firstswap to swap the characters

Ah yes - gotcha.  *Almost* the same :)

Doug

We'll say that 2 strings "match" if they are non-empty and their first chars are the same. Loop over and then return the given array of non-empty strings as follows: if a string matches an earlier string in the array, swap the 2 strings in the array. When a position in the array has been swapped, it no longer matches anything. Using a map, this can be solved making just one pass over the array. More difficult than it looks.

i was not clear on description

if string matches earlier string means whole string or some characters? what it means swapped when match? how many times we can swap.

please advise

The match is just "their first chars are the same".  So just the first character.

And then to swap means if we had strings:

A, B, C, D

and we swapped B and D then we'd have

A, D, C, B

sample input and outputs

firstSwap(["ab", "ac"]) → ["ac", "ab"]
firstSwap(["ax", "bx", "cx", "cy", "by", "ay", "aaa", "azz"]) → ["ay", "by", "cy", "cx", "bx", "ax", "aaa", "azz"]
firstSwap(["ax", "bx", "ay", "by", "ai", "aj", "bx", "by"]) → ["ay", "by", "ax", "bx", "ai", "aj", "bx", "by"]
firstSwap(["ax", "bx", "cx", "ay", "cy", "aaa", "abb"]) → ["ay", "bx", "cy", "ax", "cx", "aaa", "abb"]
firstSwap(["easy", "does", "it", "every", "ice", "eaten"]) → ["every", "does", "ice", "easy", "it", "eaten"]
firstSwap(["list", "of", "words", "swims", "over", "lily", "water", "wait"]) → ["lily", "over", "water", "swims", "of", "list", "words", "wait"]
firstSwap(["4", "8", "15", "16", "23", "42"]) → ["42", "8", "16", "15", "23", "4"]
firstSwap(["aaa"]) → ["aaa"]
firstSwap([]) → []
firstSwap(["a", "b", "c", "xx", "yy", "zz"]) → ["a", "b", "c", "xx", "yy", "zz"]

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firstSwap(["ab", "ac"]) → ["ac", "ab"]

since a match in both above 2 string swap 2nd character to get ac, ab (instead of ab, ac) and with other inputs?
is my understanding is correct?
how this firstswap different from other challenge allswap?

can we do without using map?
don't we need to use 2 for loops for this

I started some thing like below but not able to proceed

public String[] firstSwap(String[] strings) {
  for(int i=0;i<strings.length;i++){
    
    
    for(int j;j<strings.length;j++){
      if(strings.indexOf(0)
    }
    
  }
}

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please advise

this one swaps only first occurrences, not all

firstSwap(["ax", "bx", "cx", "ay", "cy", "aaa", "abb"]) → ["ay", "bx", "cy", "ax", "cx", "aaa", "abb"]

allSwap(["ax", "bx", "cx", "ay", "cy", "aaa", "abb"]) → ["ay", "bx", "cy", "ax", "cx", "abb", "aaa"]

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here, "ax" and "ay" are swapped but not "aaa" or "abb"

also, in the page you provided it says

Using a map, this can be solved making just one pass over the array.

this is one pass solution which is efficient than 2 loops inside...

this is one pass solution which is efficient than 2 loops inside.

can you show with 2 for loops then i can see map example

so do you want me write multiple solutions to a given problem and then compare :)

it will be something like this:

public String[] firstSwap(String[] strings) {
  Map<Character, Integer> map = new HashMap<>();
  
  for (int i=0; i<strings.length-1; i++){
    Character firstChar = strings[i].charAt(0);
    if (!map.containsKey(firstChar)){
      for (int j=i+1; j<strings.length; j++)  {
        if (firstChar==strings[j].charAt(0)){
          map.put(firstChar, 1);
          String t = strings[i];
          strings[i] = strings[j];
          strings[j] = t;
          break;
        }
      }
    }
  }

  return strings;
}

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number of iteration in this code will be

i=0 > j=1..n > total n-1
i=1 > j=2..n > total n-2
...
i=n-1 > j=n..n > total 1

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if we sum all,

(n-1) + (n-2) + (n-3) + ... + 1 = n*(n-1)/2

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for n=6, we will have 6*5/2=15 iteration (worst case, all  items are starting with different character)...

like this one:

firstSwap(["a", "b", "c", "xx", "yy", "zz"]) → ["a", "b", "c", "xx", "yy", "zz"]

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to test number of iterations, check this code

public String[] firstSwap(String[] strings) {
  Map<Character, Integer> map = new HashMap<>();
  int n=0;
  for (int i=0; i<strings.length-1; i++){
    Character firstChar = strings[i].charAt(0);
    if (!map.containsKey(firstChar)){
      for (int j=i+1; j<strings.length; j++)  {
        n++;
        if (firstChar==strings[j].charAt(0)){
          map.put(firstChar, 1);
          String t = strings[i];
          strings[i] = strings[j];
          strings[j] = t;
          break;
        }
      }
    }
  }
  //strings[0] = Integer.toString(n);
  return strings;
}

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uncomment line 19 to test...

let me digest this

[quote]this one swaps only first occurrences, not all

firstSwap(["ax", "bx", "cx", "ay", "cy", "aaa", "abb"]) → ["ay", "bx", "cy", "ax", "cx", "aaa", "abb"]

allSwap(["ax", "bx", "cx", "ay", "cy", "aaa", "abb"]) → ["ay", "bx", "cy", "ax", "cx", "abb", "aaa"]

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here, "ax" and "ay" are swapped but not "aaa" or "abb"[/quote]

now i understood diference between first swap and all swap challenge question.

what is the difference in solution for these two challenge questions?

