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pre { line-height: 125%; }
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span.linenos { color: inherit; background-color: transparent; padding-left: 5px; padding-right: 5px; }
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.highlight .il { color: #0000DD; font-weight: bold } /* Literal.Number.Integer.Long */
#import "/template.typ": highlight-output
#highlight-output[```
$ java StringOperations 
Menu-driven program for string operations
Enter a string: A quick fox jumps over a lazy dog.
Options:
 1. Re-enter a string
 2. Display the string
 3. Count words in the string
 4. Reverse the string
 5. Case-insensitively check whether the string is palindrome or not
 6. Exit

Enter your choice: 2
The string is: A quick brown fox jumps over a lazy dog.
Options:
 1. Re-enter a string
 2. Display the string
 3. Count words in the string
 4. Reverse the string
 5. Case-insensitively check whether the string is palindrome or not
 6. Exit

Enter your choice: 3
The string has 8 words.
Options:
 1. Re-enter a string
 2. Display the string
 3. Count words in the string
 4. Reverse the string
 5. Case-insensitively check whether the string is palindrome or not
 6. Exit

Enter your choice: 4
The given string reversed is: .god yzal eht revo spmuj xof nworb kciuq A
Options:
 1. Re-enter a string
 2. Display the string
 3. Count words in the string
 4. Reverse the string
 5. Case-insensitively check whether the string is palindrome or not
 6. Exit

Enter your choice: 5
The given string isn't case-insensitively palindrome.
Options:
 1. Re-enter a string
 2. Display the string
 3. Count words in the string
 4. Reverse the string
 5. Case-insensitively check whether the string is palindrome or not
 6. Exit

Enter your choice: 1
Enter a string: Aa bB C bB Aa
Options:
 1. Re-enter a string
 2. Display the string
 3. Count words in the string
 4. Reverse the string
 5. Case-insensitively check whether the string is palindrome or not
 6. Exit

Enter your choice: 5
The given string is case-insensitively palindrome.
Options:
 1. Re-enter a string
 2. Display the string
 3. Count words in the string
 4. Reverse the string
 5. Case-insensitively check whether the string is palindrome or not
 6. Exit

