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Interleaving Strings

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.dynamic_programming;

import java.util.*;

/ * Problem: Interleaving Strings * Group: 11. Dynamic Programming */ / * ============================================================================================ * PROBLEM: Interleaving Strings * ============================================================================================ * * --------------------- * 1. Problem Description * --------------------- * Given s1, s2, s3. Find whether s3 is formed by the interleaving of s1 and s2. * Interleaving means relative order of chars in s1 and s2 is preserved. * * --------------------- * 2. Logical Breakdown * --------------------- * Let dp[i][j] be a boolean value indicating if s3[0...i+j-1] can be formed by * s1[0...i-1] and s2[0...j-1]. * * Transitions: * To form the prefix of length (i+j) in s3, the last character s3[i+j-1] must come from * either s1 or s2. * * 1. From s1: * If s1[i-1] == s3[i+j-1], then valid if dp[i-1][j] is true. * * 2. From s2: * If s2[j-1] == s3[i+j-1], then valid if dp[i][j-1] is true. * * dp[i][j] = (MatchS1 && dp[i-1][j]) || (MatchS2 && dp[i][j-1]) * * --------------------- * 3. Complexity * --------------------- * Time Complexity: O(N * M) * Space Complexity: O(N * M) * * ============================================================================================ */ public class InterleavingStrings { public int solve(int A) { // TODO: Implement solution return 0; }

public boolean isInterleave(String aabcc, String dbbca, String aadbbcbcac) {
    return true;
}

}