POTD Solutions | 10 Nov’ 23 | Number following a pattern
Last Updated : 28 Apr, 2025
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POTD Solutions 10 November 2023
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POTD 10 November: Number following a pattern
Given a pattern containing only I's and D's. I for increasing and D for decreasing. Devise an algorithm to print the minimum number following that pattern. Digits from 1-9 and digits can't repeat.
Examples:
Input: D
Output: 21
Input: I
Output: 12
Input: DD
Output: 321
We can that when we encounter I, we got numbers in increasing order but if we encounter ‘D’, we want to have numbers in decreasing order. Length of the output string is always one more than the input string. So the loop is from 0 to the length of the string. We have to take numbers from 1-9 so we always push (i+1) to our stack. Then we check what is the resulting character at the specified index.So, there will be two cases which are as follows:
Case 1: If we have encountered I or we are at the last character of input string, then pop from the stack and add it to the end of the output string until the stack gets empty.
Case 2: If we have encountered D, then we want the numbers in decreasing order. so we just push (i+1) to our stack.
Below is the implmentation of the above approach:
C++ class Solution{ public: string printMinNumberForPattern(string seq){ // result store output string string result; // create an empty stack of integers stack<int> stk; // run n+1 times where n is length of input sequence for (int i = 0; i <= seq.length(); i++) { // push number i+1 into the stack stk.push(i + 1); // if all characters of the input sequence are // processed or current character is 'I' // (increasing) if (i == seq.length() || seq[i] == 'I') { // run till stack is empty while (!stk.empty()) { // remove top element from the stack and // add it to solution result += to_string(stk.top()); stk.pop(); } } } return result ; } }; class Solution{ static String printMinNumberForPattern(String seq){ // result store output string StringBuilder result = new StringBuilder(); // create an empty stack of integers Stack<Integer> stk = new Stack<>(); // run n+1 times where n is the length of the input sequence for (int i = 0; i <= seq.length(); i++) { // push number i+1 into the stack stk.push(i + 1); // if all characters of the input sequence are // processed or the current character is 'I' (increasing) if (i == seq.length() || seq.charAt(i) == 'I') { // run until the stack is empty while (!stk.isEmpty()) { // remove the top element from the stack and // add it to the solution result.append(stk.pop()); } } } return result.toString(); } } class Solution: def printMinNumberForPattern(ob,seq): # result stores the output string result = [] # create an empty stack of integers stk = [] # run n+1 times where n is the length of the input sequence for i in range(len(seq) + 1): # push number i+1 into the stack stk.append(i + 1) # if all characters of the input sequence are # processed or the current character is 'I' (increasing) if i == len(seq) or seq[i] == 'I': # run until the stack is empty while stk: # remove the top element from the stack and # add it to the solution result.append(str(stk.pop())) return ''.join(result)
Time Complexity: O(n)
Auxiliary Space: O(n), since n extra space has been taken.
- Since we have to find a minimum number without repeating digits, maximum length of output can be 9 (using each 1-9 digits once)
- Length of the output will be exactly one greater than input length.
- The idea is to iterate over the string and do the following if current character is ‘I’ or string is ended.
- Assign count in increasing order to each element from current-1 to the next left index of ‘I’ (or starting index is reached).
- Increase the count by 1.
Below is the implmentation of the above approach:
C++ class Solution{ public: string printMinNumberForPattern(string seq){ int n = seq.length(); if (n >= 9) return "-1"; string result(n+1, ' '); int count = 1; // The loop runs for each input character as well as // one additional time for assigning rank to remaining characters for (int i = 0; i <= n; i++) { if (i == n || seq[i] == 'I') { for (int j = i - 1 ; j >= -1 ; j--) { result[j + 1] = '0' + count++; if(j >= 0 && seq[j] == 'I') break; } } } return result; } }; class Solution{ static String printMinNumberForPattern(String seq){ int n = seq.length(); if (n >= 9) { return "-1"; } char[] result = new char[n + 1]; int count = 1; // The loop runs for each input character as well as // one additional time for assigning rank to remaining characters for (int i = 0; i <= n; i++) { if (i == n || seq.charAt(i) == 'I') { for (int j = i - 1; j >= -1; j--) { result[j + 1] = (char) ('0' + count++); if (j >= 0 && seq.charAt(j) == 'I') { break; } } } } return new String(result); } } class Solution: def printMinNumberForPattern(ob,seq): # result stores the output string n = len(seq) if n >= 9: return "-1" result = [' '] * (n + 1) count = 1 # The loop runs for each input character as well as # one additional time for assigning rank to remaining characters for i in range(n + 1): if i == n or seq[i] == 'I': j = i - 1 while j >= -1: result[j + 1] = str(count) count += 1 if j >= 0 and seq[j] == 'I': break j -= 1 return ''.join(result)
Time Complexity: O(N)
Auxiliary Space: O(N), since N extra space has been taken.