Scramble String

class Solution {
public:
    bool canScramble(string str1, string str2) {
        int len = str1.length();
        vector<vector<vector<int>>> memo(len + 1, vector<vector<int>>(len, vector<int>(len)));
        for (int i = 0; i < len; i++) {
            for (int j = 0; j < len; j++) {
                memo[1][i][j] = str1[i] == str2[j];
            }
        }
        for (int l = 2; l <= len; l++) {
            for (int i = 0; i <= len - l; i++) {
                for (int j = 0; j <= len - l; j++) {
                    for (int k = 1; k < l; k++) {
                        const auto& firstHalf = memo[k][i];
                        const auto& secondHalf = memo[l - k][i + k];
                        memo[l][i][j] |= firstHalf[j] && secondHalf[j + k];
                        memo[l][i][j] |= 
                            firstHalf[j + l - k] && secondHalf[j];
                    }
                }
            }
        }
        return memo[len][0][0];
    }
};

This solution tackles the problem of determining if one string can be transformed into another by a series of scramble operations. In C++, the implementation uses dynamic programming to efficiently decide if the second string is a scrambled version of the first.

• A 3D vector memo is initialized to store the boolean results of subproblems. Each element memo[l][i][j] represents whether a substring of str1 starting at index i with length l can be transformed into a substring of str2 starting at index j of the same length.
• The base case is set immediately for substrings of length 1, comparing characters at corresponding positions in str1 and str2.
• For substrings longer than 1, the solution combines results from smaller substrings. This involves checking both scenarios where a string might be directly scrambled or split and swapped.
• Continuous updates to memo occur by iterating over the substring length l, the starting indices i and j, and split positions k within each length l.
• Ultimately, the result is found in memo[len][0][0] where len is the length of str1, indicating whether the entire str1 can be scrambled into str2.

Make sure to understand the memoization structure and how the indices and loops interact since correct indexing is crucial for this solution to work. Furthermore, consider effective debugging and testing with different string lengths and contents to fully assess the implementation's accuracy and efficiency.

class Solution {
    public boolean canScramble(String str1, String str2) {
        int len = str1.length();
        boolean scrambled[][][] = new boolean[len + 1][len][len];
        for (int index1 = 0; index1 < len; index1++) {
            for (int index2 = 0; index2 < len; index2++) {
                scrambled[1][index1][index2] = str1.charAt(index1) == str2.charAt(index2);
            }
        }
        for (int l = 2; l <= len; l++) {
            for (int i = 0; i < len + 1 - l; i++) {
                for (int j = 0; j < len + 1 - l; j++) {
                    for (int splitPoint = 1; splitPoint < l; splitPoint++) {
                        boolean[] scramblePart1 = scrambled[splitPoint][i];
                        boolean[] scramblePart2 = scrambled[l - splitPoint][i + splitPoint];
                        scrambled[l][i][j] |= scramblePart1[j] && scramblePart2[j + splitPoint];
                        scrambled[l][i][j] |= 
                        scramblePart1[j + l - splitPoint] && scramblePart2[j];
                    }
                }
            }
        }
        return scrambled[len][0][0];
    }
}

In the provided Java solution for determining if one string is a scramble of another, the core approach involves dynamic programming to efficiently break down and solve the problem.

• The function canScramble accepts two strings str1 and str2 as input.
• A 3-dimensional boolean array scrambled is declared to store intermediate results, which spans across possible string lengths, and the start indices of substrings in both str1 and str2.
• Initialization of scrambled involves setting true when characters directly match between str1 and str2 at each position for substrings of length 1.
• The solution then considers substrings of increasing length. For each length l, and starting indices i in str1 and j in str2, it checks if there exists a split point splitPoint that validates the scramble condition based on previously computed results.
• Scrambling conditions for both possible segment divisions (keeping the order and swapping) are checked using logical OR (|=), updating the scrambled array.
• Finally, the result for the full length of the strings starting from index 0 in both strings indicates if str2 is a scramble of str1.

This method leverages dynamic programming's overlapping subproblem property to effectively manage the complexity and avoid redundant calculations. By storing intermediate results and referring back to them, it optimizes both time and space compared to a straightforward recursive approach. The end result is determined by checking the top-level entry in the scrambled array for the full string.

bool scrambledStrings(char* str1, char* str2) {
    int len = strlen(str1);
    bool scramble_dp[len + 1][len][len];
    memset(scramble_dp, false, sizeof(scramble_dp));
    for (int idx = 0; idx < len; idx++) {
        for (int jdx = 0; jdx < len; jdx++) {
            scramble_dp[1][idx][jdx] = str1[idx] == str2[jdx];
        }
    }
    for (int segment = 2; segment <= len; segment++) {
        for (int idx = 0; idx <= len - segment; idx++) {
            for (int jdx = 0; jdx <= len - segment; jdx++) {
                for (int k = 1; k < segment; k++) {
                    bool* left = scramble_dp[k][idx];
                    bool* right = scramble_dp[segment - k][idx + k];
                    scramble_dp[segment][idx][jdx] =
                        scramble_dp[segment][idx][jdx] || (left[jdx] && right[jdx + k]);
                    scramble_dp[segment][idx][jdx] = scramble_dp[segment][idx][jdx] ||
                                       (left[jdx + segment - k] && right[jdx]);
                }
            }
        }
    }
    return scramble_dp[len][0][0];
}

This solution implements a function to determine if one string is a scrambled version of another using dynamic programming. The function scrambledStrings takes two character pointers str1 and str2 as inputs and returns a boolean value indicating whether str2 is a scrambled string of str1.

