Developers flex on each with the strangest of ways. How green is your github commit history. Have you submitted a PR to a open source project. How well can you solve the silliest questions on leetcode / geeksforgeeks / hackerrank?
While normies use money and status as a measure of success. Coders have incelled their way into a paradise where the glow of the monitor is all you need.
Pray for death all you want, it’s not coming. Man up and start grinding.
You are given a 0-indexed string pattern of length n consisting of the characters 'I' meaning increasing and 'D' meaning decreasing.
A 0-indexed string num of length n + 1 is created using the following conditions:
numconsists of the digits'1'to'9', where each digit is used at most once.- If
pattern[i] == 'I', thennum[i] < num[i + 1]. - If
pattern[i] == 'D', thennum[i] > num[i + 1].
Return the lexicographically smallest possible string num that meets the conditions.
Example 1:
Input: pattern = "IIIDIDDD" Output: "123549876" Explanation: At indices 0, 1, 2, and 4 we must have that num[i] < num[i+1]. At indices 3, 5, 6, and 7 we must have that num[i] > num[i+1]. Some possible values of num are "245639871", "135749862", and "123849765". It can be proven that "123549876" is the smallest possible num that meets the conditions. Note that "123414321" is not possible because the digit '1' is used more than once.
Example 2:
Input: pattern = "DDD" Output: "4321" Explanation: Some possible values of num are "9876", "7321", and "8742". It can be proven that "4321" is the smallest possible num that meets the conditions.
Constraints:
1 <= pattern.length <= 8patternconsists of only the letters'I'and'D'.
class Solution {
public:
string smallestNumber(string pattern) {
string result;
stack<int> st;
for( int i = 0; i<= pattern.size(); i++ ){
st.push(i+1);
if ( i == pattern.length() || pattern[i] == 'I' )
while(!st.empty()){
result+=to_string( st.top() );
st.pop();
}
}
return result;
}
};