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Special Subsequences AG

Logical Breakdown

  • [ ] Subproblem Identification:
  • [ ] Optimal Substructure:
  • [ ] Constraint Handling: (e.g., Modulo \(10^9 + 7\))
  • [ ] Optimization: (Matrix Exponentiation if 'A' is huge)

Mathematical Rigor

State Definition

Let \(dp[i]\) be the state for the \(i\)-th subproblem.

Recurrence Relation

\[ dp[i] = ... \]

Visualization

graph TD A["Current State"] --> B["Previous State 1"] A --> C["Previous State 2"]

Complexity Analysis

Approach Time Complexity Space Complexity
Iterative DP \(O(N)\) \(O(1)\)

Code Reference

package com.dsa.recursion;

import java.util.*;

/ * Problem: Special Subsequences AG * Group: 07. Recursion */ / * ============================================================================================ * PROBLEM: Special Subsequences "AG" * ============================================================================================ * * --------------------- * 1. Problem Description * --------------------- * Given a string A consisting of uppercase English letters, count the total number of * times the subsequence "AG" appears in the string. * * A subsequence is formed by deleting zero or more characters from the original string. * Specifically, we are looking for indices (i, j) such that: * 1. i < j * 2. A[i] == 'A' * 3. A[j] == 'G' * * NOTE: Return the answer modulo 10^9 + 7. * * --------------------- * 2. Logical Breakdown & Optimization (Carry Forward) * --------------------- * Naive Approach: * Iterate through every character. If it is 'A', iterate through the rest of the string * to count 'G's. * Time Complexity: O(N^2). For N=10^5, this is too slow (TLE). * * Optimized Approach (Carry Forward): * We can solve this in a single pass O(N) by maintaining a running count of 'A's. * * Algorithm: * 1. Initialize countA = 0 and ans = 0. * 2. Iterate through the string from left to right: * - If current char is 'A': Increment countA. * - If current char is 'G': It means this 'G' can form a pair with all previous countA 'A's. * Add countA to ans. * - Apply modulo (ans % 10^9 + 7) at each addition to prevent overflow. * * Alternative (Suffix Sum): * Iterate from right to left, counting 'G's. When an 'A' is found, add the count of 'G's to ans. * Both approaches are mathematically equivalent. * * --------------------- * 3. Constraints & Complexity Analysis * --------------------- * Constraints: * 1 <= length(A) <= 10^5 * * Time Complexity: * O(N) - Single pass through the string. * * Space Complexity: * O(1) - Using only integer variables for counters. * * --------------------- * 4. Input / Output Format * --------------------- * Input: * - A single string A. * * Output: * - Return an integer denoting the count of "AG" subsequences modulo 10^9 + 7. * * --------------------- * 5. Examples * --------------------- * Input 1: A = "ABCGAG" * Traversal: * - 'A': countA = 1 * - 'B': ignore * - 'C': ignore * - 'G': ans += 1 (Total 1) * - 'A': countA = 2 * - 'G': ans += 2 (Total 3) * Output 1: 3 * * Input 2: A = "GAB" * Traversal: * - 'G': countA = 0 -> ans += 0 * - 'A': countA = 1 * - 'B': ignore * Output 2: 0 * * ============================================================================================ */ public class SpecialSubsequencesAG { public int solve(String A) { // TODO: Implement solution return 0; } }