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GeezerVison is showing github.com — CodingBat-Java-Solutions/RecursionOne.java at master · TechnoBlogger14o3/CodingBat-Java-Solutions Expand file tree/ Copy pathRecursionOne.javaMore file actions 531 lines (437 loc) · 16.3 KB / Copy pathRecursionOne.javaFile metadata and controls531 lines (437 loc) · 16.3 KB 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342 343 344 345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360 361 362 363 364 365 366 367 368 369 370 371 372 373 374 375 376 377 378 379 380 381 382 383 384 385 386 387 388 389 390 391 392 393 394 395 396 397 398 399 400 401 402 403 404 405 406 407 408 409 410 411 412 413 414 415 416 417 418 419 420 421 422 423 424 425 426 427 428 429 430 431 432 433 434 435 436 437 438 439 440 441 442 443 444 445 446 447 448 449 450 451 452 453 454 455 456 457 458 459 460 461 462 463 464 465 466 467 468 469 470 471 472 473 474 475 476 477 478 479 480 481 482 483 484 485 486 487 488 489 490 491 492 493 494 495 496 497 498 499 500 501 502 503 504 505 506 507 508 509 510 511 512 513 514 515 516 517 518 519 520 521 522 523 524 525 526 527 528 529 530 531 /** * */ package coding.bat.solutions; /** * @author Aman Shekhar * */ public class RecursionOne { /** * @param args */ public static void main(String[] args) { // TODO Auto-generated method stub } // -------------------------------------------------------------------------------------------- // Given n of 1 or more, return the factorial of n, which is n * (n-1) * (n-2) // ... 1. Compute the result recursively (without loops). // // // factorial(1) → 1 // factorial(2) → 2 // factorial(3) → 6 public int factorial(int n) { return (n < 1) ? 1 : n * factorial(n - 1); } // -------------------------------------------------------------------------------------------- // We have a number of bunnies and each bunny has two big floppy ears. We want // to compute the total number of ears across all the bunnies recursively // (without loops or multiplication). // // // bunnyEars(0) → 0 // bunnyEars(1) → 2 // bunnyEars(2) → 4 public int bunnyEars(int bunnies) { return (bunnies == 0) ? 0 : 2 + bunnyEars(bunnies - 1); } // -------------------------------------------------------------------------------------------- // The fibonacci sequence is a famous bit of mathematics, and it happens to have // a recursive definition. The first two values in the sequence are 0 and 1 // (essentially 2 base cases). Each subsequent value is the sum of the previous // two values, so the whole sequence is: 0, 1, 1, 2, 3, 5, 8, 13, 21 and so on. // Define a recursive fibonacci(n) method that returns the nth fibonacci number, // with n=0 representing the start of the sequence. // // // fibonacci(0) → 0 // fibonacci(1) → 1 // fibonacci(2) → 1 // public int fibonacci(int n) { if (n == 0) return 0; return (n < 3) ? 1 : fibonacci(n - 1) + fibonacci(n - 2); } // -------------------------------------------------------------------------------------------- // We have bunnies standing in a line, numbered 1, 2, ... The odd bunnies (1, 3, // ..) have the normal 2 ears. The even bunnies (2, 4, ..) we'll say have 3 // ears, because they each have a raised foot. Recursively return the number of // "ears" in the bunny line 1, 2, ... n (without loops or multiplication). // // // bunnyEars2(0) → 0 // bunnyEars2(1) → 2 // bunnyEars2(2) → 5 public int bunnyEars2(int bunnies) { if (bunnies == 0) return 0; if (bunnies % 2 == 1) return 2 + bunnyEars2(bunnies - 1); return 3 + bunnyEars2(bunnies - 1); } // -------------------------------------------------------------------------------------------- // We have triangle made of blocks. The topmost row has 1 block, the next row // down has 2 blocks, the next row has 3 blocks, and so on. Compute recursively // (no loops or multiplication) the total number of blocks in such a triangle // with the given number of rows. // // // triangle(0) → 0 // triangle(1) → 1 // triangle(2) → 3 public int triangle(int rows) { if (rows < 2) return rows; return rows + triangle(rows - 1); } // -------------------------------------------------------------------------------------------- // Given a non-negative int n, return the sum of its digits recursively (no // loops). Note