Count Palindromic Strings with Constraints

Problem Statement

Given two integers n and k, consider an alphabet consisting of the first k lowercase English letters.

We need to find the number of palindromic strings whose length is less than or equal to n, subject to the following conditions:

Every character must belong to the first k lowercase English letters.
No character can appear more than twice.
Return the answer modulo 10^9 + 7.

Example 1

Input:
n = 3
k = 2

Output:
6

The valid strings are:

a
b
aa
bb
aba
bab

Therefore, the answer is 6.

Example 2

Input:
n = 4
k = 3

Output:
18

The valid strings are:

a
b
c

aa
bb
cc

aba
aca
bab
bcb
cac
cbc

abba
acca
baab
bccb
caac
cbbc

Therefore, the answer is 18.

Important Observation

The most important part of this problem is understanding the structure of a palindrome.

A palindrome reads the same from left to right and right to left.

For example:

aba
abba
abcba

Because every character can appear at most twice, a character used in a palindrome can generally appear:

Twice, as a mirrored pair.
Once, only when the palindrome has odd length, in the middle.

This allows us to divide the problem into two cases:

Even-length palindromes
Odd-length palindromes

Case 1: Even-Length Palindromes

Suppose the length is:

2 * m

For example:

abba

The characters are:

a b b a

Here, a appears twice and b appears twice.

Because no character can appear more than twice, all m characters in the first half must be different.

For example, with k = 3:

a b

produces:

abba

The first half determines the second half.

If we choose:

a, b

the palindrome is:

abba

If we choose:

a, c

the palindrome is:

acca

Therefore, for an even-length palindrome of length 2m, we need to select and arrange m distinct characters from k characters.

This is a permutation.

The number of possibilities is:

P(k, m)

or:

k * (k - 1) * (k - 2) * ... * (k - m + 1)

Example

For:

k = 3
m = 2

we have:

P(3, 2)
= 3 * 2
= 6

The six palindromes are:

abba
acca
baab
bccb
caac
cbbc

Case 2: Odd-Length Palindromes

Now consider a palindrome of length:

2 * m + 1

For example:

aba

The structure is:

a b a

The middle character can occur only once.

The remaining m characters form pairs.

For example:

abcba

has the structure:

a b c b a

Here:

a appears twice
b appears twice
c appears once

The middle character can be selected in k ways.

After selecting the middle character, we need to select and arrange m different characters from the remaining k - 1 characters.

Therefore:

Count = k * P(k - 1, m)

Example of Odd-Length Calculation

Suppose:

n = 5
k = 3

For length 3:

m = 1

Formula:

k * P(k - 1, 1)

= 3 * 2

= 6

The palindromes are:

aba
aca
bab
bcb
cac
cbc

For length 5:

m = 2

Formula:

3 * P(2, 2)

= 3 * 2 * 1

= 6

The palindromes are:

abcba
acbca
bacab
bcacb
cabac
cbabc

Combining Both Cases

We need all palindromes whose length is at most n.

Therefore, we add:

Length 1
Length 2
Length 3
Length 4
...
Length n

But instead of calculating every palindrome explicitly, we use the formulas.

For even length:

2m

Count = P(k, m)

For odd length:

2m + 1

Count = k * P(k - 1, m)

Java Solution

class Solution {
    static final long MOD = 1000000007L;

    public int palindromicStrings(int n, int k) {
        long ans = 0;

        // Even length: 2 * len
        // Count = P(k, len)
        long perm = 1;

        for (int len = 1; len <= k && 2 * len <= n; len++) {
            perm = (perm * (k - len + 1)) % MOD;
            ans = (ans + perm) % MOD;
        }

        // Odd length: 2 * len + 1
        // Count = k * P(k - 1, len)
        perm = 1;

        for (int len = 0; len <= k - 1 && 2 * len + 1 <= n; len++) {

            if (len > 0) {
                perm = (perm * (k - len)) % MOD;
            }

            long count = (k * perm) % MOD;
            ans = (ans + count) % MOD;
        }

        return (int) ans;
    }
}

Detailed Code Explanation

1. Modulo Value

static final long MOD = 1000000007L;

The number of possible strings can become very large.

The problem asks us to return the answer modulo:

10^9 + 7

So we perform modulo after every multiplication and addition.

2. Store the Answer

long ans = 0;

ans stores the total number of valid palindromic strings.

We use long because multiplication can temporarily produce values larger than the range of int.

3. Calculate Even-Length Palindromes

long perm = 1;

for (int len = 1; len <= k && 2 * len <= n; len++) {
    perm = (perm * (k - len + 1)) % MOD;
    ans = (ans + perm) % MOD;
}

Here:

2 * len

represents the palindrome length.

