Smallest Number With No Adjacent Duplicates - Google Top Interview Questions


Problem Statement :


You are given a string s containing "1", "2", "3" and "?". Given that you can replace any “?” with "1", "2" or "3", return the smallest number you can make as a string such that no two adjacent digits are the same.

Constraints

n ≤ 100,000 where n is the length of s

Example 1

Input

s = "3?2??"

Output

"31212"

Example 2

Input

s = "???"

Output

"121"



Solution :



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                        Solution in C++ :

string solve(string s) {
    for (int i = 0; i < s.size(); i++) {
        if (s[i] == '?') {
            bool one = true, two = true;
            if (i > 0) {
                if (s[i - 1] - '0' == 1) one = false;
                if (s[i - 1] - '0' == 2) two = false;
            }
            if (i < s.size() - 1) {
                if (s[i + 1] - '0' == 1) one = false;
                if (s[i + 1] - '0' == 2) two = false;
            }
            if (one) {
                s[i] = '1';
            } else if (two)
                s[i] = '2';
            else
                s[i] = '3';
        }
    }
    return s;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public String solve(String s) {
        char[] c = s.toCharArray();
        int N = s.length();
        for (int i = 0; i < N; i++) {
            if (c[i] != '?')
                continue;
            boolean[] b = new boolean[3];
            if (i > 0)
                b[c[i - 1] - '1'] = true;
            if (i < N - 1 && c[i + 1] != '?')
                b[c[i + 1] - '1'] = true;
            for (int j = 0; j < 3; j++) {
                if (!b[j]) {
                    c[i] = (char) (j + '1');
                    break;
                }
            }
        }
        return new String(c);
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, s):
        A = list(s)
        for i, c in enumerate(A):
            if c == "?":
                for d in "123":
                    A[i] = d
                    if i - 1 >= 0 and A[i] == A[i - 1]:
                        continue
                    if i + 1 < len(A) and A[i] == A[i + 1]:
                        continue
                    break

        return "".join(A)
                    


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