Length of the smallest substring which contains all vowels

Given string str consisting of only lowercase English alphabets, the task is to find the substring of the smallest length which contains all the vowels. If no such substring is found, print -1.

Example:

Input: str = “babeivoucu” 
Output: 7 
Explanation: Smallest substring which contains each vowel atleast once is “abeivou” of length 7.

Input: str = “abcdef” 
Output: -1 
Explanation: No such substring is found. 

Two Pointer Approach:

• Store the frequencies of each vowel and the indices at which the vowels are present.
• If all the vowels are not present, straightaway print -1.
• Take two pointers i and j at the first and last indices containing a vowel.
• If the frequencies of vowel at i^th and j^th index exceed 1, decrease the count of that vowel and shift i and j to the right and left respectively to the next index containing a vowel.
• If the frequencies of vowel at i^th or j^th index is equal to 1, set flag1 or flag2 respectively and continue.
• Once both flag1 and flag2 are set, the length of the substring cannot be further minimized. The distance between the current indices pointed by the two pointers denotes the result.

The below code is the implementation of the above approach: 

C++

// C++ Program to find the
// length of the smallest
// substring of which
// contains all vowels

#include <bits/stdc++.h>
using namespace std;

int findMinLength(string s)
{
    int n = s.size();

    // Map to store the
    // frequency of vowels
    map<char, int> counts;

    // Store the indices
    // which contains
    // the vowels
    vector<int> indices;

    for (int i = 0; i < n; i++) {

        if (s[i] == 'a' || s[i] == 'e'
            || s[i] == 'o'
            || s[i] == 'i'
            || s[i] == 'u') {

            counts[s[i]]++;
            indices.push_back(i);
        }
    }

    // If all vowels are not
    // present in the string
    if (counts.size() < 5)
        return -1;

    int flag1 = 0, flag2 = 0;
    int i = 0;
    int j = indices.size() - 1;

    while ((j - i) >= 4) {

        // If the frequency of the
        // vowel at i-th index
        // exceeds 1
        if (!flag1
            && counts[s[indices[i]]] > 1) {

            // Decrease the frequency
            // of that vowel
            counts[s[indices[i]]]--;

            // Move to the left
            i++;
        }

        // Otherwise set flag1
        else
            flag1 = 1;

        // If the frequency of the
        // vowel at j-th index
        // exceeds 1
        if (!flag2
            && counts[s[indices[j]]] > 1) {

            // Decrease the frequency
            // of that vowel
            counts[s[indices[j]]]--;
            // Move to the right
            j--;
        }

        // Otherwise set flag2
        else
            flag2 = 1;

        // If both flag1 and flag2
        // are set, break out of the
        // loop as the substring
        // length cannot be minimized
        if (flag1 && flag2)
            break;
    }

    // Return the length of the substring
    return (indices[j] - indices[i] + 1);
}

int main()
{

    string s = "aaeebbeaccaaoiuooooooooiuu";
    cout << findMinLength(s);

    return 0;
}

Java

// Java program to find the length of
// the smallest substring of which
// contains all vowels
import java.util.*;
class GFG{

public static int findMinLength(String s)
{
    int n = s.length();

    // Map to store the
    // frequency of vowels
    HashMap<Character,
            Integer> counts = new HashMap<>();

    // Store the indices
    // which contains
    // the vowels
    Vector<Integer> indices = new Vector<Integer>();

    for(int i = 0; i < n; i++)
    {
        if (s.charAt(i) == 'a' ||
            s.charAt(i) == 'e' ||
            s.charAt(i) == 'o' ||
            s.charAt(i) == 'i' ||
            s.charAt(i) == 'u')
        {
            if (counts.containsKey(s.charAt(i)))
            {
                counts.replace(s.charAt(i),
                    counts.get(s.charAt(i)) + 1);
            }
            else
            {
                counts.put(s.charAt(i), 1);
            }
            indices.add(i);
        }
    }

    // If all vowels are not
    // present in the string
    if (counts.size() < 5)
        return -1;

    int flag1 = 0, flag2 = 0;
    int i = 0;
    int j = indices.size() - 1;

    while ((j - i) >= 4)
    {

        // If the frequency of the
        // vowel at i-th index
        // exceeds 1
        if (flag1 == 0 &&
            counts.get(s.charAt(
                indices.get(i))) > 1)
        {

