Number of Strings Which Can Be Rearranged to Contain Substring

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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Understanding the Visualization

Input

Length n for string generation

Requirement

Must contain letters to form "leet"

Count

Total valid rearrangements

Key Takeaway

🎯 Key Insight: Use inclusion-exclusion to count strings with required letter frequencies

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

Approach	Time	Space	Notes
✓ Inclusion-Exclusion Principle	O(1)	O(1)	Count valid strings using mathematical formula with inclusion-exclusion
Brute Force Enumeration	O(26^n)	O(n)	Generate all possible strings and check if each can form "leet"

Inclusion-Exclusion Principle — Algorithm Steps

Calculate total possible strings: 26^n
Subtract strings missing at least one required letter type
Add back over-subtracted intersections
Apply inclusion-exclusion formula

Visualization

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Step-by-Step Walkthrough

Total Strings

Start with all 26^n possible strings

Subtract Invalid

Remove strings missing required letters

Add Back

Correct for over-subtraction using inclusion-exclusion

Code -

solution.c — C

#include <stdio.h>

const int MOD = 1000000007;

long long power(long long base, long long exp, long long mod) {
    long long result = 1;
    base = base % mod;
    while (exp > 0) {
        if (exp % 2 == 1) {
            result = (result * base) % mod;
        }
        exp = exp >> 1;
        base = (base * base) % mod;
    }
    return result;
}

int max(int a, int b) {
    return a > b ? a : b;
}

int solution(int n) {
    if (n < 4) return 0;
    
    long long total = power(26, n, MOD);
    long long missing_l = power(25, n, MOD);
    long long missing_t = power(25, n, MOD);
    long long missing_2e = (power(25, n, MOD) + ((long long)n * power(25, max(0, n-1), MOD)) % MOD) % MOD;
    
    long long missing_l_t = power(24, n, MOD);
    long long missing_l_2e = (power(24, n, MOD) + ((long long)n * power(24, max(0, n-1), MOD)) % MOD) % MOD;
    long long missing_t_2e = (power(24, n, MOD) + ((long long)n * power(24, max(0, n-1), MOD)) % MOD) % MOD;
    
    long long missing_l_t_2e = (power(23, n, MOD) + ((long long)n * power(23, max(0, n-1), MOD)) % MOD) % MOD;
    
    long long result = (total - missing_l - missing_t - missing_2e + missing_l_t + missing_l_2e + missing_t_2e - missing_l_t_2e) % MOD;
    
    return (result + MOD) % MOD;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(1)

Direct mathematical formula calculation with modular exponentiation

✓ Linear Growth

Space Complexity

O(1)

Only stores constant variables for calculation

✓ Linear Space

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Medium Frequency

~25 min Avg. Time

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Inclusion-Exclusion Principle — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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