Process Summary

1. The function begins by determining the length of str1 and initializing a 3D boolean array scramble_dp to keep track of subproblem solutions. The dimensions of this array depend on the length of the string, with each entry scramble_dp[segment][idx][jdx] indicating whether the segment of str1 starting at idx and the segment of str2 starting at jdx with length segment are scrambles of each other.

2. Initially, for segments of length 1, the function sets scramble_dp[1][idx][jdx] to true if the characters at str1[idx] and str2[jdx] are the same.

3. For longer segments, the function looks at all possible splits of the segment into two parts. For each possible split, it checks two conditions using the already computed values in scramble_dp:

   • Whether one part of str1 is a scramble of one part of str2 and the other part of str1 is a scramble of the other part of str2.
   • Whether one part of str1 is a scramble of the reversed part of str2 and vice versa.

4. This is done in a nested loop structure where the outer loop iterates over all possible segment lengths (from 2 to len), and the inner two loops iterate over possible start indices for these segments in both strings.

5. The final outcome, whether the entire str1 is a scrambled version of str2, is stored in scramble_dp[len][0][0].

The algorithm efficiently solves the problem by building solutions for small segments and using them to solve for larger segments, ensuring that all combinations of segments are considered.

var scrambleCheck = function (str1, str2) {
    const len = str1.length;
    let memo = new Array(len + 1)
        .fill(0)
        .map(() => new Array(len).fill(0).map(() => new Array(len).fill(false)));
    for (let i = 0; i < len; i++) {
        for (let j = 0; j < len; j++) {
            memo[1][i][j] = str1.charAt(i) == str2.charAt(j);
        }
    }
    for (let segmentLength = 2; segmentLength <= len; segmentLength++) {
        for (let i = 0; i < len + 1 - segmentLength; i++) {
            for (let j = 0; j < len + 1 - segmentLength; j++) {
                for (let partLen = 1; partLen < segmentLength; partLen++) {
                    const part1 = memo[partLen][i];
                    const part2 = memo[segmentLength - partLen][i + partLen];
                    memo[segmentLength][i][j] |= part1[j] && part2[j + partLen];
                    memo[segmentLength][i][j] |= part1[j + segmentLength - partLen] && part2[j];
                }
            }
        }
    }
    return memo[len][0][0];
};

This article guides you through the JavaScript implementation of a function that checks if two strings are scramble strings of each other.

The function scrambleCheck is defined to take two strings, str1 and str2, as inputs and returns a boolean indicating whether the two strings can be transformed into one another by a series of scrambles. Scrambling involves breaking a string into two non-empty parts and swapping them.

• First, the function calculates the length of str1 and uses this length to initialize a 3D array memo. This array stores boolean values representing the scramble status of different segments of the strings.
• The base case is set up where single characters in str1 are checked against characters in str2. If they match, the corresponding memo value is set to true.
• The function then considers segment lengths from 2 up to the length of the strings. For each segment length, all possible starting indices for the segments in str1 and str2 are evaluated.
• For every possible partition of the segment (partLen), the function checks two conditions:
  • Scrambling without crossing: where the two parts before and after the partition in str1 directly match the segments in str2
  • Scrambling with crossing: where the first part of str1 matches the second part of str2 and vice versa
• These checks update the memo array for larger segments based on the results of smaller segments.
• Finally, the function returns the value from the memo array that indicates whether the entire str1 can be scrambled to match str2.

Using dynamic programming, this solution efficiently determines the scramble status between the two strings by building solutions for smaller segments and combining them to address larger segments. This avoids redundant calculations seen in naive recursive approaches, leading to improved performance.

class Solution:
    def checkScramble(self, str1: str, str2: str) -> bool:
        limit = len(str1)
        memo = [
            [[False for x in range(limit)] for y in range(limit)] for z in range(limit + 1)
        ]
        for x in range(limit):
            for y in range(limit):
                memo[1][x][y] = str1[x] == str2[y]
        for sz in range(2, limit + 1):
            for x in range(limit + 1 - sz):
                for y in range(limit + 1 - sz):
                    for split in range(1, sz):
                        subMemo1 = memo[split][x]
                        subMemo2 = memo[sz - split][x + split]
                        memo[sz][x][y] |= subMemo1[y] and subMemo2[y + split]
                        memo[sz][x][y] |= (
                            subMemo1[y + sz - split] and subMemo2[y]
                        )
        return memo[limit][0][0]

The provided code defines a function to determine if one string is a scramble of another. Scrambling here involves potentially splitting the string into two at any point, and then swapping these two parts, recursively.

To solve this problem, the function checkScramble uses a dynamic programming approach. The key elements in the implementation are:

• memo: A 3D list that stores boolean values, initialized to track the scramble status for substrings of various lengths and starting positions.
• Initialization: Set each element memo[1][x][y] to True if the characters at position x in str1 and y in str2 are the same. This addresses the base case where substrings are of length 1.
• Nested loops are used to build up the solution for strings from size 2 up to the full length of str1. For each size (sz), starting indices (x) in str1 and str2 (y) are considered.
• For each possible substring size and starting index, consider all possible splits. This is facilitated by a further nested loop over split, indicating where the split occurs.
• Conditions evaluated in these loops check whether making a particular split and potentially swapping leads to both parts matching corresponding parts in the other string.
• The computed results of shorter substrings are reused to determine the result for larger substrings, culminating in whether the total strings str1 and str2 can be considered scrambles of each other, stored in memo[limit][0][0].

This method provides a clear and efficient way to recursively decide if two strings can be transformed into one another by this specific scrambling process. By leveraging dynamic programming, the function efficiently subdivides and conquers the problem, using previously calculated results to speed up the solution. This method is quite effective for large strings as it avoids redundant recalculations found in plain recursive approaches.