = “geeksforgeeks”, patt = “set”
Output: e
Both e and s of patt are present in str,
but e is present at minimum index, which is 1.Input: str = “adcffaet”, patt = “onkl”
Output: No character present
Source: OLA Interview Experience | Set 12.
Naive Approach: Using two loops, find the first index of each character of patt in str. Print the character having the minimum index. If no character of patt is present in str then print “No character present”.
Implementation:
// C++ implementation to find the character in // first string that is present at minimum index // in second string #include <bits/stdc++.h> using namespace std;
// function to find the minimum index character void printMinIndexChar(string str, string patt)
{ // to store the index of character having
// minimum index
int minIndex = INT_MAX;
// lengths of the two strings
int m = str.size();
int n = patt.size();
// traverse 'patt'
for ( int i = 0; i < n; i++) {
// for each character of 'patt' traverse 'str'
for ( int j = 0; j < m; j++) {
// if patt[i] is found in 'str', check if
// it has the minimum index or not. If yes,
// then update 'minIndex' and break
if (patt[i] == str[j] && j < minIndex) {
minIndex = j;
break ;
}
}
}
// print the minimum index character
if (minIndex != INT_MAX)
cout << "Minimum Index Character = "
<< str[minIndex];
// if no character of 'patt' is present in 'str'
else
cout << "No character present" ;
} // Driver program to test above int main()
{ string str = "geeksforgeeks" ;
string patt = "set" ;
printMinIndexChar(str, patt);
return 0;
} |
// Java implementation to find the character in // first string that is present at minimum index // in second string public class GFG
{ // method to find the minimum index character
static void printMinIndexChar(String str, String patt)
{
// to store the index of character having
// minimum index
int minIndex = Integer.MAX_VALUE;
// lengths of the two strings
int m = str.length();
int n = patt.length();
// traverse 'patt'
for ( int i = 0 ; i < n; i++) {
// for each character of 'patt' traverse 'str'
for ( int j = 0 ; j < m; j++) {
// if patt.charAt(i)is found in 'str', check if
// it has the minimum index or not. If yes,
// then update 'minIndex' and break
if (patt.charAt(i)== str.charAt(j) && j < minIndex) {
minIndex = j;
break ;
}
}
}
// print the minimum index character
if (minIndex != Integer.MAX_VALUE)
System.out.println( "Minimum Index Character = " +
str.charAt(minIndex));
// if no character of 'patt' is present in 'str'
else
System.out.println( "No character present" );
}
// Driver Method
public static void main(String[] args)
{
String str = "geeksforgeeks" ;
String patt = "set" ;
printMinIndexChar(str, patt);
}
} |
# Python3 implementation to find the character in # first that is present at minimum index # in second String # function to find the minimum index character def printMinIndexChar( Str , patt):
# to store the index of character having
# minimum index
minIndex = 10 * * 9
# lengths of the two Strings
m = len ( Str )
n = len (patt)
# traverse 'patt'
for i in range (n):
# for each character of 'patt' traverse 'Str'
for j in range (m):
# if patt[i] is found in 'Str', check if
# it has the minimum index or not. If yes,
# then update 'minIndex' and break
if (patt[i] = = Str [j] and j < minIndex):
minIndex = j
break
# print the minimum index character
if (minIndex ! = 10 * * 9 ):
print ( "Minimum Index Character = " , Str [minIndex])
# if no character of 'patt' is present in 'Str'
else :
print ( "No character present" )
# Driver code Str = "geeksforgeeks"
patt = "set"
printMinIndexChar( Str , patt)
# This code is contributed by mohit kumar 29 |
// C# implementation to find the character in // first string that is present at minimum index // in second string using System;
class GFG
{ // method to find the minimum index character
static void printMinIndexChar(String str, String patt)
{
// to store the index of character having
// minimum index
int minIndex = int .MaxValue;
// lengths of the two strings
int m = str.Length;
int n = patt.Length;
// traverse 'patt'
for ( int i = 0; i < n; i++) {
// for each character of 'patt' traverse 'str'
for ( int j = 0; j < m; j++) {
// if patt.charAt(i)is found in 'str', check if
// it has the minimum index or not. If yes,
// then update 'minIndex' and break
if (patt[i]== str[j] && j < minIndex) {
minIndex = j;
break ;
