Numbers With Repeated Digits - Digit DP [JS]

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Description 

Solution: Digit DP

Memoize each dp(i, mask, state, hasRepeat), where

  • i = the ith digit

  • mask = bitmask which indicates which digit we have already used

  • state = indicates whether the current number is tracking smaller, equal, or greater than n

    • 0 = smaller
    • 1 = equal
    • 2 = greater

  • hasRepeat = whether we have a repeated digit

If hasRepeat is true (1), count it as 1 way.
For each state, count the total number of ways after appending each digit (0 - 9).

state:

  • If digit < n[index], update state to 0 (smaller) if state is currently 1 (equal).
  • If digit === n[index], keep state the same.
  • If digit > n[index], update state to 2 (greater) if state is currently 1 (equal).

d = number of digits in n
Time Complexity: O(d * 2^10 * 3 * 2 * 10) 488ms
d * 2^10 * 3 * 2 = the number of different states we can have
10 = at each state we have 10 options for digits 0-9
Space Complexity: O(d * 2^10 * 3 * 2) 62MB

var numDupDigitsAtMostN = function(n) {
  let str = n.toString(), size = str.length, memo = new Map();
  return dp(0, 0, 1, 0);
  
  function dp(i, mask, state, hasRepeat) {
    if (i === size) return state < 2 && hasRepeat ? 1 : 0;
    let key = `${i},${mask},${state},${hasRepeat}`;
    if (memo.has(key)) return memo.get(key);
    
    let ans = hasRepeat;
    for (let digit = 0; digit <= 9; digit++) {
      if (i === 0 && digit === 0) continue;
      let newMask = mask | (1 << digit), repeat = hasRepeat || (mask === newMask ? 1 : 0);
      if (digit < Number(str[i])) {
        ans += dp(i + 1, newMask, state === 1 ? 0 : state, repeat);
      } else if (digit === Number(str[i])) {
        ans += dp(i + 1, newMask, state, repeat);
      } else {
        ans += dp(i + 1, newMask, state === 1 ? 2 : state, repeat);
      }
    }
    memo.set(key, ans);
    return ans;
  }  
};

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Location: Auckland, New Zealand

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