# Count the number of vowels occurring in all the substrings of given string

Given a string of length N of lowercase characters containing 0 or more vowels, the task is to find the count of vowels that 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(N^{3}) 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 occurrences of vowels` ` ` `return` `sum;` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `main(String args[])` ` ` `{` ` ` `String s = ` `"daceh"` `;` ` ` `System.out.println(vowel_calc(s));` ` ` `}` `}` |

## Python3

`# Python 3 implementation of` `# a more efficient approach.` `# return sum of all occurrences of all vowels` `def` `sumVowel(string):` ` ` `n ` `=` `len` `(string)` ` ` `sum` `=` `0` ` ` `string ` `=` `string.lower()` ` ` `# iterate through every character in the string` ` ` `for` `i ` `in` `range` `(` `0` `, n):` ` ` `s ` `=` `string[i]` ` ` `# checks if the character is a vowel or not` ` ` `if` `(s` `=` `=` `"a"` `or` `s ` `=` `=` `"e"` `or` `s ` `=` `=` `"i"` `or` `s ` `=` `=` `"o"` `or` `s ` `=` `=` `"u"` `):` ` ` `# uses below expression to calculate the count` ` ` `# of all occurrences of character in substrings` ` ` `# of the string` ` ` `sum` `+` `=` `((n ` `-` `i) ` `*` `(i ` `+` `1` `)) ` ` ` `# return the total sum of occurrence` ` ` `return` `sum` `#driver code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `#input string here` ` ` `string ` `=` `"abhay"` ` ` `#print returned sum` ` ` `print` `(sumVowel(string))` `# This code is contributed by` `# Abhay Subramanian K` |

## C#

`// C# implementation of the above approach` ` ` `using` `System;` ` ` `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[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 occurrences of vowels` ` ` `return` `sum;` ` ` `}` ` ` ` ` `// Driver Code` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `string` `s = ` `"daceh"` `;` ` ` `Console.Write(vowel_calc(s));` ` ` `}` `}` |

## PHP

`<?php` `// PHP implementation of the above approach` `// Returns the total sum of` `// occurrences of all vowels` `function` `vowel_calc(` `$s` `)` `{` ` ` `$n` `= ` `strlen` `(` `$s` `);` ` ` `$arr` `= ` `array` `();` ` ` `for` `(` `$i` `= 0; ` `$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] - ` `$i` `;` ` ` `}` ` ` `$sum` `= 0;` ` ` `for` `(` `$i` `= 0; ` `$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` `$s` `= ` `"daceh"` `;` `echo` `(vowel_calc(` `$s` `));` `// This code is contributed by Shivi_Aggarwal` `?>` |

## Javascript

`<script>` `// Javascript implementation of the above approach` `// Returns the total sum of` `// occurrences of all vowels` `function` `vowel_calc(s)` `{` ` ` `var` `n = s.length;` ` ` `var` `arr = [];` ` ` `for` `(` `var` `i = 0; i < n; i++) {` ` ` `if` `(i == 0)` ` ` `// No. of occurrences of 0th character` ` ` `// in all the substrings` ` ` `arr.push(n);` ` ` `else` ` ` `// No. of occurrences of the ith character` ` ` `// in all the substrings` ` ` `arr.push((n - i) + arr[i - 1] - i);` ` ` `}` ` ` `var` `sum = 0;` ` ` `for` `(` `var` `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` `var` `s = ` `"daceh"` `;` `document.write( vowel_calc(s));` `// This code is contributed by importantly.` `</script>` |

**Output**

16

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

**Another efficient approach without additional space:**

The idea is based on the number of times a particular character can occur across all the substrings. If we know a particular character occurs x times in all the substrings, then we can just check if it is a vowel, and add the count to the sum.

A particular character A[i] can be part of substrings in the following two ways:

1. Substrings starting with A[i].

2. Substrings that contain character A[i] but do not start with A[i]

**Observation: **

Let’s discuss this with an example:**s = “abcdefg”**

**Occurrences of character ‘a’:**

**Substrings starting with ‘a’:**“a”, “ab”, “abc” … “abcdefg” => 7 Count.- No Substrings has ‘a’ in the middle.

Total occurrences for a = 7 =**7 * 1**

**Occurrences of character ‘b’:**

**Substrings starting with ‘b’**: “b”, “bc”, … “bcdefg” => 6 Count.**Substrings containing ‘b’ within them:**

‘b’ can also exist within substrings starting from a. Their count will be (7 – 1) = 6

e.g.: “ab”, “abc” … “abcdefg”

Total occurrences for b = 6 + 6 =**6 * 2**

**Occurrences of character ‘c’:**

**Substrings starting with ‘c’**: “c”, “cd” … “cdefg” => 5 Count.**Substrings containing ‘c’ within them:**

‘c’ can also exist within substrings**starting from a**. e.g. “abc”, “abcd” … “abcdefg” (except ‘a’ and ‘ab’) = (7 – 2) = 5

‘c’ can also exist within substrings**starting from b**. e.g. “bc”, “bcd” … “bcdefg” (except ‘b’) = (6 – 1) = 5

Total occurrences for c = 5 + 5 + 5 =**5 * 3**

Here, a pattern in the number of occurrences can be observed, in which the total occurrences of the character will be equal to the product of the number of characters on the right side with the number of characters on the left side (including the character itself). So, whenever a vowel is encountered we can add the product of the count of right and left characters in the answer.

Below is the implementation of the above approach:

## Java

`// JAVA Code to implement the above approach` `import` `java.io.*;` `import` `java.util.*;` `public` `class` `Gfg {` ` ` `// Function to return the total sum of` ` ` `// occurrences of all vowels` ` ` `static` `int` `vowel_calc(String s)` ` ` `{` ` ` `int` `n = s.length();` ` ` `// Variable to store the answer` ` ` `int` `totalOccurrences = ` `0` `;` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) {` ` ` `// Count of total occurrences of` ` ` `// current character` ` ` `int` `totalOccurrenceForCharInAllSubstrings` ` ` `= (i + ` `1` `) * (n - i);` ` ` `char` `ch = s.charAt(i);` ` ` `// if the current character is a vowel` ` ` `if` `(ch == ` `'a'` `|| ch == ` `'e'` `|| ch == ` `'i'` ` ` `|| ch == ` `'o'` `|| ch == ` `'u'` `) {` ` ` `// Add the occurrences of current` ` ` `// character to the answer` ` ` `totalOccurrences` ` ` `+= totalOccurenceForCharInAllSubstrings;` ` ` `}` ` ` `}` ` ` `// Return the total sum` ` ` `// of occurrences of vowels` ` ` `return` `totalOccurrences;` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `main(String args[])` ` ` `{` ` ` `String s = ` `"abcde"` `;` ` ` ` ` `// Function Call` ` ` `System.out.println(vowel_calc(s));` ` ` `}` `}` |

**Output**

10

**Time complexity:** O(N)**Auxiliary Space:** O(1)