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.
- 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 ith and jth 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 ith or jth 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 vowelsimport 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 defaultdictdef 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 Codes = "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 vowelsusing 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 vowelsfunction 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)
- 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 countint 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 lengthint 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 codeint main(){ string s = "aaeebbeaccaaoiuooooooooiuu"; cout << findMinLength(s); return 0;} |
Java
// Java program to find the length // of the smallest subString of // which contains all vowelsimport java.util.*;class GFG{// Function to return the// index for respective// vowels to increase their countstatic 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 lengthstatic 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 codepublic 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 countdef 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 lengthdef 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 codeif __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 vowelsusing System;class GFG{// Function to return the// index for respective// vowels to increase their countstatic 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 lengthstatic 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 codepublic 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 countfunction 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 lengthfunction 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 codelet s = "aaeebbeaccaaoiuooooooooiuu";document.write(findMinLength(s));// This code is contributed by avanitrachhadiya2155</script> |
Output:
9
Time Complexity: O(N)
Auxiliary Space: O(N)
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