that mod (%) by 10 yields the rightmost digit (126 % 10 is 6), // while divide (/) by 10 removes the rightmost digit (126 / 10 is 12). // // // sumDigits(126) → 9 // sumDigits(49) → 13 // sumDigits(12) → 3 public int sumDigits(int n) { if (n < 10) return n; return sumDigits(n / 10) + n % 10; } // -------------------------------------------------------------------------------------------- // Given a non-negative int n, return the count of the occurrences of 7 as a // digit, so for example 717 yields 2. (no loops). Note that mod (%) by 10 // yields the rightmost digit (126 % 10 is 6), while divide (/) by 10 removes // the rightmost digit (126 / 10 is 12). // // // count7(717) → 2 // count7(7) → 1 // count7(123) → 0 public int count7(int n) { if (n == 0) return 0; if (n % 10 == 7) return 1 + count7(n / 10); return count7(n / 10); } // -------------------------------------------------------------------------------------------- // Given a non-negative int n, compute recursively (no loops) the count of the // occurrences of 8 as a digit, except that an 8 with another 8 immediately to // its left counts double, so 8818 yields 4. Note that mod (%) by 10 yields the // rightmost digit (126 % 10 is 6), while divide (/) by 10 removes the rightmost // digit (126 / 10 is 12). // // // count8(8) → 1 // count8(818) → 2 // count8(8818) → 4 public int count8(int n) { if (n == 0) return 0; if (n % 10 == 8) { if (n / 10 % 10 == 8) return 2 + count8(n / 10); return 1 + count8(n / 10); } return count8(n / 10); } // -------------------------------------------------------------------------------------------- // Given base and n that are both 1 or more, compute recursively (no loops) the // value of base to the n power, so powerN(3, 2) is 9 (3 squared). // // // powerN(3, 1) → 3 // powerN(3, 2) → 9 // powerN(3, 3) → 27 public int powerN(int base, int n) { if (n == 1) return base; return base * powerN(base, n - 1); } // -------------------------------------------------------------------------------------------- // Given a string, compute recursively (no loops) the number of lowercase 'x' // chars in the string. // // // countX("xxhixx") → 4 // countX("xhixhix") → 3 // countX("hi") → 0 public int countX(String str) { if (str.length() == 0) return 0; if (str.charAt(0) == 'x') return 1 + countX(str.substring(1)); return countX(str.substring(1)); } // -------------------------------------------------------------------------------------------- // Given a string, compute recursively (no loops) the number of times lowercase // "hi" appears in the string. // // // countHi("xxhixx") → 1 // countHi("xhixhix") → 2 // countHi("hi") → 1 public int countHi(String str) { if (str.length() < 2) return 0; if (str.charAt(0) == 'h' && str.charAt(1) == 'i') return 1 + countHi(str.substring(2)); return countHi(str.substring(1)); } // -------------------------------------------------------------------------------------------- // Given a string, compute recursively (no loops) a new string where all the // lowercase 'x' chars have been changed to 'y' chars. // // // changeXY("codex") → "codey" // changeXY("xxhixx") → "yyhiyy" // changeXY("xhixhix") → "yhiyhiy" public String changeXY(String str) { char ch; if (str.length() == 0) return str; ch = str.charAt(0); if (ch == 'x') return 'y' + changeXY(str.substring(1)); return ch + changeXY(str.substring(1)); } // -------------------------------------------------------------------------------------------- // Given a string, compute recursively (no loops) a new string where all // appearances of "pi" have been replaced by "3.14". // // // changePi("xpix") → "x3.14x" // changePi("pipi") → "3.143.14" // changePi("pip") → "3.14p" public String changePi(String str) { if (str.length() < 2) return str; if (str.substring(0, 2).equals("pi")) return "3.14" + changePi(str.substring(2)); return str.charAt(0) + changePi(str.substring(1)); } // -------------------------------------------------------------------------------------------- // Given a string, compute recursively a new string where all the 'x' chars have // been removed. // // // noX("xaxb") → "ab" // noX("abc") → "abc" // noX("xx") → "" public String noX(String str) { char ch; if (str.length() == 0) return str; ch = str.charAt(0); if (ch == 'x') return noX(str.substring(1)); return ch + noX(str.substring(1)); } // -------------------------------------------------------------------------------------------- // Given an array of ints, compute recursively if the array contains a 6. We'll // use the convention of considering only the part of the array that begins at // the given index. In this way, a recursive call can pass index+1 to move down // the array. The initial call will pass in index as 0. // // // array6([1, 6, 4], 0) → true // array6([1, 4], 0) → false // array6([6], 0) → true public boolean array6(int[] nums, int index) { if (index == nums.length) return false; if (nums[index] == 6) return true; return array6(nums, index + 1); } // -------------------------------------------------------------------------------------------- // Given an array of ints, compute recursively the number of times that the // value 11 appears in the array. We'll use the convention of considering only // the part of the array that begins at the given index. In this way, a // recursive call can pass index+1 to move down the array. The initial call will // pass in index as 0. public int array11(int[] nums, int index) { if (index == nums.length) return 0; if (nums[index] == 11) return 1 + array11(nums, index + 1); return array11(nums, index + 1); } // -------------------------------------------------------------------------------------------- // Given an array of ints, compute recursively if the array contains somewhere a // value followed in the array by that value times 10. We'll use the convention // of considering only the part of the array that begins at the given index. In // this way, a recursive call can pass index+1 to move down the array. The // initial call will pass in index as 0. public boolean array220(int[] nums, int index) { if (index >= nums.length - 1) return false; if (nums[index] * 10 == nums[index + 1]) return true; return array220(nums, index + 1); } // -------------------------------------------------------------------------------------------- // Given a string, compute recursively a new string where all the adjacent chars // are now separated by a "*". public String allStar(String str) { if (str.length() < 2) return str; return str.charAt(0) + "*" + allStar(str.substring(1)); } // -------------------------------------------------------------------------------------------- // Given a string, compute recursively a new string where identical chars that // are adjacent in the original string are separated from each other by a "*". public String pairStar(String str) { if (str.length() < 2) return str; if (str.charAt(0) == str.charAt(1)) return str.charAt(0) + "*" + pairStar(str.substring(1)); return str.charAt(0) + pairStar(str.substring(1)); } // -------------------------------------------------------------------------------------------- // Given a string, compute recursively a new string where all the lowercase 'x' // chars have been moved to the end of the string. public String endX(String str) { if (str.length() == 0) return str; if (str.charAt(0) == 'x') return endX(str.substring(1)) + 'x'; return str.charAt(0) + endX(str.substring(1)); } // -------------------------------------------------------------------------------------------- // We'll say that a "pair" in a string is two instances of a char separated by a // char. So "AxA" the A's make a pair. Pair's can overlap, so "AxAxA" contains 3 // pairs -- 2 for A and 1 for x. Recursively compute the number of pairs in the // given string. public int countPairs(String str) { if (str.length() < 3) return 0; if (str.charAt(0) == str.charAt(2)) return 1 + countPairs(str.substring(1)); return countPairs(str.substring(1)); } // -------------------------------------------------------------------------------------------- // Count recursively the total number of "abc" and "aba" substrings that appear // in the given string. public int countAbc(String str) { String left; if (str.length() < 3) return 0; left = str.substring(0, 3); if (left.equals("abc")) return 1 + countAbc(str.substring(3)); if (left.equals("aba")) return 1 + countAbc(str.substring(2)); return countAbc(str.substring(1)); } // -------------------------------------------------------------------------------------------- // Given