Definition
- Counting variant increments a running total each time a range in the DP table is marked as a palindrome.
- Same recurrence as the longest palindromic substring:
dp[i][j] = (s[i]==s[j]) && (j-i<2 || dp[i+1][j-1]). - Baseline count is n because every single character is already a palindrome.
- Maximum count is n(n+1)/2 when all characters are identical.
- O(n squared) time and space for the DP table approach.
The National Bank of India measures how much symmetric structure a batch file identifier contains by counting every symmetric fragment inside it. A high count of overlapping symmetric fragments suggests a machine-generated identifier rather than a human-assigned one.
Understanding the Concepts
What is the problem?
Given a string, count how many contiguous substrings are palindromes, counting each position separately even if the text is identical. For "aaa", the answer is 6: three singles, two pairs, and one triple. This is the counting companion to the longest palindromic substring.
package com.bank.dsa;
public class PalindromicSubstringsDP {
public static int countSubstrings(String s) {
int n = s.length();
boolean[][] dp = new boolean[n][n];
int count = 0;
for (int i = n - 1; i >= 0; i--) {
dp[i][i] = true;
count++;
for (int j = i + 1; j < n; j++) {
if (s.charAt(i) == s.charAt(j) && (j - i < 2 || dp[i + 1][j - 1])) {
dp[i][j] = true;
count++;
}
}
}
return count;
}
}
Why does this matter in banking?
The security team at the National Bank of India computes a symmetry score for batch file identifiers. A high palindrome count relative to the string length signals a highly structured, likely machine-generated identifier. Rajesh flags identifiers where this ratio exceeds a threshold for manual review.
package com.bank.service;
import com.bank.dsa.PalindromicSubstringsDP;
public class SymmetryScorer {
public void score(String id) {
int count = PalindromicSubstringsDP.countSubstrings(id);
System.out.println("ID: " + id + ", Palindrome count: " + count);
}
public static void main(String[] args) {
new SymmetryScorer().score("ABCBA");
new SymmetryScorer().score("AAAA");
new SymmetryScorer().score("ABCD");
}
}
ID: ABCBA, Palindrome count: 7
ID: AAAA, Palindrome count: 10
ID: ABCD, Palindrome count: 4
How does the DP counting work?
Reuse the same boolean palindrome table from the longest palindromic substring. Each time dp[i][j] is set to true, increment the counter. The recurrence is identical: a range is a palindrome when the end characters match and either the range is short or the inner range is already a palindrome.
package com.bank.dsa;
public class CountingDemo {
public static void main(String[] args) {
String s = "ABA";
int n = s.length();
boolean[][] dp = new boolean[n][n];
int count = 0;
for (int i = n - 1; i >= 0; i--) {
dp[i][i] = true;
count++;
System.out.println("Palindrome [" + i + "," + i + "]: " + s.charAt(i) + " count=" + count);
for (int j = i + 1; j < n; j++) {
if (s.charAt(i) == s.charAt(j) && (j - i < 2 || dp[i + 1][j - 1])) {
dp[i][j] = true;
count++;
System.out.println("Palindrome [" + i + "," + j + "]: " + s.substring(i, j + 1) + " count=" + count);
}
}
}
System.out.println("Total: " + count);
}
}
Palindrome [2,2]: A count=1
Palindrome [1,1]: B count=2
Palindrome [0,0]: A count=3
Palindrome [0,2]: ABA count=4
Total: 4
How does this compare to the expand-around-center approach?
The expand-around-center approach (Q1448) counts palindromes in O(n squared) time with O(1) space, while the DP approach uses O(n squared) space. The DP table is useful when other queries need to check arbitrary ranges. For standalone counting, expand-around-center is more space-efficient.
package com.bank.dsa;
public class DPvsExpandDemo {
public static void main(String[] args) {
String s = "ABCBA";
System.out.println("DP count: " + PalindromicSubstringsDP.countSubstrings(s));
System.out.println("Expand count: " + PalindromicSubstringsCount.countSubstrings(s));
}
}
DP count: 7
Expand count: 7
What are the edge cases?
A single character has exactly 1 palindrome. An empty string has 0. Two identical characters have 3 (two singles plus one pair). All distinct characters have exactly n palindromes (only the singles).
package com.bank.dsa;
public class EdgeCaseDemo {
public static void main(String[] args) {
System.out.println(PalindromicSubstringsDP.countSubstrings("A")); // 1
System.out.println(PalindromicSubstringsDP.countSubstrings("")); // 0
System.out.println(PalindromicSubstringsDP.countSubstrings("AA")); // 3
System.out.println(PalindromicSubstringsDP.countSubstrings("AB")); // 2
}
}
1
0
3
2
What is the time and space complexity?
Time is O(n squared) because each cell in the n-by-n table takes O(1) to compute. Space is O(n squared) for the boolean table. Manacher's algorithm can compute palindrome radii in O(n) and sum them for O(n) total counting.
What are common mistakes?
Forgetting to count single-character palindromes as base cases. Using an integer DP table (for subsequence length) instead of a boolean table (for substring palindrome check). Confusing the substring count with the subsequence count.
What are common interview follow-up questions?
- How would you count distinct palindromic substrings?
- How would you count palindromic subsequences instead of substrings?
- How would you use this palindrome table for the minimum-cut partitioning problem?
- What is the relationship between the palindrome count and string complexity?
Visual Overview
String: "ABA" (n=3)
DP Table (true = palindrome):
j=0 j=1 j=2
i=0 [ T T ] ← A, ABA
i=1 [ T ] ← B
i=2 [ T ] ← A
Count: 4 palindromes
1. [0,0] "A"
2. [1,1] "B"
3. [2,2] "A"
4. [0,2] "ABA"
For "AAAA" (n=4):
Count = 4+3+2+1 = 10 = n(n+1)/2
Key Takeaway
Counting palindromic substrings via DP reuses the same boolean palindrome table as the longest palindromic substring. Each time a range is marked as a palindrome, increment a running counter. The recurrence is identical: a range is a palindrome when its end characters match and the inner range is a palindrome or the range is short enough.
The baseline count is always n because every single character is a palindrome. The maximum count is n(n+1)/2 when all characters are identical, giving every possible contiguous range as a palindrome. Rajesh at the National Bank of India uses this count as a symmetry score for batch file identifiers, where a high score flags machine-generated patterns.
The DP approach uses O(n squared) time and space. The expand-around-center approach from Q1448 achieves the same count in O(n squared) time with O(1) space, making it preferable for standalone counting. The DP table is valuable when other problems, like palindromic partitioning, need O(1) palindrome queries on arbitrary ranges.
Manacher's algorithm can compute per-center palindrome radii in O(n) time and sum them for O(n) total counting, which is optimal. However, in interviews the expand-around-center approach is the standard expectation.
Remember one sentence: reuse the palindrome DP table and increment a counter for every range marked true, starting from the baseline of n single-character palindromes.
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