combinations.js

//You have an array of characters (string) that may be '1', '0' o '*'. e.g. 10*00*0. The program needs to
//generate an output of all the possible combinations by replacing * with an 0 and 1. I.e. input> 10**0
//output> 10000, 10010, 10100, 10110. Input > *0 output > 00, 10.

function findOcurrencesOf(character = "*", array) {
  let occ = [];
  for (let i = 0; i < array.length; i++) {
    if (array[i] === character) {
      occ.push(i);
    }
  }
  return occ;
}

function produceCombinations(array) {
  let occ = findOcurrencesOf("*", array);
  console.log(occ)
  let combos = binaryCombos(occ.length);
  // Create a copy of the array
  let result = [];

  combos.forEach((combo, index) => {
     let arrayCopy = [...array];
     for (let i = 0; i < occ.length;i++) {
      // console.log(combo);
      arrayCopy[occ[i]] = combo.shift();
    }
    result.push(arrayCopy);
  });

  console.log(result);
}

//Returns an array of boolean values representing
//every possible combination of n binary digits
function binaryCombos(n){
    let result = [];
    for(let y=0; y<Math.pow(2,n); y++){
        let combo = [];
        for(x=0; x<n; x++){
            //shift bit and and it with 1 so that basically we're moving a "1" from right to left 
            if((y >> x) & 1) {
                combo.push("1");
            }
             else {
                combo.push("0");
             }
        }
        result.push(combo);
    }
    return result;
}

// Complexity is O(n!)

//  Test Cases
produceCombinations(["1","0","*","*","0","*","0"]);
produceCombinations(["1","0","*","*","0"]);
produceCombinations(["*","0"]);