a string, compute recursively (no loops) the number of "11" substrings // in the string. The "11" substrings should not overlap. public int count11(String str) { if (str.length() < 2) return 0; if (str.substring(0, 2).equals("11")) return 1 + count11(str.substring(2)); return count11(str.substring(1)); } // -------------------------------------------------------------------------------------------- // Given a string, return recursively a "cleaned" string where adjacent chars // that are the same have been reduced to a single char. So "yyzzza" yields // "yza". public String stringClean(String str) { if (str.length() < 2) return str; if (str.charAt(0) == str.charAt(1)) return stringClean(str.substring(1)); return str.charAt(0) + stringClean(str.substring(1)); } // -------------------------------------------------------------------------------------------- // Given a string, compute recursively the number of times lowercase "hi" // appears in the string, however do not count "hi" that have an 'x' immediately // before them. public int countHi2(String str) { if (str.length() < 2) return 0; if (str.length() == 2) return (str.equals("hi")) ? 1 : 0; if (str.charAt(0) == 'x') { if (str.substring(1, 3).equals("hi")) return countHi2(str.substring(3)); return countHi2(str.substring(1)); } if (str.substring(0, 2).equals("hi")) return 1 + countHi2(str.substring(2)); if (str.substring(1, 3).equals("hi")) return 1 + countHi2(str.substring(3)); return countHi2(str.substring(2)); } // -------------------------------------------------------------------------------------------- // Given a string that contains a single pair of parenthesis, compute // recursively a new string made of only of the parenthesis and their contents, // so "xyz(abc)123" yields "(abc)". public String parenBit(String str) { int len = str.length(); if (str.charAt(0) != '(') { if (str.charAt(len - 1) != ')') return parenBit(str.substring(1, len - 1)); return parenBit(str.substring(1)); } if (str.charAt(len - 1) != ')') return parenBit(str.substring(0, len - 1)); return str; } // -------------------------------------------------------------------------------------------- // Given a string, return true if it is a nesting of zero or more pairs of // parenthesis, like "(())" or "((()))". Suggestion: check the first and last // chars, and then recur on what's inside them. public boolean nestParen(String str) { int len = str.length(); if (len == 0) return true; if (str.charAt(0) == '(' && str.charAt(len - 1) == ')') return nestParen(str.substring(1, len - 1)); return false; } // -------------------------------------------------------------------------------------------- // Given a string and a non-empty substring sub, compute recursively the number // of times that sub appears in the string, without the sub strings overlapping. public int strCount(String str, String sub) { int sLen = sub.length(); if (str.length() < sLen) return 0; if (str.substring(0, sLen).equals(sub)) return 1 + strCount(str.substring(sLen), sub); return strCount(str.substring(1), sub); } // -------------------------------------------------------------------------------------------- // Given a string and a non-empty substring sub, compute recursively if at least // n copies of sub appear in the string somewere, possibly with overlapping. N // will be non-negative. public boolean strCopies(String str, String sub, int n) { if (n == 0) return true; if (str.length() < sub.length()) return false; if (str.substring(0, sub.length()).equals(sub)) return strCopies(str.substring(1), sub, n - 1); return strCopies(str.substring(1), sub, n); } // -------------------------------------------------------------------------------------------- // Given a string and a non-empty substring sub, compute recursively the largest // substring which starts and ends with sub and return its length. public int strDist(String str, String sub) { int stLen = str.length(); int sbLen = sub.length(); if (stLen < sbLen) return 0; if (str.substring(0, sbLen).equals(sub)) { if (str.substring(stLen - sbLen, stLen).equals(sub)) return stLen; return strDist(str.substring(0, stLen - 1), sub); } return strDist(str.substring(1), sub); } } |
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