title: Number Of Substrings With Odd Decimal Value In A Binary String
signature: fun numberOfSubstringsWithOddDecimalValueInABinaryString(s: String): 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: String, 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("OGiOkJF", 0)
      println("TEST-0...${result}")
      
      result = driver("517376", 1)
      println("TEST-1...${result}")
      
      result = driver("11", 3)
      println("TEST-2...${result}")
      
      result = driver("Ze", 0)
      println("TEST-3...${result}")
      
      result = driver("8763644247018", 1)
      println("TEST-4...${result}")
      
      result = driver("00111010001", 15)
      println("TEST-5...${result}")
      
      result = driver("HGwkBKUOVu", 0)
      println("TEST-6...${result}")
      
      result = driver("652", 0)
      println("TEST-7...${result}")
      
      result = driver("101000011110", 21)
      println("TEST-8...${result}")
      
      result = driver("kvdpG ", 0)
      println("TEST-9...${result}")
      
          
  }
  
  main()
source_py: def number_of_substrings_with_odd_decimal_value_in_a_binary_string(s):
      n = len(s)
      auxArr = [0 for i in range(n)]
      if (s[0] == '1'):
          auxArr[0] = 1
      for i in range(0, n):
          if (s[i] == '1'):
              auxArr[i] = (auxArr[(i - 1)] + 1)
          else:
              auxArr[i] = auxArr[(i - 1)]
      count = 0
      for i in range((n - 1), (- 1), (- 1)):
          if (s[i] == '1'):
              count += auxArr[i]
      return count
source_cpp: int number_of_substrings_with_odd_decimal_value_in_a_binary_string(string s) {
    int n = s . length ( );
    int auxArr [ n ] = {
      0 };
      if ( s [ 0 ] == '1' ) auxArr [ 0 ] = 1;
      for ( int i = 1;
      i < n;
      i ++ ) {
        if ( s [ i ] == '1' ) auxArr [ i ] = auxArr [ i - 1 ] + 1;
        else auxArr [ i ] = auxArr [ i - 1 ];
      }
      int count = 0;
      for ( int i = n - 1;
      i >= 0;
      i -- ) if ( s [ i ] == '1' ) count += auxArr [ i ];
      return count;
    }