title: Efficient Search In An Array Where Difference Between Adjacent Is 1
signature: function efficient_search_in_an_array_where_difference_between_adjacent_is_1(arr, n, x)
test_code: using Printf
  
  # SOLUTION CODE
  # ============================================
  PLACEHOLDER_CODE_BODY
  
  # TESTING CODE 
  # ============================================
  function validate_solution(actual, expected)
      return actual == expected
  end
  
  function driver(arr, n, x, expected)
      try
  
          if validate_solution(PLACEHOLDER_FN_NAME(arr, n, x), expected)
              return "PASSED";
          end
          return "FAILED";
      catch exception_obj
          return string(Base.typename(typeof(exception_obj)).wrapper);
      end
  end
  
  function main()
      
      result = driver([5, 4, 5, 6, 5, 4, 3, 2], 29, 6, 3);
      @printf("TEST-0...%s\n",result);
      
      result = driver([5, 4, 5, 6, 5, 4, 3, 2], 19, 3, 6);
      @printf("TEST-1...%s\n",result);
      
      result = driver([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], 37, 1, 14);
      @printf("TEST-2...%s\n",result);
      
      result = driver([50, 51, 52, 51, 50, 49, 48], 17, 49, 5);
      @printf("TEST-3...%s\n",result);
      
      result = driver([-86, -68, -32, -6, 6, 10, 30, 34, 58, 92], 6, 6, -1);
      @printf("TEST-4...%s\n",result);
      
      result = driver([1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 0], 27, 22, -1);
      @printf("TEST-5...%s\n",result);
      
      result = driver([58], 0, 0, -1);
      @printf("TEST-6...%s\n",result);
      
      result = driver([-64, 78, 58, 36, 48, 80, -80, 74, -98, 46, -48, 24, 80, 72, 90, -46, 14, 68, 38, 58, -54, 80, 44, -62, 34, -28, 92, 84, 90, 44, -26, 88, 18, 22, -32, 54, 58, 92], 24, 34, -1);
      @printf("TEST-7...%s\n",result);
      
      result = driver([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], 35, 1, 21);
      @printf("TEST-8...%s\n",result);
      
      result = driver([5], 0, 0, -1);
      @printf("TEST-9...%s\n",result);
      
  end
  
  main()
source_py: def efficient_search_in_an_array_where_difference_between_adjacent_is_1(arr, n, x):
      i = 0
      while (i <= (n - 1)):
          if (arr[i] == x):
              return i
          i += abs((arr[i] - x))
      return (- 1)
source_cpp: int efficient_search_in_an_array_where_difference_between_adjacent_is_1(vector<int> arr, int n, int x) {
    int i = 0;
    while ( i <= n - 1 ) {
      if ( arr [ i ] == x ) return i;
      i += abs ( arr [ i ] - x );
    }
    return - 1;
  }