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

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++

// 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);
    }
}

Python3

# 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#

// 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

Javascript

<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>

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++.

The below image is a dry run of the above approach: 

Below is the implementation of the above approach:

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 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

// 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

# 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#

// 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

Javascript

<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>

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.