Float number and operators -

FloatNumberAndOperators java

For example, given the array [1, 12, -3], the maximum number 33 can be found using 1 - 12 * (-3), since all operators have flat preference, so 1 - 12 * (-3) will be handled from left to right, first operation is 1 - 12 with value -11, and then second operation is -11 * (-3) = 33.
  1 12
  13
  -11
  12
  1/12
  
  13*-3
  13/-3
  13 - -3
  13 + -3
  
[1, 12, -3]

4 number 
for each number,you have 4 options with -3 
  
when you get maximum,you have to get 16 numbers 
  
double evaluateMaximize(float[] terms) { // [1, 12, -3], 33, first two numbers 1 - 12 = -11
  int N = terms.length;
  if (N > 16)
    throw new IllegalArgumentException("Too complex for this solution");
  
  double[] intemediate = new float[1 << (2 * N)];
  int iInterm = 0;
  double maximized = 0;
  intermediate[iInterm++] = terms[0];
  for (int iT = 1; iT < N; iT++) {
    int iMaxInterm = iInterm - 1;
    double m = 0;  // 
    for (int iPrev = iMaxInterm; iPrev >= 0; iPrev--) { //<--
      double prev = intermediate[iPrev];
      intermediate[iPrev * 4 + 0] = prev * terms[iT];
      intermediate[iPrev * 4 + 1] = prev / terms[iT];
      intermediate[iPrev * 4 + 2] = prev - terms[iT];
      intermediate[iPrev * 4 + 3] = prev + terms[iT];
      if (iT != N - 1) // skip if not the last term
        continue;

      m = Math.max(m, intermediate[iPrev * 4 + 3]);
      m = Math.max(m, intermediate[iPrev * 4 + 2]);
      m = Math.max(m, intermediate[iPrev * 4 + 1]);
      m = Math.max(m, intermediate[iPrev * 4 + 0]);
    }

    maximized = m;
    iInterm *= 4;
  }
  
  return maximized;
}