Count Substrings Starting and Ending with Given Character

Count Substrings Starting and Ending with Given Character - Problem

Given a string s and a character c, you need to find the total number of substrings that start and end with the character c.

A substring is a contiguous sequence of characters within a string. For example, in the string "hello", "ell" is a substring but "hlo" is not.

Key Points:

Both single character c counts as a valid substring (starts AND ends with c)
The substring must be contiguous
We're looking for ALL possible substrings, not just unique ones

Example: If s = "abcab" and c = 'a', the valid substrings are: "a" (index 0), "abca" (index 0-3), "a" (index 3). So the answer is 3.

Input & Output

example_1.py — Basic Case

$ Input: s = "ababa", c = 'a'

› Output: 6

💡 Note: The substrings starting and ending with 'a' are: "a" (index 0), "aba" (index 0-2), "ababa" (index 0-4), "a" (index 2), "aba" (index 2-4), "a" (index 4). Total = 6.

example_2.py — Single Character

$ Input: s = "aa", c = 'a'

› Output: 3

💡 Note: The substrings are: "a" (index 0), "aa" (index 0-1), "a" (index 1). With 2 occurrences of 'a', we get 2×3/2 = 3 substrings.

example_3.py — No Matches

$ Input: s = "xyz", c = 'a'

› Output: 0

💡 Note: There are no occurrences of 'a' in the string, so no substrings can start and end with 'a'. The result is 0.

Constraints

1 ≤ s.length ≤ 10⁵
s consists of lowercase English letters
c is a lowercase English letter
Time limit: 1 second

Visualization

Tap to expand

Understanding the Visualization

Identify all positions

Mark every position where character 'c' appears in the string

Count total positions

Let n be the total number of positions where 'c' appears

Apply combination formula

Calculate n×(n+1)/2 which gives us all possible pairs including self-pairs

Key Takeaway

🎯 Key Insight: Instead of checking all possible substrings, we use combinatorics. If there are n occurrences of character c, there are exactly n×(n+1)/2 valid substrings - this includes both single characters and all possible pairs between positions.

Asked in

G Google 15 a Amazon 12 f Meta 8 ⊞ Microsoft 6

The optimal solution uses a mathematical approach: count the occurrences of character c (let's call it n), then apply the formula n×(n+1)/2 to get the total number of valid substrings. This works because we're essentially choosing pairs from n positions with repetition allowed, which is a classic combinatorics problem.

Common Approaches

Approach	Time	Space	Notes
✓ Mathematical Formula (Optimal)	O(n)	O(1)	Count occurrences of character c, then use formula n×(n+1)/2 for total combinations
Brute Force (Nested Loops)	O(n²)	O(1)	Check every possible substring to see if it starts and ends with the target character

Mathematical Formula (Optimal) — Algorithm Steps

Count total occurrences of character c in the string
Apply the combination formula: n×(n+1)/2
This counts all ways to choose 2 positions from n positions (with repetition allowed)
Return the calculated result

Visualization

Tap to expand

Step-by-Step Walkthrough

Count target characters

Find all positions where character c appears

Apply combination formula

Calculate n×(n+1)/2 where n is the count

Get result instantly

No need to check individual substrings

Code -

solution.c — C

#include <stdio.h>
#include <string.h>

long long countSubstrings(char* s, char c) {
    // Count occurrences of character c
    long long count = 0;
    int n = strlen(s);
    
    for (int i = 0; i < n; i++) {
        if (s[i] == c) {
            count++;
        }
    }
    
    // Use formula: n*(n+1)/2 for combinations
    return count * (count + 1) / 2;
}

int main() {
    char s[100001], c_str[10];
    fgets(s, sizeof(s), stdin);
    fgets(c_str, sizeof(c_str), stdin);
    
    // Remove quotes and newlines
    int len = strlen(s);
    if (s[len-1] == '\n') s[len-1] = '\0';
    memmove(s, s+1, strlen(s));
    s[strlen(s)-1] = '\0';
    
    char c = c_str[1];
    
    printf("%lld\n", countSubstrings(s, c));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the string to count occurrences of character c

✓ Linear Growth

Space Complexity

O(1)

Only storing the count variable

✓ Linear Space

23.4K Views

Medium Frequency

~15 min Avg. Time

890 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Mathematical Formula (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler