Как рассчитать сложность времени для этих алгоритмов возврата и имеют ли они одинаковую сложность времени? Если отличается как? Пожалуйста, объясните подробно и спасибо за помощь.

1. Hamiltonian cycle:

        bool hamCycleUtil(bool graph[V][V], int path[], int pos) {
            /* base case: If all vertices are included in Hamiltonian Cycle */
            if (pos == V) {
                // And if there is an edge from the last included vertex to the
                // first vertex
                if ( graph[ path[pos-1] ][ path[0] ] == 1 )
                    return true;
                else
                    return false;
            }

            // Try different vertices as a next candidate in Hamiltonian Cycle.
            // We don't try for 0 as we included 0 as starting point in in hamCycle()
            for (int v = 1; v < V; v++) {
                /* Check if this vertex can be added to Hamiltonian Cycle */
                if (isSafe(v, graph, path, pos)) {
                    path[pos] = v;

                    /* recur to construct rest of the path */
                    if (hamCycleUtil (graph, path, pos+1) == true)
                        return true;

                    /* If adding vertex v doesn't lead to a solution, then remove it */
                    path[pos] = -1;
                }
            }

            /* If no vertex can be added to Hamiltonian Cycle constructed so far, then return false */
            return false;
        }

2. Word break:

       a. bool wordBreak(string str) {
            int size = str.size();

            // Base case
            if (size == 0)
                return true;

            // Try all prefixes of lengths from 1 to size
            for (int i=1; i<=size; i++) {
                // The parameter for dictionaryContains is str.substr(0, i)
                // str.substr(0, i) which is prefix (of input string) of
                // length 'i'. We first check whether current prefix is in
                // dictionary. Then we recursively check for remaining string
                // str.substr(i, size-i) which is suffix of length size-i
                if (dictionaryContains( str.substr(0, i) ) && wordBreak( str.substr(i, size-i) ))
                    return true;
            }

            // If we have tried all prefixes and none of them worked
            return false;
        }
    b. String SegmentString(String input, Set<String> dict) {
           if (dict.contains(input)) return input;
           int len = input.length();
           for (int i = 1; i < len; i++) {
               String prefix = input.substring(0, i);
               if (dict.contains(prefix)) {
                   String suffix = input.substring(i, len);
                   String segSuffix = SegmentString(suffix, dict);
                   if (segSuffix != null) {
                       return prefix + " " + segSuffix;
                   }
               }
           }
           return null;
      }


3. N Queens:

        bool solveNQUtil(int board[N][N], int col) {
            /* base case: If all queens are placed then return true */
            if (col >= N)
                return true;

            /* Consider this column and try placing this queen in all rows one by one */
            for (int i = 0; i < N; i++) {
                /* Check if queen can be placed on board[i][col] */
                if ( isSafe(board, i, col) ) {
                    /* Place this queen in board[i][col] */
                    board[i][col] = 1;

                    /* recur to place rest of the queens */
                    if ( solveNQUtil(board, col + 1) == true )
                        return true;

                    /* If placing queen in board[i][col] doesn't lead to a solution then remove queen from board[i][col] */
                    board[i][col] = 0; // BACKTRACK
                }
            }
        }

Я на самом деле немного смущен, так как для Word Break (б) сложность O (2)n), но для гамильтонова цикла он разный и поэтому для печати различных перестановок одной и той же строки, а затем снова для решения проблемы n королев.