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Min Deletions To Obtain String in Right Format Microsoft OA 2023

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Min Deletions To Obtain String in Right Format

You are given a string s consisting only of characters ‘a’ and ‘b’​​​​. You can delete any number of characters in s to make s balanced. s is balanced if there is no pair of indices (i,j) such that i < j and s[i] = ‘b’ and s[j]= ‘a’. Return the minimum number of deletions needed to make s balanced.

Example 1:

Input: s = “aababbab”

Output: 2

Explanation: You can either:
Delete the characters at 0-indexed positions 2 and 6 (“aababbab” -> “aaabbb”), or
Delete the characters at 0-indexed positions 3 and 6 (“aababbab” -> “aabbbb”).

Example 2:

Input: s = “bbaaaaabb”

Output: 2

Explanation: The only solution is to delete the first two characters.

Constraints:

  • 1 <= s.length <= 105
  • s[i] is ‘a’ or ‘b’​​.

Also See: Amazon OA Online Assessment 2023 Questions and Answers

SOLUTION

Program: Min Deletions To Obtain String in Right Format in Python

Two pointer Python O(N) time and O(1) space solution

The idea is to use two pointers i and j where i traverses from left to right and j traverses from right to left.
While traversing right, get to the first position where s[i] == ‘b’ and similarly while traversing left, get to the first position where s[j] == ‘a’. We also keep on adjusting the count of ‘a’ and ‘b’ accordingly. Now we reach to a point where s[0:i] is all ‘a’ and s[j+1:] is all ‘b’. At this point we need to decide whether we want to delete ‘a’ or ‘b’. So we go greedy and delete that char whose count is less. So if count_a < count_b, we delete char ‘a’ else we delete char ‘b’.

def get_a_count(word):
    count = 0
    for ch in word:
        if ch == 'a':
            count += 1
    return count        
    
def get_minimum_deletions(word):
    result = 0
    i = 0
    j = len(word)-1
    count_a = get_a_count(word)
    count_b = len(word)-count_a
    while i < j:
	    # get to the point where word[i] = 'b'
        while i < j and word[i] == 'a':
            i += 1
            count_a -= 1
		# get to the point where word[j] = 'a'	
        while i < j and word[j] == 'b':
            j -= 1
            count_b -= 1
            
		# we go greedy here and delete that char whose count is less
        if count_a != 0 and count_b != 0:
            if count_a < count_b:
                j -= 1  # simulates deletion of 'a'
                result += 1
                count_a -= 1
            else:
                i += 1  # simulates deletion of 'b'
                result += 1
                count_b -= 1
    return result

Program: Min Deletions To Obtain String in Right Format in C++

Logic:

Scan the string from left to right

  • 1. drop all sequences of the same chars longer than 2;
  • 2. count length of sequences of different chars;
  • 3. save length of the longest sequence.

Algorithm:

We should have a counter of the same chars and index which points to the last two chars of this sequence. Lets call them “count” and “start”. And we should have two variable where we will save length of the longest sequence of different chars and index which points to the beginning of this sequence. Lets call them “max_length” and “start_ml”.

So the algorithm in general.

Scan the string from left to right:

  • 1. Take next char of the string;
    • 1.1 If next char is the same as the previous one increase counter of the same chars “count”;
    • 1.2 If next char is different drop counter of the same chars “count” to 1;
  • 2 If number of previous the same chars is:
    • 2.1 More than 2. Move pointer to the beginning of the current sequence of different chars “start” to the previous char before the current one, to keep only two the same chars at the beginning of the sequence.
    • drop counter of the same chars “count” to 2.
    • 2.2. Less or equal 2. Check if current sequence of different chars is longer than current maximum. If yes — update maximum to the current length. Save pointer to the beginning of the new longest sequence.
  • 3. Go to phase 0.
#include <iostream>
#include <vector>
using namespace std;
string solution(const string &s) {
    int s_size = s.size();
    // start position and length of the longest sequence
    // which doesn't contain 3 contiguous occurrences of a and b
    int start_ml = 0, max_length = 0;
    int start = 0; // start of current processing string of the same letters.
    int count = 1; // length of current processing string of the same letters.
    for (int i = 1; i < s_size; ++i) {
        if (s[i] == s[i - 1]) {
            // if we met two the same letters increase the counter of the same letters
            count++;
        }
        else {
            // if next letter is different drop the counter to 1
            count = 1;
        }
        if (count <= 2) {
            // if the sequence of different letters continuing, set it's current length as
            // max_length if it is bigger than current max_length
            // "i - start + 1" is length of the current processed sequence
            if (i - start + 1 > max_length) {
                max_length = i - start + 1;
                start_ml = start;
            }
        }
        else {
            // if the sequence of the same letters continuing,
            // move the pointer to points to the last two chars of this sequence
            // drop the count to 2
            start = i - 1;
            count = 2;
        }
    }
    return s.substr(start_ml, max_length);
}
int main() {
    cout << "Result: " << solution("aabbaabbaabbaa") << " Expected: aabbaabbaabbaa" << endl;
    cout << "Result: " << solution("aabbaaaaabb") << " Expected: aabbaa" << endl;
    return 0;
}

Program: Min Deletions To Obtain String in Right Format in Java

class Solution {
    public int minimumDeletions(String s) {
      int n=s.length();
      int []dp=new int[n+1];
       int bcount=0;
         for(int i=0;i<n;i++){
             char c=s.charAt(i);
             if(c=='a'){
               dp[i+1]=Math.min(dp[i]+1,bcount);  
             }else{
                 dp[i+1]=dp[i];
                 bcount++;
             }
         }
        return dp[n];
        
    }
}

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