Numbers At Most N Given Digit Set.java

class Solution {
    
    
    // DIGIT DP IS LOVE 
    Integer[][][] digitdp;
    public int solve(String num, int pos, boolean bound, Integer[] dig,boolean lead) {

        if (pos == num.length()) {
            return 1;
        }

        int maxDigit = -1;
        
        if(digitdp[pos][(bound==true)?1:0][(lead==true)?1:0] !=null) return digitdp[pos][(bound==true)?1:0][(lead==true)?1:0];

        if (bound) {
            maxDigit = num.charAt(pos) - '0';
        } else {
            maxDigit = 9;
        }

        int ans = 0;
        for (int i = 0; i <=maxDigit; i++) {
            
            // 0 can only be leading 
            if(i==0 && lead){

                ans += solve(num,pos+1,false,dig,lead);
                
            }else{
                
              int res =  Arrays.binarySearch(dig,i);

              if(res>=0){
                 // lead = false; // now it is not possible to 0 to be in lead any more once any other call has been made
                 ans += solve(num, pos + 1, bound & (i == num.charAt(pos)-'0'), dig,false);
              }
                
            }
        }

        return digitdp[pos][(bound==true)?1:0][(lead==true)?1:0] = ans;

    }

    public int atMostNGivenDigitSet(String[] digits, int n) {

        String num = n + "";
        Integer[] dig = new Integer[digits.length];
        for(int i=0;i<dig.length;i++){
            dig[i] = Integer.parseInt(digits[i]);
        }
        
        digitdp = new Integer[20][2][2];
        return solve(num, 0, true, dig,true)-1;
    }
}