Explore the wonders of generating number permutations using JavaScript recursion!

Question

Explore the wonders of generating number permutations using JavaScript recursion!

After providing the input of 85 to this function, I noticed that it only returns 85. I am confused as to why it is not recursively calling itself again with 5 as the first number.

console.log(PermutationStep(85));

function PermutationStep(num) { 
  var permutations = [];

  recursive(String(num), String(num).length, [], '');
  return permutations;

  function recursive(num, numLength, used, currentPermutation) {
    console.log(currentPermutation);
    if (currentPermutation.length === numLength) {
      permutations.push(num);
    }
    for (var j=0; j<numLength; j++) {
      if (used[j]) continue;
      else {
        used[j]=true;
        recursive(num, numLength, used, currentPermutation+num[j]);
      }
    }
  }
}

javascript recursion permutation

Answer 1

Answer №1

When the length is reached, remember to push currPerm rather than num.
Ensure that used[j] is reset once it has been tracked.
Be cautious of generating duplicate results, such as 885; always check for repetitions before proceeding.

console.log(PermutationStep(85));
console.log(PermutationStep(885));

function PermutationStep(num) { 
  var perms = [];
  
  // Convert number to char array.
  var str = ('' + num).split('');
  var length = str.length;
  // Sort in alphabetical order.
  str.sort(function(a, b) {
    return a - b;
  });

  // Create used map.
  var used = str.map(function() {
    return false;
  });

  rec(str, '');
  return perms;

  // Note that JavaScript can access values from outer function scope,
  // So there's no need to pass length and used each time, similar to how you handle perms.
  function rec(num, currPerm) {
    console.log(currPerm);
    if (currPerm.length === length) {
      // Insert the constructed string here, not num
      perms.push(currPerm);
    }
    var prev = null, ch;
    for (var j = 0; j < length; j++) {
      if (used[j]) {
        continue;
      }
      ch = str[j];
      if (prev === ch) { 
        // If the current character matches the previous one, meaning there are at least 2 characters in the string with the same value,
        // Skip to prevent redundant results.
        continue;
      }
      used[j] = true;
      rec(num, currPerm + ch);
      // Reset the usability of the number.
      used[j]=false;
      prev = ch;
    }
  }
}

Answer 2

When the length is reached, remember to push currPerm rather than num.
Ensure that used[j] is reset once it has been tracked.
Be cautious of generating duplicate results, such as 885; always check for repetitions before proceeding.

console.log(PermutationStep(85));
console.log(PermutationStep(885));

function PermutationStep(num) { 
  var perms = [];
  
  // Convert number to char array.
  var str = ('' + num).split('');
  var length = str.length;
  // Sort in alphabetical order.
  str.sort(function(a, b) {
    return a - b;
  });

  // Create used map.
  var used = str.map(function() {
    return false;
  });

  rec(str, '');
  return perms;

  // Note that JavaScript can access values from outer function scope,
  // So there's no need to pass length and used each time, similar to how you handle perms.
  function rec(num, currPerm) {
    console.log(currPerm);
    if (currPerm.length === length) {
      // Insert the constructed string here, not num
      perms.push(currPerm);
    }
    var prev = null, ch;
    for (var j = 0; j < length; j++) {
      if (used[j]) {
        continue;
      }
      ch = str[j];
      if (prev === ch) { 
        // If the current character matches the previous one, meaning there are at least 2 characters in the string with the same value,
        // Skip to prevent redundant results.
        continue;
      }
      used[j] = true;
      rec(num, currPerm + ch);
      // Reset the usability of the number.
      used[j]=false;
      prev = ch;
    }
  }
}

Explore the wonders of generating number permutations using JavaScript recursion!

Answer №1

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