            // Decrease the frequency
            // of that vowel
            counts.replace(s.charAt(indices.get(i)),
                counts.get(s.charAt(indices.get(i))) - 1);

            // Move to the left
            i++;
        }

        // Otherwise set flag1
        else
            flag1 = 1;

        // If the frequency of the
        // vowel at j-th index
        // exceeds 1
        if (flag2 == 0 &&
            counts.get(s.charAt(
                indices.get(j))) > 1)
        {

            // Decrease the frequency
            // of that vowel
            counts.replace(s.charAt(indices.get(j)),
                counts.get(s.charAt(indices.get(j))) - 1);

            // Move to the right
            j--;
        }

        // Otherwise set flag2
        else
            flag2 = 1;

        // If both flag1 and flag2
        // are set, break out of the
        // loop as the substring
        // length cannot be minimized
        if (flag1 == 1 && flag2 == 1)
            break;
    }

    // Return the length of the substring
    return (indices.get(j) - indices.get(i) + 1);
}

// Driver Code
public static void main(String[] args)
{
    String s = "aaeebbeaccaaoiuooooooooiuu";

    System.out.print(findMinLength(s));
}
}

// This code is contributed by divyeshrabadiya07

Python3

# Python3 program to find the
# length of the smallest
# substring of which
# contains all vowels
from collections import defaultdict

def findMinLength(s):

    n = len(s)

    # Map to store the
    # frequency of vowels
    counts = defaultdict(int)

    # Store the indices
    # which contains
    # the vowels
    indices = []

    for i in range(n):

        if (s[i] == 'a' or s[i] == 'e' or
            s[i] == 'o' or s[i] == 'i' or
            s[i] == 'u'):

            counts[s[i]] += 1
            indices.append(i)

    # If all vowels are not
    # present in the string
    if len(counts) < 5:
        return -1

    flag1 = 0
    flag2 = 0
    i = 0
    j = len(indices) - 1

    while (j - i) >= 4:

        # If the frequency of the
        # vowel at i-th index
        # exceeds 1
        if (~flag1 and
             counts[s[indices[i]]] > 1):

            # Decrease the frequency
            # of that vowel
            counts[s[indices[i]]] -= 1

            # Move to the left
            i += 1

        # Otherwise set flag1
        else:
            flag1 = 1

        # If the frequency of the
        # vowel at j-th index
        # exceeds 1
        if (~flag2 and
             counts[s[indices[j]]] > 1):

            # Decrease the frequency
            # of that vowel
            counts[s[indices[j]]] -= 1

            # Move to the right
            j -= 1

        # Otherwise set flag2
        else:
            flag2 = 1

        # If both flag1 and flag2
        # are set, break out of the
        # loop as the substring
        # length cannot be minimized
        if (flag1 and flag2):
            break

    # Return the length of the substring
    return (indices[j] - indices[i] + 1)

# Driver Code
s = "aaeebbeaccaaoiuooooooooiuu"
print(findMinLength(s))

# This code is contributed by
# divyamohan123

C#

// C# program to find the length of
// the smallest substring of which
// contains all vowels
using System;
using System.Collections.Generic;
class GFG{

public static int findMinLength(String s)
{
  int n = s.Length;
  int i = 0;

  // Map to store the
  // frequency of vowels
  Dictionary<char,
             int> counts = new Dictionary<char,
                                          int>();

  // Store the indices
  // which contains
  // the vowels
  List<int> indices = new List<int>();

  for(i = 0; i < n; i++)
  {
    if (s[i] == 'a' ||
        s[i] == 'e' ||
        s[i] == 'o' ||
        s[i] == 'i' ||
        s[i] == 'u')
    {
      if (counts.ContainsKey(s[i]))
      {
        counts[s[i]] = counts[s[i]] + 1;
      }
      else
      {
        counts.Add(s[i], 1);
      }
      indices.Add(i);
    }
  }

