Find frequency of all characters across all substrings of given string

Given a string S containing all lowercase characters and its length N. Find frequency of all characters across all substrings of the given string.

Examples:

Input: N = 3, S = "aba"
Output:
a 6
b 4
Explanation: The substrings are: a, b, a, ab, ba, aba. The frequency of each character: a = 6, b = 4. Hence the output is "a 6" and "b 4".

Input: N=10, S="ccdbccbbca"
Output:
a 10
b 80
c 106
d 24

Input: N=8, S="dddbabcd"
Output:
a 20
b 38
c 14
d 48

Approach: To solve the problem, follow the below idea:

The problem can be solved by finding the contribution of each character across all substrings. Let's say, a character ch is present at any index idx then this character will occur in all the substrings that start at index <= idx and end at index >= idx. So, this character will contribute one occurrence in all these subtrings.

The count of such substrings can be calculated using Combinatorics. For the starting index, we have total (idx + 1) choices ranging from (0 to idx) and for the ending index, we have total (N - idx) choices ranging from (idx, N - 1). So, the total number of such substrings will be: (idx + 1) * (N - idx). In all these substrings, character ch will occur at least once. Similarly, we can find the contribution of all the characters to get the final answer.

Step-by-step algorithm:

• Run a loop to iterate over the entire string.
• For each index idx, count the number of starting indices for subarrays that contain S[idx] in them, that is (idx + 1).
• For each index idx, count the number of ending indices for subarrays that contain S[idx] in them, that is (N - idx).
• In order to get all the possible subarrays multiply the number of possible starting indices with number of possible ending indices.
• Similarly, count the contribution of all characters at every index to get the final answer.

import java.util.Arrays;

public class AllCharOccurencesInAllSubstrings {

    // function to count the contribution of every character
    static void allCharOccurrencesInAllSubstrings(
            String S, int N, long[] freq) {
        for (int idx = 0; idx < N; idx++) {
            char ch = S.charAt(idx);
            // count subarrays containing the character at index
            // idx
            long count = (idx + 1L) * (N - idx);
            freq[ch - 'a'] += count;
        }
    }

    public static void main(String[] args) {
        // Sample Input
        String S = "ccdbccbbca";
        long[] freq = new long[26];
        Arrays.fill(freq, 0);
        allCharOccurrencesInAllSubstrings(S, S.length(), freq);
        for (char ch = 'a'; ch <= 'z'; ch++) {
            if (freq[ch - 'a'] > 0)
                System.out.println(ch + "  " + freq[ch - 'a']);
        }
    }
}

// This code is contributed by shivamgupta0987654321

// Function to count the contribution of every character
function allCharOccurrencesInAllSubstrings(S, N, freq) {
    for (let idx = 0; idx < N; idx++) {
        let ch = S.charAt(idx);
        // count subarrays containing the character at index idx
        let count = (idx + 1) * (N - idx);
        freq[ch.charCodeAt(0) - 'a'.charCodeAt(0)] += count;
    }
}

// Main function
function main() {
    // Sample Input
    let S = "ccdbccbbca";
    let freq = new Array(26).fill(0);
    allCharOccurrencesInAllSubstrings(S, S.length, freq);
    for (let ch = 'a'.charCodeAt(0); ch <= 'z'.charCodeAt(0); ch++) {
        if (freq[ch - 'a'.charCodeAt(0)] > 0)
            console.log(String.fromCharCode(ch) + "  " + freq[ch - 'a'.charCodeAt(0)]);
    }
}

// Call the main function
main();

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

// function to count the contribution of every character
void AllCharOccurencesInAllSubtrings(
    string S, int N, vector<long long>& freq)
{
    for (int idx = 0; idx < N; idx++) {
        char ch = S[idx];
        // count subarrays containing the character at index
        // idx
        long long count = (idx + 1) * (N - idx);
        freq[ch - 'a'] += count;
    }
}

int main()
{
    // Sample Input
    string S = "ccdbccbbca";
    vector<long long> freq(26, 0);
    AllCharOccurencesInAllSubtrings(S, S.length(), freq);
    for (char ch = 'a'; ch <= 'z'; ch++) {
        if (freq[ch - 'a'] > 0)
            cout << ch << " " << freq[ch - 'a'] << endl;
    }
    return 0;
}
// this code is contributed by prophet1999

# function to count the contribution of every character
def all_char_occurrences_in_all_substrings(S, N, freq):
    for idx in range(N):
        ch = S[idx]
        # count subarrays containing the character at index idx
        count = (idx + 1) * (N - idx)
        freq[ord(ch) - ord('a')] += count

# Sample Input
S = "ccdbccbbca"
freq = [0] * 26

# Call the function to count character occurrences in all substrings
all_char_occurrences_in_all_substrings(S, len(S), freq)

# Print the result
for ch in range(ord('a'), ord('z') + 1):
    if freq[ch - ord('a')] > 0:
        print(chr(ch), freq[ch - ord('a')])

a  10
b  80
c  106
d  24

Time Complexity: O(N), where N is the length of input string S.
Auxiliary Space: O(1)

prophet1999

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Article Tags :

• Strings
• Combinatorial
• DSA

Practice Tags :

• Combinatorial
• Strings