Given a string of length N of lowercase characters containing 0 or more vowels, the task is to find the count of vowels occurred in all the substrings of the given string.
Examples:
Input: str = “abc”
Output: 3
The given string “abc” contains only one vowel = ‘a’
Substrings of “abc” are = {“a”, “b”, “c”, “ab”, “bc, “abc”}
Hence, the sum of occurrences of the vowel in these strings = 3.(‘a’ occurred 3 times)Input: str = “daceh”
Output: 16
Naive Approach: Given a string of length N, the number of substrings that can be formed=N(N+1)/2. A simple solution is for each substring, we count the occurrences of the vowels and add them to get the result. The time complexity of this approach is O(N3) which is not suitable for large values of N.
Efficient Approach: The idea is to use a prefix sum array-based technique where we store the occurrences of each character in all the substrings concatenated.
- For the first character,
no. of occurrences = no. of substrings starting with the first character = N.
- For each of the following characters, we store the
no. of substrings starting with that character + the number of substrings formed by the previous characters containing this character – the number of substrings formed by the previous characters only.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std; // Returns the total sum of // occurrences of all vowels int vowel_calc(string s) { int n = s.length(); vector<int> arr; for (int i = 0; i < n; i++) { if (i == 0) // No. of occurrences of 0th character // in all the substrings arr.push_back(n); else // No. of occurrences of the ith character // in all the substrings arr.push_back((n - i) + arr[i - 1] - i); } int sum = 0; for (int i = 0; i < n; i++) // Check if ith character is a vowel if (s[i] == 'a' || s[i] == 'e' || s[i] == 'i' || s[i] == 'o' || s[i] == 'u') sum += arr[i]; // Return the total sum // of occurrences of vowels return sum; } // Driver code int main() { string s = "daceh"; cout << vowel_calc(s) << endl; return 0; } |
Java
// Java implementation of the above approach import java.io.*; import java.util.*; public class Gfg { // Returns the total sum of // occurrences of all vowels static int vowel_calc(String s) { int n = s.length(); int arr[] = new int[n]; for (int i = 0; i < n; i++) { if (i == 0) // No. of occurrences of 0th character // in all the substrings arr[i] = n; else // No. of occurrences of ith character // in all the substrings arr[i] = (n - i) + arr[i - 1] - i; } int sum = 0; for (int i = 0; i < n; i++) { char ch = s.charAt(i); // Check if ith character is a vowel if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u') sum += arr[i]; } // Return the total sum // of occurences of vowels return sum; } // Driver Code public static void main(String args[]) { String s = "daceh"; System.out.println(vowel_calc(s)); } } |
16
Time Complexity: O(N)
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Improved By : SouravAChowdhury_97