}
}
}
// print the minimum index character
if (minIndex != int .MaxValue)
Console.WriteLine( "Minimum Index Character = " +
str[minIndex]);
// if no character of 'patt' is present in 'str'
else
Console.WriteLine( "No character present" );
}
// Driver Method
public static void Main()
{
String str = "geeksforgeeks" ;
String patt = "set" ;
printMinIndexChar(str, patt);
}
} // This code is contributed by Sam007 |
<script> // Javascript implementation to find the character in // first string that is present at minimum index // in second string // method to find the minimum index character
function printMinIndexChar(str,patt)
{
// to store the index of character having
// minimum index
let minIndex = Number.MAX_VALUE;
// lengths of the two strings
let m = str.length;
let n = patt.length;
// traverse 'patt'
for (let i = 0; i < n; i++) {
// for each character of 'patt' traverse 'str'
for (let j = 0; j < m; j++) {
// if patt.charAt(i)is found in 'str', check if
// it has the minimum index or not. If yes,
// then update 'minIndex' and break
if (patt[i]== str[j] && j < minIndex) {
minIndex = j;
break ;
}
}
}
// print the minimum index character
if (minIndex != Number.MAX_VALUE)
document.write( "Minimum Index Character = " +
str[minIndex]);
// if no character of 'patt' is present in 'str'
else
document.write( "No character present" );
}
// Driver Method
let str = "geeksforgeeks" ;
let patt = "set" ;
printMinIndexChar(str, patt);
//This code is contributed by rag2127
</script> |
Minimum Index Character = e
Time Complexity: O(mn), where m and n are the lengths of the two strings.
Auxiliary Space: O(1)
Method 2 Efficient Approach(Hashing):
- Create a hash table with (key, value) tuple represented as (character, index) tuple.
- Store the first index of each character of str in the hash table.
- Now, for each character of patt check if it is present in the hash table or not.
- If present then get its index from the hash table and update minIndex(minimum index encountered so far).
- For no matching character print “No character present”.
Hash table is implemented using unordered_set in C++.
The below image is a dry run of the above approach:
Below is the implementation of the above approach:
// C++ implementation to find the character in first // string that is present at minimum index in second // string #include <bits/stdc++.h> using namespace std;
// function to find the minimum index character void printMinIndexChar(string str, string patt)
{ // unordered_map 'um' implemented as hash table
unordered_map< char , int > um;
// to store the index of character having
// minimum index
int minIndex = INT_MAX;
// lengths of the two strings
int m = str.size();
int n = patt.size();
// store the first index of each character of 'str'
for ( int i = 0; i < m; i++) {
if (um.find(str[i]) == um.end())
um[str[i]] = i;
}
// traverse the string 'patt'
for ( int j = 0; j < n; j++) {
// if patt[i] is found in 'um', check if
// it has the minimum index or not accordingly
// update 'minIndex'
if (um.find(patt[j]) != um.end()
&& um[patt[j]] < minIndex)
minIndex = um[patt[j]];
}
// print the minimum index character
if (minIndex != INT_MAX)
cout << "Minimum Index Character = "
<< str[minIndex];
// if no character of 'patt' is present in 'str'
else
cout << "No character present" ;
} // Driver program to test above int main()
{ string str = "geeksforgeeks" ;
string patt = "set" ;
printMinIndexChar(str, patt);
return 0;
} |
// Java implementation to find the character in // first string that is present at minimum index // in second string import java.util.HashMap;
public class GFG
{ // method to find the minimum index character
static void printMinIndexChar(String str, String patt)
{
// map to store the first index of each character of 'str'
HashMap<Character, Integer> hm = new HashMap<>();
// to store the index of character having
// minimum index
int minIndex = Integer.MAX_VALUE;
// lengths of the two strings
int m = str.length();
int n = patt.length();
// store the first index of each character of 'str'
for ( int i = 0 ; i < m; i++)
if (!hm.containsKey(str.charAt(i)))
hm.put(str.charAt(i),i);
// traverse the string 'patt'
for ( int i = 0 ; i < n; i++)
// if patt[i] is found in 'um', check if
// it has the minimum index or not accordingly
// update 'minIndex'
if (hm.containsKey(patt.charAt(i)) &&
hm.get(patt.charAt(i)) < minIndex)
