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| 1 | +/** |
| 2 | + * Booth's Algorithm finds the lexicographically minimal rotation of a string. |
| 3 | + * Time Complexity: O(n) - Linear time where n is the length of input string |
| 4 | + * Space Complexity: O(n) - Linear space for failure function array |
| 5 | + * For More Visit - https://en.wikipedia.org/wiki/Booth%27s_multiplication_algorithm |
| 6 | + * @example |
| 7 | + * Input: "baca" |
| 8 | + * All possible rotations: |
| 9 | + * - "baca" |
| 10 | + * - "acab" |
| 11 | + * - "caba" |
| 12 | + * - "abac" |
| 13 | + * Output: "abac" (lexicographically smallest) |
| 14 | + * |
| 15 | + * How it works: |
| 16 | + * 1. Doubles the input string to handle all rotations |
| 17 | + * 2. Uses failure function (similar to KMP) to find minimal rotation |
| 18 | + * 3. Maintains a pointer to the start of minimal rotation found so far |
| 19 | + * @param {string} str - Input string to find minimal rotation |
| 20 | + * @returns {string} - Lexicographically minimal rotation of the input string |
| 21 | + * @throws {Error} - If input is not a string or is empty |
| 22 | + */ |
| 23 | +export function findMinimalRotation(str) { |
| 24 | + if (typeof str !== 'string') { |
| 25 | + throw new Error('Input must be a string') |
| 26 | + } |
| 27 | + |
| 28 | + if (str.length === 0) { |
| 29 | + throw new Error('Input string cannot be empty') |
| 30 | + } |
| 31 | + |
| 32 | + // Double the string for rotation comparison |
| 33 | + // This allows us to check all rotations by just sliding a window |
| 34 | + const s = str + str |
| 35 | + const n = s.length |
| 36 | + |
| 37 | + // Initialize failure function array |
| 38 | + const f = new Array(n).fill(-1) |
| 39 | + let k = 0 // Starting position of minimal rotation |
| 40 | + |
| 41 | + //Algorithm's implementation |
| 42 | + // Iterate through the doubled string |
| 43 | + // j is the current position we're examining |
| 44 | + for (let j = 1; j < n; j++) { |
| 45 | + // i is the length of the matched prefix in the current candidate |
| 46 | + // Get the failure function value for the previous position |
| 47 | + let i = f[j - k - 1] |
| 48 | + // This loop handles the case when we need to update our current minimal rotation |
| 49 | + // It compares characters and finds if there's a better (lexicographically smaller) rotation |
| 50 | + while (i !== -1 && s[j] !== s[k + i + 1]) { |
| 51 | + // If we find a smaller character, we've found a better rotation |
| 52 | + // Update k to the new starting position |
| 53 | + if (s[j] < s[k + i + 1]) { |
| 54 | + // j-i-1 gives us the starting position of the new minimal rotation |
| 55 | + k = j - i - 1 |
| 56 | + } |
| 57 | + // Update i using the failure function to try shorter prefixes |
| 58 | + i = f[i] |
| 59 | + } |
| 60 | + |
| 61 | + // This block updates the failure function and handles new character comparisons |
| 62 | + if (i === -1 && s[j] !== s[k + i + 1]) { |
| 63 | + // If current character is smaller, update the minimal rotation start |
| 64 | + if (s[j] < s[k + i + 1]) { |
| 65 | + k = j |
| 66 | + } |
| 67 | + //If no match found,mark failure function accordingly |
| 68 | + f[j - k] = -1 |
| 69 | + } else { |
| 70 | + //If match found, extend the matched length |
| 71 | + f[j - k] = i + 1 |
| 72 | + } |
| 73 | + } |
| 74 | + // After finding k (the starting position of minimal rotation): |
| 75 | + // 1. slice(k): Take substring from position k to end |
| 76 | + // 2. slice(0, k): Take substring from start to position k |
| 77 | + // 3. Concatenate them to get the minimal rotation |
| 78 | + return str.slice(k) + str.slice(0, k) |
| 79 | +} |
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