  // If all vowels are not
  // present in the string
  if (counts.Count < 5)
    return -1;

  int flag1 = 0, flag2 = 0;
  i = 0;
  int j = indices.Count - 1;

  while ((j - i) >= 4)
  {
    // If the frequency of the
    // vowel at i-th index
    // exceeds 1
    if (flag1 == 0 &&
        counts[s[indices[i]]] > 1)
    {
      // Decrease the frequency
      // of that vowel
      counts[s[indices[i]]]=
             counts[s[indices[i]]] - 1;

      // Move to the left
      i++;
    }

    // Otherwise set flag1
    else
      flag1 = 1;

    // If the frequency of the
    // vowel at j-th index
    // exceeds 1
    if (flag2 == 0 &&
        counts[s[indices[j]]] > 1)
    {
      // Decrease the frequency
      // of that vowel
      counts[s[indices[j]]] =
             counts[s[indices[j]]] - 1;

      // Move to the right
      j--;
    }

    // Otherwise set flag2
    else
      flag2 = 1;

    // If both flag1 and flag2
    // are set, break out of the
    // loop as the substring
    // length cannot be minimized
    if (flag1 == 1 && flag2 == 1)
      break;
  }

  // Return the length of the substring
  return (indices[j] - indices[i] + 1);
}

// Driver Code
public static void Main(String[] args)
{
  String s = "aaeebbeaccaaoiuooooooooiuu";
  Console.Write(findMinLength(s));
}
}

// This code is contributed by shikhasingrajput

Javascript

<script>

// JavaScript Program to find the
// length of the smallest
// substring of which
// contains all vowels

function findMinLength(s)
{
    var n = s.length;

    // Map to store the
    // frequency of vowels
    var counts = new Map();

    // Store the indices
    // which contains
    // the vowels
    var indices = [];

    for (var i = 0; i < n; i++) {

        if (s[i] == 'a' || s[i] == 'e'
            || s[i] == 'o'
            || s[i] == 'i'
            || s[i] == 'u') {

                if(counts.has(s[i]))
                    counts.set(s[i], counts.get(s[i])+1)
                else
                    counts.set(s[i], 1)
            indices.push(i);
        }
    }

    // If all vowels are not
    // present in the string
    if (counts.size < 5)
        return -1;

    var flag1 = 0, flag2 = 0;
    var i = 0;
    var j = indices.length - 1;

    while ((j - i) >= 4) {

        // If the frequency of the
        // vowel at i-th index
        // exceeds 1
        if (!flag1
            && counts.get(s[indices[i]]) > 1) {

            // Decrease the frequency
            // of that vowel
            if(counts.has(s[indices[i]]))
                    counts.set(s[indices[i]],
                    counts.get(s[indices[i]])-1)


            // Move to the left
            i++;
        }

        // Otherwise set flag1
        else
            flag1 = 1;

        // If the frequency of the
        // vowel at j-th index
        // exceeds 1
        if (!flag2
            && counts.get(s[indices[j]]) > 1) {

            // Decrease the frequency
            // of that vowel
            if(counts.has(s[indices[j]]))
                    counts.set(s[indices[j]],
                    counts.get(s[indices[j]])-1)
            // Move to the right
            j--;
        }

        // Otherwise set flag2
        else
            flag2 = 1;

        // If both flag1 and flag2
        // are set, break out of the
        // loop as the substring
        // length cannot be minimized
        if (flag1 && flag2)
            break;
    }

    // Return the length of the substring
    return (indices[j] - indices[i] + 1);
}

var s = "aaeebbeaccaaoiuooooooooiuu";
document.write( findMinLength(s));


</script>

Output:

9

Time Complexity: O(N) 
Auxiliary Space: O(N)

Sliding Window Approach:  

• Maintain an array count to store the frequency of each vowel.
• Maintain a start variable to store the starting index of the current substring.
• Iterate over the string and do the following: 
  • Increase the frequency of the character if it is a vowel.
  • Move start as right as possible based on the condition that either the character at the start is not a vowel or it is a vowel with a frequency greater than 1 which means that it has appeared previously.
  • Once the start is moved as far as possible, check if all vowels are present in the substring between start and the current index.
  • Whenever the above condition satisfies, update the minimum length of such substring obtained
• Print the minimum length of substring obtained or print -1 if no such substring is obtained

The below code implements the above approach: 

C++

// C++ Program to find the
// length of the smallest
// substring of which
// contains all vowels

#include <bits/stdc++.h>
using namespace std;

