title: Count Strings Adjacent Characters Difference One
signature: function countStringsAdjacentCharactersDifferenceOne(n) {
test_code: 
  
  // SOLUTION CODE
  // ============================================
  PLACEHOLDER_CODE_BODY
  
  // TESTING CODE 
  // ============================================
  function convertToString(value) {
      return JSON.stringify(value, function (key, val) {
          if (val instanceof Set) {
              return Array.from(val);
          } return val;});
  }
  
  function validateSolution(actual, expected){
      return convertToString(actual)===convertToString(expected);
  }
  
  function driver(n, expected){
      try{
          if (validateSolution(PLACEHOLDER_FN_NAME(n),expected)){
              return "PASSED";
          }
          return "FAILED";
      } catch (error) {
        return error.name;
      }
      
  }
      
  function main(){
    let result = '';
    result = driver(7, 1468);
    console.log(`TEST-0...${result}`);
  
    result = driver(47, 1134264891541288);
    console.log(`TEST-1...${result}`);
  
    result = driver(56, 544505079089235836);
    console.log(`TEST-2...${result}`);
  
    result = driver(22, 41103600);
    console.log(`TEST-3...${result}`);
  
    process.exit(0);
  };
  
  if (require.main === module) {
    main();
  }
source_py: def count_strings_adjacent_characters_difference_one(n):
      dp = [[0 for j in range(27)] for i in range((n + 1))]
      for i in range(0, 26):
          dp[1][i] = 1
      for i in range(2, (n + 1)):
          for j in range(0, 26):
              if (j == 0):
                  dp[i][j] = dp[(i - 1)][(j + 1)]
              else:
                  dp[i][j] = (dp[(i - 1)][(j - 1)] + dp[(i - 1)][(j + 1)])
      sum = 0
      for i in range(0, 26):
          sum = (sum + dp[n][i])
      return sum
source_cpp: int count_strings_adjacent_characters_difference_one(int n) {
    long int dp [ n + 1 ] [ 27 ];
    memset ( dp, 0, sizeof ( dp ) );
    for ( int i = 0;
    i <= 25;
    i ++ ) dp [ 1 ] [ i ] = 1;
    for ( int i = 2;
    i <= n;
    i ++ ) {
      for ( int j = 0;
      j <= 25;
      j ++ ) if ( j == 0 ) dp [ i ] [ j ] = dp [ i - 1 ] [ j + 1 ];
      else dp [ i ] [ j ] = ( dp [ i - 1 ] [ j - 1 ] + dp [ i - 1 ] [ j + 1 ] );
    }
    long int sum = 0;
    for ( int i = 0;
    i <= 25;
    i ++ ) sum = ( sum + dp [ n ] [ i ] );
    return sum;
  }