Little Ponny 2 Subsequence
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
Visualization
Complexity Analysis
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Iterative DP | \(O(N)\) | \(O(1)\) |
Code Reference
package com.dsa.recursion;
import java.util.*;
/ * Problem: Little Ponny 2 Subsequence * Group: 07. Recursion */ / * ============================================================================================ * PROBLEM: Little Ponny and 2-Subsequence * ============================================================================================ * * --------------------- * 1. Problem Description * --------------------- * Little Ponny wants to find the lexicographically minimum subsequence of string A * with a size >= 2. * * Definition: * - A subsequence is derived by deleting zero or more characters from the original string. * - String 'a' is lexicographically smaller than string 'b' if at the first position where * they differ, the character in 'a' comes before the character in 'b' in the alphabet. * * --------------------- * 2. Logical Breakdown & Greedy Strategy * --------------------- * Goal: Minimize the resulting string S. * * Observation 1: Optimal Length * To minimize a string lexicographically, shorter is generally better if the prefixes match. * However, the primary factor is the first character. "a" < "ab" < "b". * Since we need a subsequence of size >= 2, the shortest possible valid size is 2. * Any valid subsequence of length > 2 (e.g., "abc") will be lexicographically larger than * its own prefix of length 2 ("ab"). * Therefore, the optimal answer will always have a length of exactly 2. * * Observation 2: Greedy Selection * To form the smallest string S = c1 + c2: * 1. c1 must be the smallest possible character available in A, such that there is at least * one character remaining after it (to serve as c2). * -> We search for the minimum char in A[0 ... N-2]. * * 2. c2 must be the smallest possible character occurring after the chosen c1. * -> Once we find the index of c1, we search for the minimum char in A[index of c1 + 1 ... N-1]. * * Why the first occurrence of c1? * If the minimal c1 appears multiple times, picking the first occurrence leaves the largest * possible suffix. A larger suffix strictly increases the probability (or maintains it) of * finding a smaller c2. * * --------------------- * 3. Algorithm * --------------------- * 1. Iterate from index 0 to N-2 to find the minimum character 'minChar1'. * Keep track of its first index 'idx1'. * 2. Iterate from 'idx1 + 1' to N-1 to find the minimum character 'minChar2'. * 3. Result = minChar1 + "" + minChar2. * * --------------------- * 4. Constraints & Complexity * --------------------- * Constraints: * 1 <= |A| <= 10^5 * * Time Complexity: * O(N) - Two passes over the string (or parts of it). * * Space Complexity: * O(1) - Only storing character variables. * * --------------------- * 5. Examples * --------------------- * Input 1: A = "abcdsfhjagj" * - Range for c1 (0 to N-2): "abcdsfhjag". Min is 'a' at index 0. * - Range for c2 (1 to N-1): "bcdsfhjagj". Min is 'a' at index 8. * Output 1: "aa" * * Input 2: A = "ksdjgha" * - Range for c1 (0 to N-2): "ksdjgh". Min is 'd' at index 2. * - Range for c2 (3 to N-1): "jgha". Min is 'a' at index 6. * Output 2: "da" * * ============================================================================================ */ public class LittlePonny2Subsequence { public String solve(String A) { // TODO: Implement solution return ""; } }