Today I have done Letter Combinations of a Phone Number and Transpose File on leetcode and leetcode's December LeetCoding Challenge with cpp.

Letter Combinations of a Phone Number

Description

Given a string containing digits from 2-9 inclusive, return all possible letter combinations that the number could represent. Return the answer in any order.

A mapping of digit to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.

Example 1:

Input: digits = "23"
Output: ["ad","ae","af","bd","be","bf","cd","ce","cf"]

Example 2:

Input: digits = ""
Output: []

Example 3:

Input: digits = "2"
Output: ["a","b","c"]

Constraints:

• 0 <= digits.length <= 4
• digits[i] is a digit in the range ['2', '9'].

Solution

nothing to say.

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        if(digits.empty()) return vector<string>();
        vector<string> mp = {"abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        vector<string> nxt;
        vector<string> current{""};
        for(auto c : digits) {
            for(auto &s : current) {
                for(auto letter : mp[c-'2']) {
                    nxt.push_back(s+letter);
                }
            }
            current = nxt;
            nxt.clear();
        }
        return current;
    }
};

Transpose File

Description

Given a text file file.txt, transpose its content.

You may assume that each row has the same number of columns and each field is separated by the ' ' character.

Example:

If file.txt has the following content:

name age
alice 21
ryan 30

Output the following:

name alice ryan
age 21 30

Solution

awk solution

# Read from the file file.txt and print its transposed content to stdout.
awk '
{ 
    for (i=1; i<=NF; i++)  {
        a[NR,i] = $i
    }
}
NF>p { p = NF }
END {    
    for(j=1; j<=p; j++) {
        str=a[1,j]
        for(i=2; i<=NR; i++){
            str=str" "a[i,j];
        }
        print str
    }
}' file.txt

December LeetCoding Challenge 18

Description

Increasing Triplet Subsequence

Given an integer array nums, return true if there exists a triple of indices (i, j, k) such that i < j < k and nums[i] < nums[j] < nums[k]. If no such indices exists, return false.

Example 1:

Input: nums = [1,2,3,4,5]
Output: true
Explanation: Any triplet where i < j < k is valid.

Example 2:

Input: nums = [5,4,3,2,1]
Output: false
Explanation: No triplet exists.

Example 3:

Input: nums = [2,1,5,0,4,6]
Output: true
Explanation: The triplet (3, 4, 5) is valid because nums[3] == 0 < nums[4] == 4 < nums[5] == 6.

Constraints:

• 1 <= nums.length <= 105
• -231 <= nums[i] <= 231 - 1

Follow up: Could you implement a solution that runs in O(n) time complexity and O(1) space complexity?

Solution

could be more percise.

class Solution {
public:
    bool increasingTriplet(vector<int>& nums) {
        bool hasMin = false, hasMiddle = false;
        int min, middle;
        for(auto i : nums) {
            if(!hasMin) {
                min = i;
                hasMin = true;
                continue;
            }
            if(i < min) {
                min = i;
            } else if(!hasMiddle && i > min) {
                middle = i;
                hasMiddle = true;
            } else if(hasMiddle && i > min && i < middle) {
                middle = i;
            }
            if(hasMiddle && i > middle) return true;
        }
        return false;
    }
};