i was not clear on overall code flow and specifically below 3 lines




// We've seen this same first letter and havent swapped yet, so swap the two entries
      String swap = strings[i];
      strings[i] = strings[match];
      strings[match] = swap;

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  map.put(firstChar, -1) ;

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what is meaning of above line?

how this firstSwap code diferent from allSwap code

public String[] allSwap(String[] strings) {
  // The map is from the first character in the string to its position in the array.
  Map<Character, Integer> map = new HashMap<>() ;
  
  for (int i = 0 ; i < strings.length ; i++) {
    Character firstChar = strings[i].charAt(0) ;
    Integer match = map.get(firstChar) ;

    if (match == null) {
       // This is the first time we've seen this first letter
      map.put(firstChar, i) ;
    } else {
      // We've seen this same first letter before so swap the two entries
      String swap = strings[i] ;
      strings[i] = strings[match] ;
      strings[match] = swap ;

      // And they say once you've done a swap you can't swap
      // again so remove it from the map.
      map.remove(firstChar) ;
    }
  }
  
  return strings;
}

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map.remove(firstChar) ;
how above different from below
  map.put(firstChar, -1) ;

i think i got it instead if removing a,0 we are making a, -1

for ab, ac, ax, ay example
once ab ac swapped to ac ab
then ax turn comes
first letter in ax a and map.get a is -1 since else condition checks if not -1 then only goes inside and swaps so ax remains as it is
simiary ay 4th iteration turn comes
first letter in ay a and map.get a is -1 since else condition checks if not -1 then only goes inside and swaps so ax remains as it is

any other wqy of doing than map?

public String[] firstSwap(String[] strings) {
  Map<Character, Integer> map = new HashMap<>();
  
  for (int i=0; i<strings.length-1; i++){
    Character firstChar = strings[i].charAt(0);
    if (!map.containsKey(firstChar)){
      for (int j=i+1; j<strings.length; j++)  {
        if (firstChar==strings[j].charAt(0)){
          map.put(firstChar, 1);
          String t = strings[i];
          strings[i] = strings[j];
          strings[j] = t;
          break;
        }
      }
    }
  }

  return strings;
}

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number of iteration in this code will be

i=0 > j=1..n > total n-1
i=1 > j=2..n > total n-2
...
i=n-1 > j=n..n > total 1

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not clear on n(n-1)/2 iternations here? how that many iterations? how to count them?

If strings[] contains ["0", "1", "2", "3", "4"]
How many times will you execute line 8 when i=0?

i is index of array or string array element?

  
  for (int i=0; i<strings.length-1; i++){
    Character firstChar = strings[i].charAt(0);
    if (!map.containsKey(firstChar)){
      for (int j=i+1; j<strings.length; j++)  {
        if (firstChar==strings[j].charAt(0)){

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line 8 has j right not i? i am confused. please advise

i is this i  for (int i=0; i<strings.length-1; i++){
line 8 in https:#a42351839 was if (firstChar==strings[j].charAt(0)){

line 8 in https:#a42351839 was if (firstChar==strings[j].charAt(0)){

for me it is showing as line 5

public String[] firstSwap(String[] strings) {
  Map<Character, Integer> map = new HashMap<>();
  
  for (int i=0; i<strings.length-1; i++){
    Character firstChar = strings[i].charAt(0);

If strings[] contains ["0", "1", "2", "3", "4"]
How many times will you execute line 8 when i=0?// for i =0 we run once then increment i to 1 right?
How many times will you execute line 8 when i=1?for i =1 we run once then increment i to 2 right?same way others
How many times will you execute line 8 when i=2?
How many times will you execute line 8 when i=3?
How many times will you execute line 8 when i=4?
What is the total?//5 times?

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for me, https://www.experts-exchange.com/viewCodeSnippet.jsp?codeSnippetId=20-42351839-1 displays as

 1: public String[] firstSwap(String[] strings) {
 2:   Map<Character, Integer> map = new HashMap<>();
 3:
 4:   for (int i=0; i<strings.length-1; i++){
 5:     Character firstChar = strings[i].charAt(0);
 6:     if (!map.containsKey(firstChar)
 7:       for (int j=i+1; j<strings.length; j++) {
 8:         if (firstChar==strings[j].charAt(0)){
 9:           map.put(firstChar, 1);
10:           String t = strings[i];
11:           strings[i] = strings[j];
12:           strings[j] = t;
13:           break;
14:         }
15:       }
16:     }
17:   }
18:
19:   return strings;
20: }
21:
22: Select all
23:
24: Open in new window
25:
26: number of iteration in this code will be
27:
28: i=0 > j=1..n > total n-1
29: i=1 > j=2..n > total n-2
30: ...
31: i=n-1 > j=n..n > total 1

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public String[] firstSwap(String[] strings) {
 2:   Map<Character, Integer> map = new HashMap<>();
 3:
 4:   for (int i=0; i<strings.length-1; i++){
 5:     Character firstChar = strings[i].charAt(0);
 6:     if (!map.containsKey(firstChar)
 7:       for (int j=i+1; j<strings.length; j++) {
 8:         if (firstChar==strings[j].charAt(0)){

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line 8 is if condition within iner for loop whose counter is j ??

Yes. How many times does that condition get tested?

i*j times? just guessing

for i=0, j goes on and on until strings.length
for (int i=0; i<strings.length-1; i++){
 5:     Character firstChar = strings[i].charAt(0);
 6:     if (!map.containsKey(firstChar)
 7:       for (int j=i+1; j<strings.length; j++) {
 8:         if (firstChar==strings[j].charAt(0)){

then i=1 again j goes on and on until strings.length

until i<strings.length again j goes on and on until strings.length

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Think we covered this pretty well :)