* k) worst case.
   */
  private void trim() {
    int left = 0;
    int right = 2 * k - 1;

    int minThresholdPosition = 0;
    // The leftmost position at which the greatest of the k lower elements
    // -- the new value of threshold -- might be found.

    int iterations = 0;
    int maxIterations = IntMath.log2(right - left, RoundingMode.CEILING) * 3;

    }
    return "";
  }

  /**
   * Identify and mark the longest run of zeroes in an IPv6 address.
   *
   * <p>Only runs of two or more hextets are considered. In case of a tie, the leftmost run wins. If
   * a qualifying run is found, its hextets are replaced by the sentinel value -1.
   *
   * @param hextets {@code int[]} mutable array of eight 16-bit hextets
   */

    }
    return "";
  }

  /**
   * Identify and mark the longest run of zeroes in an IPv6 address.
   *
   * <p>Only runs of two or more hextets are considered. In case of a tie, the leftmost run wins. If
   * a qualifying run is found, its hextets are replaced by the sentinel value -1.
   *
   * @param hextets {@code int[]} mutable array of eight 16-bit hextets
   */

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