# Find the character in first string that is present at minimum index in second string

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
```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Naive Approach: Using two lopp, 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”.

## C++

```// 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

```// 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);
}
}
```

Output:

```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++.

## C++

```// 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 charcter 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 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 (um.find(patt[i]) != um.end() &&
um[patt[i]] < minIndex)
minIndex = um[patt[i]];

// 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

```// 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);
}
}
```

Output:

```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.

This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

# GATE CS Corner    Company Wise Coding Practice

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.
2 Average Difficulty : 2/5.0
Based on 5 vote(s)

Writing code in comment? Please use ide.geeksforgeeks.org, generate link and share the link here.