// Function to return the
// index for respective
// vowels to increase their count
int get_index(char ch)
{
    if (ch == 'a')
        return 0;
    else if (ch == 'e')
        return 1;
    else if (ch == 'i')
        return 2;
    else if (ch == 'o')
        return 3;
    else if (ch == 'u')
        return 4;

    // Returns -1 for consonants
    else
        return -1;
}

// Function to find the minimum length
int findMinLength(string s)
{
    int n = s.size();
    int ans = n + 1;

    // Store the starting index
    // of the current substring
    int start = 0;

    // Store the frequencies of vowels
    int count[5] = { 0 };

    for (int x = 0; x < n; x++) {

        int idx = get_index(s[x]);

        // If the current character
        // is a vowel
        if (idx != -1) {

            // Increase its count
            count[idx]++;
        }

        // Move start as much
        // right as possible
        int idx_start
            = get_index(s[start]);

        while (idx_start == -1
               || count[idx_start] > 1) {

            if (idx_start != -1) {

                count[idx_start]--;
            }

            start++;
            if (start < n)
                idx_start
                    = get_index(s[start]);
        }

        // Condition for valid substring
        if (count[0] > 0 && count[1] > 0
            && count[2] > 0 && count[3] > 0
            && count[4] > 0) {
            ans = min(ans, x - start + 1);
        }
    }

    if (ans == n + 1)
        return -1;

    return ans;
}

// Driver code
int main()
{
    string s = "aaeebbeaccaaoiuooooooooiuu";
    cout << findMinLength(s);

    return 0;
}

Java

// Java program to find the length
// of the smallest subString of
// which contains all vowels
import java.util.*;

class GFG{

// Function to return the
// index for respective
// vowels to increase their count
static int get_index(char ch)
{
    if (ch == 'a')
        return 0;
    else if (ch == 'e')
        return 1;
    else if (ch == 'i')
        return 2;
    else if (ch == 'o')
        return 3;
    else if (ch == 'u')
        return 4;

    // Returns -1 for consonants
    else
        return -1;
}

// Function to find the minimum length
static int findMinLength(String s)
{
    int n = s.length();
    int ans = n + 1;

    // Store the starting index
    // of the current subString
    int start = 0;

    // Store the frequencies of vowels
    int count[] = new int[5];

    for(int x = 0; x < n; x++)
    {
       int idx = get_index(s.charAt(x));

       // If the current character
       // is a vowel
       if (idx != -1)
       {

           // Increase its count
           count[idx]++;
       }

       // Move start as much
       // right as possible
       int idx_start = get_index(s.charAt(start));

       while (idx_start == -1 ||
              count[idx_start] > 1)
       {
           if (idx_start != -1)
           {
               count[idx_start]--;
           }
           start++;
           if (start < n)
               idx_start = get_index(s.charAt(start));
       }

       // Condition for valid subString
       if (count[0] > 0 && count[1] > 0 &&
           count[2] > 0 && count[3] > 0 &&
           count[4] > 0)
       {
           ans = Math.min(ans, x - start + 1);
       }
    }
    if (ans == n + 1)
        return -1;

    return ans;
}

// Driver code
public static void main(String[] args)
{
    String s = "aaeebbeaccaaoiuooooooooiuu";

    System.out.print(findMinLength(s));
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 Program to find the
# length of the smallest
# substring of which
# contains all vowels

# Function to return the
# index for respective
# vowels to increase their count
def get_index(ch):

    if (ch == 'a'):
        return 0
    elif (ch == 'e'):
        return 1
    elif (ch == 'i'):
        return 2
    elif (ch == 'o'):
        return 3
    elif (ch == 'u'):
        return 4

    # Returns -1 for consonants
    else:
        return -1

# Function to find the minimum length
def findMinLength(s):

    n = len(s)
    ans = n + 1

    # Store the starting index
    # of the current substring
    start = 0

    # Store the frequencies
    # of vowels
    count = [0] * 5

    for x in range (n):

        idx = get_index(s[x])

        # If the current character
        # is a vowel
        if (idx != -1):

            # Increase its count
            count[idx] += 1

        # Move start as much
        # right as possible
        idx_start = get_index(s[start])

        while (idx_start == -1 or
               count[idx_start] > 1):

            if (idx_start != -1):

                count[idx_start] -= 1

            start += 1
            if (start < n):
                idx_start = get_index(s[start])

