The Indian Engineer

Problem 2000 Reverse Prefix of Word

Posted on 2 mins

Cpp String

Table of Contents

Problem Statement

Link - Problem 2000

Question

Given a 0-indexed string word and a character ch, reverse the segment of word that starts at index 0 and ends at the index of the first occurrence of ch (inclusive). If the character ch does not exist in word, do nothing.

For example, if word = "abcdefd" and ch = "d", then you should reverse the segment that starts at 0 and ends at 3 (inclusive). The resulting string will be "dcbaefd".

Return the resulting string.

Example 1 

Input: word = "abcdefd", ch = "d"
Output: "dcbaefd"
Explanation: The first occurrence of "d" is at index 3.
Reverse the part of word from 0 to 3 (inclusive), the resulting string is "dcbaefd".

Example 2 

Input: word = "xyxzxe", ch = "z"
Output: "zxyxxe"
Explanation: The first and only occurrence of "z" is at index 3.
Reverse the part of word from 0 to 3 (inclusive), the resulting string is "zxyxxe".

Example 3 

Input: word = "abcd", ch = "z"
Output: "abcd"
Explanation: "z" does not exist in word.
You should not do any reverse operation, the resulting string is "abcd".

Constraints

Solution 

class Solution {
public:
    string reversePrefix(string word, char ch) {
        std::ios::sync_with_stdio(false);
        cin.tie(nullptr);
        cout.tie(nullptr);

        int prefixIndex = INT_MIN;

        for(int i = 0;i<word.size();i++)
        {
            if(word[i]==ch)
            {
                prefixIndex = i;
                break;
            }
        }
        if(prefixIndex == INT_MIN)\
            return word;
        string ans = "";
        for(int i = prefixIndex;i>=0;i--)
            ans += word[i];

        for(int i = prefixIndex+1;i<word.size();i++)
            ans += word[i];

        return ans;
    }
};

Complexity Analysis

Explanation

  1. Iterate through the string and check if the character ch exists in word.
  2. If not return the word as it is.
  3. If it exisits swap construct a new string ans in the following fashion.
  4. Append characters from prefixIndex to 0, then append characters from prefixIndex +1 to word.size().
  5. Return the ans string.

Note 

class Solution {
public:
    string reversePrefix(string word, char ch) {
        int j = word.find(ch);
        if (j != -1) {
            reverse(word.begin(), word.begin() + j + 1);
        }
        return word;
    }
};

Showcases the string manipulation.