Successfully counted each letter but letter order can't display sequentially as in word. For example- Input word-bbinood Output=b2i1n1o2d1
Here my program in qbasic:
INPUT "Enter the string:", A$
n = LEN(A$)
FOR i = 97 TO 122
FOR j = 1 TO n
IF CHR$(i) = MID$(A$, j, 1) THEN
count = count + 1
END IF
NEXT
FOR j = 1 TO n
IF (MID$(A$, j, 1) = CHR$(i)) THEN
PRINT CHR$(i), count
j = n
END IF
NEXT
count = 0
NEXT
There are two ways to answer this.
The first is a simple run-length encoding scheme, where the input cabc results in c1 a1 b1 c1. This has the benefit that you can often output things immediately with a fairly low memory requirement to boot:
input-string := Get-Input
(* nil is the representation of no characters here. *)
last-char := nil
count := 0
For Each (char c) In input-string
If c = last-char Then
count := count + 1
Else
If last-char = nil Then
count := 1
last-char := c
Else
Display last-char, count
count := 1
last-char := c
End If
End If
Loop
If count != 0 Then
Display last-char, count
End If
The other solution I conceived will preserve the order and determine the counts of all unique letters in the string, yielding c2 a1 b1 for an input of cabc. This solution is a bit more complex and requires more memory and often more execution time, but it results in more compact output due to the lack of repeated letters:
input-string := Get-Input
(* 26 is the number of letters a-z. *)
counts := Create-Array 26
order-string := ""
For Each (char c) In input-string
i := Locate-Char order-string, c
If i = 0 Then
order-string := order-string + c
counts [Length order-string] := 1
Else
counts [i] := counts [i] + 1
End If
Loop
For i := 1 To (Length order-string)
Display (Char-At order-string, i), counts [i]
Loop
The first should be straightforward to convert to QBASIC, but you might need to use the help file to learn about the DIM keyword and how to use it to create an array. This algorithm assumes that arrays start at 1, not 0.
Okay, here's code that should work in Qbasic.
DEFINT A-Z
DIM char(1 TO 255) AS STRING * 1
DIM outp(1 TO 255) AS STRING
INPUT "Type your string: ", inp$
'**** Comment out the following line if you want upper and lower cases
'**** treated separately:
inp$ = LCASE$(inp$)
FOR i = 1 TO LEN(inp$)
char(i) = MID$(inp$, i, 1)
NEXT i
l = 0
FOR i = 1 TO LEN(inp$)
k = 1
FOR j = 1 TO i - 1
IF char(j) = char(i) THEN GOTO skplet
NEXT j
l = l + 1
FOR j = i + 1 TO LEN(inp$)
IF char(j) = char(i) THEN k = k + 1
NEXT j
outp(l) = char(i) + LTRIM$(STR$(k))
skplet:
NEXT i
FOR i = 1 TO l
PRINT outp(i);
NEXT i
Note that, as remarked in the comment, upper and lower case will be treated as the same letters as this answer stands. If you want them to be treated separately, simply delete or comment out the inp$ = LCASE$(inp$) line. Hope this helps!
suggest 2 arrays, each 26 (if only considering alphabetic letters) that array
size_t counts[26] = {0};
then a second array `
char order[26] = {'\0'};
then for each letter in the input string.
if( isalpha( str[i]) )
{
letter = tolower( str[i];
counts[ letter - 'a' ]++;
then loop through order if match found the do nothing, else replace '\0' with letter
printing out would be loop through order[] until either '\0' encountered or all elements examined.
At each order[] != '\0':
putc( order[ element ]);
printf( "%d", counts[ element-'\a' ] );