        # Condition for valid substring
        if (count[0] > 0 and count[1] > 0 and
            count[2] > 0 and count[3] > 0 and
            count[4] > 0):
            ans = min(ans, x - start + 1)

    if (ans == n + 1):
        return -1
    return ans

# Driver code
if __name__ == "__main__":

    s = "aaeebbeaccaaoiuooooooooiuu"
    print(findMinLength(s))

# This code is contributed by Chitranayal

C#

// C# program to find the length
// of the smallest subString of
// which contains all vowels
using System;
class GFG{

// Function to return the
// index for respective
// vowels to increase their count
static int get_index(char ch)
{
    if (ch == 'a')
        return 0;
    else if (ch == 'e')
        return 1;
    else if (ch == 'i')
        return 2;
    else if (ch == 'o')
        return 3;
    else if (ch == 'u')
        return 4;

    // Returns -1 for consonants
    else
        return -1;
}

// Function to find the minimum length
static int findMinLength(String s)
{
    int n = s.Length;
    int ans = n + 1;

    // Store the starting index
    // of the current subString
    int start = 0;

    // Store the frequencies of vowels
    int []count = new int[5];

    for(int x = 0; x < n; x++)
    {
        int idx = get_index(s[x]);

        // If the current character
        // is a vowel
        if (idx != -1)
        {

            // Increase its count
            count[idx]++;
        }

        // Move start as much
        // right as possible
        int idx_start = get_index(s[start]);

        while (idx_start == -1 ||
               count[idx_start] > 1)
        {
            if (idx_start != -1)
            {
                count[idx_start]--;
            }
            start++;
            if (start < n)
                idx_start = get_index(s[start]);
        }

        // Condition for valid subString
        if (count[0] > 0 && count[1] > 0 &&
            count[2] > 0 && count[3] > 0 &&
            count[4] > 0)
        {
            ans = Math.Min(ans, x - start + 1);
        }
    }
    if (ans == n + 1)
        return -1;

    return ans;
}

// Driver code
public static void Main(String[] args)
{
    String s = "aaeebbeaccaaoiuooooooooiuu";

    Console.Write(findMinLength(s));
}
}

// This code is contributed by sapnasingh4991

Javascript

<script>
// Javascript program to find the length
// of the smallest subString of
// which contains all vowels

// Function to return the
// index for respective
// vowels to increase their count
function get_index(ch)
{
    if (ch == 'a')
        return 0;
    else if (ch == 'e')
        return 1;
    else if (ch == 'i')
        return 2;
    else if (ch == 'o')
        return 3;
    else if (ch == 'u')
        return 4;

    // Returns -1 for consonants
    else
        return -1;
}

// Function to find the minimum length
function findMinLength(s)
{
    let n = s.length;
    let ans = n + 1;

    // Store the starting index
    // of the current subString
    let start = 0;

    // Store the frequencies of vowels
    let count = new Array(5);
     for(let i = 0; i < 5; i++)
    {
        count[i] = 0;
    }

    for(let x = 0; x < n; x++)
    {
       let idx = get_index(s[x]);

       // If the current character
       // is a vowel
       if (idx != -1)
       {

           // Increase its count
           count[idx]++;
       }

       // Move start as much
       // right as possible
       let idx_start = get_index(s[start]);

       while (idx_start == -1 ||
              count[idx_start] > 1)
       {
           if (idx_start != -1)
           {
               count[idx_start]--;
           }
           start++;
           if (start < n)
               idx_start = get_index(s[start]);
       }

       // Condition for valid subString
       if (count[0] > 0 && count[1] > 0 &&
           count[2] > 0 && count[3] > 0 &&
           count[4] > 0)
       {
           ans = Math.min(ans, x - start + 1);
       }
    }
    if (ans == n + 1)
        return -1;

    return ans;
}

// Driver code
let s = "aaeebbeaccaaoiuooooooooiuu";
document.write(findMinLength(s));

// This code is contributed by avanitrachhadiya2155
</script>

Output:

9

Time Complexity: O(N)

Auxiliary Space: O(N)