Number of Strings Which Can Be Rearranged to Contain Substring - Problem

You are given an integer n. A string s is called good if it contains only lowercase English characters and it is possible to rearrange the characters of s such that the new string contains "leet" as a substring.

For example:

The string "lteer" is good because we can rearrange it to form "leetr".
"letl" is not good because we cannot rearrange it to contain "leet" as a substring.

Return the total number of good strings of length n. Since the answer may be large, return it modulo 10^9 + 7.

A substring is a contiguous sequence of characters within a string.

Input & Output

Example 1 — Minimum Valid Length

$ Input: n = 4

› Output: 12

💡 Note: With n=4, we need exactly "leet" or rearrangements that contain these 4 letters. There are 4! = 24 arrangements of "leet", but since we have repeated 'e', we get 4!/2! = 12 unique arrangements.

Example 2 — Length 5

$ Input: n = 5

› Output: 960

💡 Note: With n=5, we can have "leet" plus one additional letter (any of 26 choices). The additional letter can be placed in any of 5 positions, giving us more arrangements.

Example 3 — Too Short

$ Input: n = 3

› Output: 0

💡 Note: With only 3 characters, it's impossible to form "leet" which requires at least 4 characters (l, e, e, t).

Constraints

1 ≤ n ≤ 10⁵

Visualization

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Asked in

M Meta 15 G Google 12 a Amazon 8

The key insight is to use inclusion-exclusion principle to count strings containing at least 1 'l', 2 'e', and 1 't'. Best approach uses mathematical formula with modular exponentiation. Time: O(1), Space: O(1)

Common Approaches

✓ Brute Force Enumeration

⏱️ Time: O(26^n) Space: O(n)

Generate every possible string of length n using lowercase letters, then check if each string contains enough characters (at least 1 'l', 2 'e', 1 't') to rearrange into containing "leet" as a substring.

Inclusion-Exclusion Principle

⏱️ Time: O(1) Space: O(1)

Use inclusion-exclusion principle to count strings with at least the required letters. Calculate total strings minus those missing required letters (l, e, e, t).

Brute Force Enumeration — Algorithm Steps

Generate all 26^n possible strings
For each string, count occurrences of 'l', 'e', 't'
Check if counts are sufficient: l≥1, e≥2, t≥1

Visualization

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Step-by-Step Walkthrough

Generate

Create all 26^n possible strings

Count Letters

Count l, e, t in each string

Validate

Check if l≥1, e≥2, t≥1

Code -

solution.c — C

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

const int MOD = 1000000007;
int count_result = 0;

void generateStrings(char* current, int pos, int remaining) {
    if (remaining == 0) {
        int lCount = 0, eCount = 0, tCount = 0;
        for (int i = 0; i < pos; i++) {
            if (current[i] == 'l') lCount++;
            else if (current[i] == 'e') eCount++;
            else if (current[i] == 't') tCount++;
        }
        
        if (lCount >= 1 && eCount >= 2 && tCount >= 1) {
            count_result = (count_result + 1) % MOD;
        }
        return;
    }
    
    for (char c = 'a'; c <= 'z'; c++) {
        current[pos] = c;
        generateStrings(current, pos + 1, remaining - 1);
    }
}

int solution(int n) {
    if (n < 4) return 0;
    count_result = 0;
    char buffer[n + 1];
    buffer[n] = '\0';
    generateStrings(buffer, 0, n);
    return count_result;
}

int main() {
    int n;
    scanf("%d", &n);
    printf("%d\n", solution(n));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(26^n)

Generate 26^n strings and check each one

✓ Linear Growth

Space Complexity

O(n)

Store current string being generated

✓ Linear Space

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Number of Strings Which Can Be Rearranged to Contain Substring - Problem

Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force Enumeration — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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