minIndex = hm.get(patt.charAt(i));
// print the minimum index character
if (minIndex != Integer.MAX_VALUE)
System.out.println( "Minimum Index Character = " +
str.charAt(minIndex));
// if no character of 'patt' is present in 'str'
else
System.out.println( "No character present" );
}
// Driver Method
public static void main(String[] args)
{
String str = "geeksforgeeks" ;
String patt = "set" ;
printMinIndexChar(str, patt);
}
} |
# Python3 implementation to # find the character in first # string that is present at # minimum index in second string import sys
# Function to find the # minimum index character def printMinIndexChar(st, patt):
# unordered_map 'um'
# implemented as hash table
um = {}
# to store the index of
# character having minimum index
minIndex = sys.maxsize
# Lengths of the two strings
m = len (st)
n = len (patt)
# Store the first index of
# each character of 'str'
for i in range (m):
if (st[i] not in um):
um[st[i]] = i
# traverse the string 'patt'
for i in range (n):
# If patt[i] is found in 'um',
# check if it has the minimum
# index or not accordingly
# update 'minIndex'
if (patt[i] in um and
um[patt[i]] < minIndex):
minIndex = um[patt[i]]
# Print the minimum index character
if (minIndex ! = sys.maxsize):
print ( "Minimum Index Character = " ,
st[minIndex])
# If no character of 'patt'
# is present in 'str'
else :
print ( "No character present" )
# Driver program to test above if __name__ = = "__main__" :
st = "geeksforgeeks"
patt = "set"
printMinIndexChar(st, patt)
# This code is contributed by Chitranayal |
// C# implementation to find the character in // first string that is present at minimum index // in second string using System;
using System.Collections.Generic;
class GFG
{ // method to find the minimum index character
static void printMinIndexChar(String str, String patt)
{
// map to store the first index of
// each character of 'str'
Dictionary< char ,
int > hm = new Dictionary< char ,
int >();
// to store the index of character having
// minimum index
int minIndex = int .MaxValue;
// lengths of the two strings
int m = str.Length;
int n = patt.Length;
// store the first index of
// each character of 'str'
for ( int i = 0; i < m; i++)
if (!hm.ContainsKey(str[i]))
hm.Add(str[i], i);
// traverse the string 'patt'
for ( int i = 0; i < n; i++)
// if patt[i] is found in 'um',
// check if it has the minimum index
// or not, accordingly update 'minIndex'
if (hm.ContainsKey(patt[i]) &&
hm[patt[i]] < minIndex)
minIndex = hm[patt[i]];
// print the minimum index character
if (minIndex != int .MaxValue)
Console.WriteLine( "Minimum Index Character = " +
str[minIndex]);
// if no character of 'patt' is present in 'str'
else
Console.WriteLine( "No character present" );
}
// Driver Code
public static void Main(String[] args)
{
String str = "geeksforgeeks" ;
String patt = "set" ;
printMinIndexChar(str, patt);
}
} // This code is contributed by Princi Singh |
<script> // Javascript implementation to find the // character in first string that is // present at minimum index in second // string // Method to find the minimum index character function printMinIndexChar(str, patt)
{ // map to store the first index of
// each character of 'str'
let hm = new Map();
// To store the index of character having
// minimum index
let minIndex = Number.MAX_VALUE;
// Lengths of the two strings
let m = str.length;
let n = patt.length;
// Store the first index of
// each character of 'str'
for (let i = 0; i < m; i++)
if (!hm.has(str[i]))
hm.set(str[i], i);
// Traverse the string 'patt'
for (let i = 0; i < n; i++)
// If patt[i] is found in 'um', check
// if it has the minimum index or not
// accordingly update 'minIndex'
if (hm.has(patt[i]) &&
hm.get(patt[i]) < minIndex)
minIndex = hm.get(patt[i]);
// Print the minimum index character
if (minIndex != Number.MAX_VALUE)
document.write( "Minimum Index Character = " +
str[minIndex]);
// If no character of 'patt' is
// present in 'str'
else
document.write( "No character present" );
} // Driver Code let str = "geeksforgeeks" ;
let patt = "set" ;
printMinIndexChar(str, patt); // This code is contributed by avanitrachhadiya2155 </script> |
Minimum Index Character = e
Time Complexity: O(m + n), where m and n are the lengths of the two strings.
Auxiliary Space: O(d), where d is the size of hash table, which is the count of distinct characters in str.