Apoorv Jain Apoorv Jain - 7 months ago 22
Ruby Question

Logic to simplify item replacement chain

I have some items that are being replaced by other items hence creating a replacement chain as:


+---------------+------------------+
| Replaced Item | Replacement Item |
+---------------+------------------+
| A | B |
| B | C |
| C | D |
| G | H |
+---------------+------------------+


If we simplify it relationship chains we found as

A -> B -> C -> D

&
G -> H

Finally, I want to achieve output as simplified table Like:


+---------------+------------------+
| Replaced Item | Replacement Item |
+---------------+------------------+
| A | D |
| B | D |
| C | D |
| G | H |
+---------------+------------------+


My question is:
Is any existing API or algorithm to solve this type of chain simplification problem in javascript/java/ruby etc.

What I tried is:
I thought i could solve it by making use of Java references.
when we assign a reference to another reference so both references will point same object, Hence Object ID will be same.
I have created several references as:

String ref1 = "A";
String ref2 = "B";
String ref3 = "C";
String ref4 = "D";
String ref5 = "G";
String ref6 = "H";


I got hashcodes from ref.hashCode() method.

//A = 65
//B = 66
//C = 67
//D = 68
//E = 71
//F = 72
//----

// Now A --> B means
ref2 = ref1;
//A = 65
//B = 65
//C = 67
//D = 68
//E = 71
//F = 72
//----

// Now B --> C means
ref3 = ref2;
//A = 65
//B = 65
//C = 65
//D = 68
//E = 71
//F = 72
//----


// Now C --> D means
ref4 = ref3;
//A = 65
//B = 65
//C = 65
//D = 65
//E = 71
//F = 72
//----

// Now C --> D means
ref6 = ref5;
//A = 65
//B = 65
//C = 65
//D = 65
//E = 71
//F = 71
//----


Now I would need to iterate through all references and put hashcodes into a set which contains unique values. so i got 65 and 71 only.

Now,
65 -> A,B,C,D and precedence-wise D is last element.
71 -> G,H precedence-wise H is last element.

so i could conclude it as:


+---------------+------------------+
| Replaced Item | Replacement Item |
+---------------+------------------+
| A | D |
| B | D |
| C | D |
| G | H |
+---------------+------------------+

Answer Source

You can get a new "map" with some simple recursion:

var replacements = {
      A: 'B',
      B: 'C',
      C: 'D',
      G: 'H'
    },
    map = {};

function replace(letter){
    if(!replacements[letter]) // If the replacements don't contain the current letter,
      return letter;          // Just use the current letter
    return replace(replacements[letter]); // Otherwise, get the replacement letter.
}

for(var key in replacements){  // 'key' being: A, B, C, G
    map[key] = replace(key);   // Assign a new replacement value in the map.
}

console.log(map);