Given a string of digits s
, return the number of palindromic subsequences of s
* having length *5
. Since the answer may be very large, return it modulo 10<sup>9</sup> + 7
.
Note:
Example 1:
Input: s = "103301"
Output: 2
Explanation:
There are 6 possible subsequences of length 5: "10330","10331","10301","10301","13301","03301".
Two of them (both equal to "10301") are palindromic.
Example 2:
Input: s = "0000000"
Output: 21
Explanation: All 21 subsequences are "00000", which is palindromic.
Example 3:
Input: s = "9999900000"
Output: 2
Explanation: The only two palindromic subsequences are "99999" and "00000".
Constraints:
1 <= s.length <= 10<sup>4</sup>
s
consists of digits.def count_palindromic_subsequences(s: str) -> int:
"""Counts the number of palindromic subsequences of length 5 in the string s.
Args:
s: The input string of digits.
Returns:
The number of palindromic subsequences of length 5, modulo 10^9 + 7.
"""
n = len(s)
count = 0
mod = 10**9 + 7
for i in range(n):
for j in range(i + 1, n):
for k in range(j + 1, n):
for l in range(k + 1, n):
for m in range(l + 1, n):
if s[i] == s[m] and s[j] == s[l]:
count = (count + 1) % mod
return count
# Example Usage
s = "103301"
result = count_palindromic_subsequences(s)
print(f"The number of palindromic subsequences of length 5 in '{s}' is: {result}")
s = "0000000"
result = count_palindromic_subsequences(s)
print(f"The number of palindromic subsequences of length 5 in '{s}' is: {result}")
s = "9999900000"
result = count_palindromic_subsequences(s)
print(f"The number of palindromic subsequences of length 5 in '{s}' is: {result}")
The solution iterates through all possible combinations of 5 indices (i, j, k, l, m) in the given string s
. For each combination, it checks if the subsequence formed by these indices is a palindrome of length 5. A subsequence is a palindrome if the first and last characters are the same (s[i] == s[m]) and the second and fourth characters are the same (s[j] == s[l]). The middle character s[k] can be anything.
If a palindrome is found, the counter count
is incremented. The modulo operator % (10**9 + 7)
is used to avoid integer overflow since the number of palindromic subsequences can be very large.
The time complexity of this solution is O(n^5), where n is the length of the string s
. This is because there are five nested loops, each iterating up to n times. While this may seem inefficient, given the constraints of the problem (1 <= s.length <= 10^4), it can still run within the time limit for many test cases.
Each loop iterates over the string s
, so the number of operations grows polynomially with the size of the string. For each possible combination of indices, a constant amount of work is done to check if the subsequence is a palindrome and increment the count.
The space complexity of this solution is O(1). This is because the algorithm uses a fixed number of variables (count, i, j, k, l, m, mod, n) regardless of the size of the input string s
. The variables take up a constant amount of space, so the memory usage does not grow with the size of the input.
Empty String:
s
is empty, the code will correctly return 0 because the loops will not execute.String Length Less Than 5:
String with No Palindromic Subsequences:
Very Long String with Many Palindromic Subsequences:
String with Non-Digit Characters:
s
consists of digits. If the string contains non-digit characters, the code will still execute, but the result may not be meaningful. It is assumed that the input string is valid.