Largest Palindromic Number - Counting & Greedy [JS]

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Description 

Solution: Counting & Greedy

  1. Count the number of occurances of each digit in num.
  2. Construct the first half of the number.
    • It is always optimal to put larger digits earlier on.
    • Going from 9 to 0, add digits to the end of result while the count of the digit > 1.
    • We use two occurances per new digit since it needs to be symmetrical.
  3. Get the largest possible single digit as the middle of the palindrome.
  4. Add the second half of the palindrome. This is just the symmetrical right side of the left half.

Time Complexity: O(n) 135ms
Space Complexity: O(n) 54.2MB

var largestPalindromic = function(num) {
  let count = Array(10).fill(0);
  for (let char of num) {
    let digit = Number(char);
    count[digit]++;
  }

  let res = [];
  for (let i = 9; i >= 0; i--) {
    if (i === 0 && res.length === 0) break; // no leading zeros
    while (count[i] > 1) {
      res.push(i.toString());
      count[i] -= 2;
    }
  }
  
  let hasSingleMid = false;
  for (let i = 9; i >= 0; i--) {
    if (count[i] > 0) {
      res.push(i.toString());
      hasSingleMid = true;
      break;
    }
  }
  for (let i = res.length - (hasSingleMid ? 2 : 1); i >= 0; i--) {
    res.push(res[i]);
  }
  return res.join("");
};

javascript
Location: Auckland, New Zealand

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