{
  "version": 3,
  "sources": ["../../src/assets/scs.ts"],
  "sourcesContent": ["/**\n * Calculates the Shortest Common Supersequence (SCS) of two sequences.\n *\n * A supersequence is a sequence that contains both input sequences as subsequences.\n * The shortest common supersequence is the shortest possible such sequence.\n *\n * This implementation uses dynamic programming with a time complexity of O(mn)\n * and space complexity of O(mn), where m and n are the lengths of sequences X and Y.\n *\n * @example\n * ```ts\n * const seq1 = [1, 3, 5];\n * const seq2 = [2, 3, 4];\n * const scs = shortestCommonSupersequence(seq1, seq2); // [1, 2, 3, 4, 5]\n * ```\n *\n * @param X       The first sequence.\n * @param Y       The second sequence.\n * @param isEqual Optional equality function to compare elements.\n *                Defaults to strict equality (===).\n * @return The shortest common supersequence of X and Y.\n */\nexport function shortestCommonSupersequence< E = unknown >(\n\tX: E[],\n\tY: E[],\n\tisEqual = ( a: E, b: E ) => a === b\n) {\n\tconst m = X.length;\n\tconst n = Y.length;\n\n\t// Create a 2D dp table where dp[i][j] is the SCS for X[0..i-1] and Y[0..j-1].\n\tconst dp: E[][][] = Array.from( { length: m + 1 }, () =>\n\t\tArray( n + 1 ).fill( null )\n\t);\n\n\t// Base cases: one of the sequences is empty.\n\tfor ( let i = 0; i <= m; i++ ) {\n\t\tdp[ i ][ 0 ] = X.slice( 0, i );\n\t}\n\tfor ( let j = 0; j <= n; j++ ) {\n\t\tdp[ 0 ][ j ] = Y.slice( 0, j );\n\t}\n\n\t// Fill in the dp table.\n\tfor ( let i = 1; i <= m; i++ ) {\n\t\tfor ( let j = 1; j <= n; j++ ) {\n\t\t\tif ( isEqual( X[ i - 1 ], Y[ j - 1 ] ) ) {\n\t\t\t\t// When X[i-1] equals Y[j-1], use the reference from X.\n\t\t\t\tdp[ i ][ j ] = dp[ i - 1 ][ j - 1 ].concat( X[ i - 1 ] );\n\t\t\t} else {\n\t\t\t\t// Choose the shorter option between appending X[i-1] or Y[j-1].\n\t\t\t\tconst option1 = dp[ i - 1 ][ j ].concat( X[ i - 1 ] );\n\t\t\t\tconst option2 = dp[ i ][ j - 1 ].concat( Y[ j - 1 ] );\n\t\t\t\tdp[ i ][ j ] =\n\t\t\t\t\toption1.length <= option2.length ? option1 : option2;\n\t\t\t}\n\t\t}\n\t}\n\n\treturn dp[ m ][ n ];\n}\n"],
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