For example:

len = 1 → length 2
len = 2 → length 4
len = 3 → length 6

The variable perm maintains:

P(k, len)

Iteration Example

Suppose:

k = 4

Initially:

perm = 1

For len = 1:

perm = 1 * 4
      = 4

So:

P(4,1) = 4

For len = 2:

perm = 4 * 3
      = 12

So:

P(4,2) = 12

For len = 3:

perm = 12 * 2
      = 24

So:

P(4,3) = 24

This avoids calculating factorials separately.

4. Calculate Odd-Length Palindromes

perm = 1;

for (int len = 0; len <= k - 1 && 2 * len + 1 <= n; len++) {

Here:

2 * len + 1

represents the odd palindrome length.

For example:

len = 0 → length 1
len = 1 → length 3
len = 2 → length 5
len = 3 → length 7

5. Calculate P(k - 1, len)

if (len > 0) {
    perm = (perm * (k - len)) % MOD;
}

For odd-length palindromes, one character is reserved for the middle.

Therefore, the remaining characters are selected from:

k - 1

characters.

The required permutation is:

P(k - 1, len)

The code calculates this incrementally.

For example, if:

k = 4

then:

P(3, 0) = 1
P(3, 1) = 3
P(3, 2) = 6
P(3, 3) = 6

6. Select the Middle Character

long count = (k * perm) % MOD;

There are k choices for the middle character.

Therefore:

Odd palindrome count
=
k * P(k - 1, len)

For example:

k = 3
len = 1

Then:

P(2,1) = 2

Therefore:

count = 3 * 2
      = 6

7. Add to the Answer

ans = (ans + count) % MOD;

We add the number of palindromes of the current length to the total answer.

Dry Run

Consider:

n = 4
k = 3

The alphabet is:

a, b, c

Length 1

Odd case:

len = 0

count = 3 * P(2,0)
      = 3 * 1
      = 3

Strings:

a
b
c

Total:

Length 2

Even case:

len = 1

count = P(3,1)
      = 3

Strings:

aa
bb
cc

Total:

3 + 3 = 6

Length 3

Odd case:

len = 1

count = 3 * P(2,1)
      = 3 * 2
      = 6

Strings:

aba
aca
bab
bcb
cac
cbc

Total:

6 + 6 = 12

Length 4

Even case:

len = 2

count = P(3,2)
      = 3 * 2
      = 6

Strings:

abba
acca
baab
bccb
caac
cbbc

Total:

12 + 6 = 18

Therefore:

Answer = 18

Why We Do Not Need Dynamic Programming

Although the problem has a Dynamic Programming tag, the constraints and palindrome structure allow us to solve it directly using combinatorics.

The important fact is:

Once the first half of the palindrome is chosen, the second half is completely determined.

For example:

First half: abc

The complete palindrome is:

abccba

Therefore, we only need to count valid ways to construct the first half.

Because each character can occur at most twice, the characters in the first half must be distinct.

This converts the problem into a permutation-counting problem.

Why `n <= 2 * k` Matters

A palindrome of even length 2m needs m different characters.

Since there are only k available characters:

m <= k

Similarly, an odd palindrome of length 2m + 1 needs m paired characters plus one middle character.

The constraint:

n <= 2 * k

ensures that we never need more than k characters in the paired portion.

Complexity Analysis

The even-length loop runs at most k times.

The odd-length loop also runs at most k times.

Therefore:

Time Complexity: O(k)

Only a few variables are used:

ans
perm
count

Therefore:

Auxiliary Space: O(1)

This is better than the expected:

Time: O(k²)
Space: O(k²)

Final Formula

For every possible palindrome length:

Even length

For length:

2m

the number of palindromes is:

P(k, m)

Odd length

For length:

2m + 1

the number of palindromes is:

k × P(k - 1, m)

Therefore, the complete answer is:

Σ P(k, m)       for every 2m <= n

+

Σ k × P(k-1, m) for every 2m+1 <= n

All calculations are performed modulo:

1,000,000,007

Key Takeaway

The trick is not to generate the strings.

Instead:

Divide the palindrome into even and odd lengths.
For an even palindrome, choose and arrange distinct characters for the first half.
For an odd palindrome, choose the middle character and then arrange distinct characters for the first half.
Use permutations instead of generating strings.
Calculate permutations incrementally to achieve O(k) time and O(1) extra space.

This turns what looks like a string-generation problem into a simple combinatorial counting problem.

Problem Statement

Example 1

Example 2

Important Observation

Case 1: Even-Length Palindromes

Example

Case 2: Odd-Length Palindromes

Example of Odd-Length Calculation

Combining Both Cases

Java Solution

Detailed Code Explanation

1. Modulo Value

2. Store the Answer

3. Calculate Even-Length Palindromes

Iteration Example

4. Calculate Odd-Length Palindromes

5. Calculate P(k - 1, len)

6. Select the Middle Character

7. Add to the Answer

Dry Run

Length 1

Length 2

Length 3

Length 4

Why We Do Not Need Dynamic Programming

Why n <= 2 * k Matters

Complexity Analysis

Final Formula

Even length

Odd length

Key Takeaway

Why `n <= 2 * k` Matters