Enter your choice: 6
Bye.
```]
nd-color: #fff0f0 } /* Literal.String.Affix */ .highlight .sb { color: #dd2200; background-color: #fff0f0 } /* Literal.String.Backtick */ .highlight .sc { color: #dd2200; background-color: #fff0f0 } /* Literal.String.Char */ .highlight .dl { color: #dd2200; background-color: #fff0f0 } /* Literal.String.Delimiter */ .highlight .sd { color: #dd2200; background-color: #fff0f0 } /* Literal.String.Doc */ .highlight .s2 { color: #dd2200; background-color: #fff0f0 } /* Literal.String.Double */ .highlight .se { color: #0044dd; background-color: #fff0f0 } /* Literal.String.Escape */ .highlight .sh { color: #dd2200; background-color: #fff0f0 } /* Literal.String.Heredoc */ .highlight .si { color: #3333bb; background-color: #fff0f0 } /* Literal.String.Interpol */ .highlight .sx { color: #22bb22; background-color: #f0fff0 } /* Literal.String.Other */ .highlight .sr { color: #008800; background-color: #fff0ff } /* Literal.String.Regex */ .highlight .s1 { color: #dd2200; background-color: #fff0f0 } /* Literal.String.Single */ .highlight .ss { color: #aa6600; background-color: #fff0f0 } /* Literal.String.Symbol */ .highlight .bp { color: #003388 } /* Name.Builtin.Pseudo */ .highlight .fm { color: #0066bb; font-weight: bold } /* Name.Function.Magic */ .highlight .vc { color: #336699 } /* Name.Variable.Class */ .highlight .vg { color: #dd7700 } /* Name.Variable.Global */ .highlight .vi { color: #3333bb } /* Name.Variable.Instance */ .highlight .vm { color: #336699 } /* Name.Variable.Magic */ .highlight .il { color: #0000DD; font-weight: bold } /* Literal.Number.Integer.Long */
#
#
#           The Nim Compiler
#        (c) Copyright 2018 Andreas Rumpf
#
#    See the file "copying.txt", included in this
#    distribution, for details about the copyright.
#

## BTree implementation with few features, but good enough for the
## Nim compiler's needs.

const
  M = 512    # max children per B-tree node = M-1
             # (must be even and greater than 2)
  Mhalf = M div 2

type
  Node[Key, Val] = ref object
    entries: int
    keys: array[M, Key]
    case isInternal: bool
    of false:
      vals: array[M, Val]
    of true:
      links: array[M, Node[Key, Val]]
  BTree*[Key, Val] = object
    root: Node[Key, Val]
    entries: int      ## number of key-value pairs

proc initBTree*[Key, Val](): BTree[Key, Val] =
  BTree[Key, Val](root: Node[Key, Val](entries: 0, isInternal: false))

template less(a, b): bool = cmp(a, b) < 0
template eq(a, b): bool = cmp(a, b) == 0

proc getOrDefault*[Key, Val](b: BTree[Key, Val], key: Key): Val =
  var x = b.root
  while x.isInternal:
    for j in 0..<x.entries:
      if j+1 == x.entries or less(key, x.keys[j+1]):
        x = x.links[j]
        break
  assert(not x.isInternal)
  for j in 0..<x.entries:
    if eq(key, x.keys[j]): return x.vals[j]

proc contains*[Key, Val](b: BTree[Key, Val], key: Key): bool =
  var x = b.root
  while x.isInternal:
    for j in 0..<x.entries:
      if j+1 == x.entries or less(key, x.keys[j+1]):
        x = x.links[j]
        break
  assert(not x.isInternal)
  for j in 0..<x.entries:
    if eq(key, x.keys[j]): return true
  return false

proc copyHalf[Key, Val](h, result: Node[Key, Val]) =
  for j in 0..<Mhalf:
    result.keys[j] = h.keys[Mhalf + j]
  if h.isInternal:
    for j in 0..<Mhalf:
      result.links[j] = h.links[Mhalf + j]
  else:
    for j in 0..<Mhalf:
      shallowCopy(result.vals[j], h.vals[Mhalf + j])

proc split[Key, Val](h: Node[Key, Val]): Node[Key, Val] =
  ## split node in half
  result = Node[Key, Val](entries: Mhalf, isInternal: h.isInternal)
  h.entries = Mhalf
  copyHalf(h, result)

proc insert[Key, Val](h: Node[Key, Val], key: Key, val: Val): Node[Key, Val] =
  #var t = Entry(key: key, val: val, next: nil)
  var newKey = key
  var j = 0
  if not h.isInternal:
    while j < h.entries:
      if eq(key, h.keys[j]):
        h.vals[j] = val
        return
      if less(key, h.keys[j]): break
      inc j
    for i in countdown(h.entries, j+1):
      shallowCopy(h.vals[i], h.vals[i-1])
    h.vals[j] = val
  else:
    var newLink: Node[Key, Val] = nil
    while j < h.entries:
      if j+1 == h.entries or less(key, h.keys[j+1]):
        let u = insert(h.links[j], key, val)
        inc j
        if u == nil: return nil
        newKey = u.keys[0]
        newLink = u
        break
      inc j
    for i in countdown(h.entries, j+1):
      h.links[i] = h.links[i-1]
    h.links[j] = newLink

  for i in countdown(h.entries, j+1):
    h.keys[i] = h.keys[i-1]
  h.keys[j] = newKey
  inc h.entries
  return if h.entries < M: nil else: split(h)

proc add*[Key, Val](b: var BTree[Key, Val]; key: Key; val: Val) =
  let u = insert(b.root, key, val)
  inc b.entries
  if u == nil: return

  # need to split root
  let t = Node[Key, Val](entries: 2, isInternal: true)
  t.keys[0] = b.root.keys[0]
  t.links[0] = b.root
  t.keys[1] = u.keys[0]
  t.links[1] = u
  b.root = t

proc toString[Key, Val](h: Node[Key, Val], indent: string; result: var string) =
  if not h.isInternal:
    for j in 0..<h.entries:
      result.add(indent)
      result.add($h.keys[j] & " " & $h.vals[j] & "\n")
  else:
    for j in 0..<h.entries:
      if j > 0: result.add(indent & "(" & $h.keys[j] & ")\n")
      toString(h.links[j], indent & "   ", result)

proc `$`[Key, Val](b: BTree[Key, Val]): string =
  result = ""
  toString(b.root, "", result)

proc hasNext*[Key, Val](b: BTree[Key, Val]; index: int): bool = index < b.entries

proc countSubTree[Key, Val](it: Node[Key, Val]): int =
  if it.isInternal:
    result = 0
    for k in 0..<it.entries:
      inc result, countSubTree(it.links[k])
  else:
    result = it.entries

proc next*[Key, Val](b: BTree[Key, Val]; index: int): (Key, Val, int) =
  var it = b.root
  var i = index
  # navigate to the right leaf:
  while it.isInternal:
    var sum = 0
    for k in 0..<it.entries:
      let c = countSubTree(it.links[k])
      inc sum, c
      if sum > i:
        it = it.links[k]
        dec i, (sum - c)
        break
  result = (it.keys[i], it.vals[i], index+1)

iterator pairs*[Key, Val](b: BTree[Key, Val]): (Key, Val) =
  var i = 0
  while hasNext(b, i):
    let (k, v, i2) = next(b, i)
    i = i2
    yield (k, v)

proc len*[Key, Val](b: BTree[Key, Val]): int {.inline.} = b.entries