title: Maximize Volume Cuboid Given Sum Sides
signature: fun maximizeVolumeCuboidGivenSumSides(s: Int): Int {
test_code: package main
  
  import kotlin.text.*
  import kotlin.math.*
  
  
  // SOLUTION CODE
  // ===================================
  PLACEHOLDER_CODE_BODY
  
  // TESTING CODE
  // ===================================
  fun validateSolution(actual:Int, expected:Int):Boolean{
    return actual.equals(expected)
  }
  fun driver(s: Int, expected:Int):String {
      try{
        if (validateSolution(PLACEHOLDER_FN_NAME(s), expected)) {
                return "PASSED"
        }
        return "FAILED"
      } catch (e: Exception){
        return e::class.simpleName as String
      }
  
  }
  
  fun main() {
      var result: String = ""
      result = driver(67, 11132)
      println("TEST-0...${result}")
      
      result = driver(48, 4096)
      println("TEST-1...${result}")
      
      result = driver(59, 7600)
      println("TEST-2...${result}")
      
      result = driver(22, 392)
      println("TEST-3...${result}")
      
      result = driver(14, 100)
      println("TEST-4...${result}")
      
      result = driver(66, 10648)
      println("TEST-5...${result}")
      
      result = driver(1, 0)
      println("TEST-6...${result}")
      
      result = driver(75, 15625)
      println("TEST-7...${result}")
      
      result = driver(58, 7220)
      println("TEST-8...${result}")
      
      result = driver(78, 17576)
      println("TEST-9...${result}")
      
          
  }
  
  main()
source_py: def maximize_volume_cuboid_given_sum_sides(s):
      maxvalue = 0
      i = 1
      for i in range((s - 1)):
          j = 1
          for j in range(s):
              k = ((s - i) - j)
              maxvalue = max(maxvalue, ((i * j) * k))
      return maxvalue
source_cpp: int maximize_volume_cuboid_given_sum_sides(int s) {
    int maxvalue = 0;
    for ( int i = 1;
    i <= s - 2;
    i ++ ) {
      for ( int j = 1;
      j <= s - 1;
      j ++ ) {
        int k = s - i - j;
        maxvalue = max ( maxvalue, i * j * k );
      }
    }
    return maxvalue;
  }