C++ || STL || standard template libraries

-
Image
                                       C++ STL  1.      Pair :  Syntax :      pair <int,char> p1;     it creates a pair {2,'c'} like this.           pair <int,int> p2;     it creates a pair {2,3} like this .  Basic Code :  #include <bits/stdc++.h> using namespace std; int main() {     // cout<<"hello world\n";     pair <int,int> p1 = {1,2};     cout<<p1.first<< " "<< p1.second<<endl;          pair <int, pair<int ,char>> p2 = {1,{2,'c'}};     cout<<p2.first<<" "<<p2.second.first<<" "<<p2.second.second<<endl;          pair <int,int> arr[] = {{2,3},{4,5},{6,7}};     cout...
Post a Comment

Merge Strings Alternatively || Leetcode Solution || Easy Approach || Simple || ✅✅✅✅💯💯💯💯🔥🔥🔥🔥

-

                Merge Strings Alternatively👇👇

1768. Merge Strings Alternately

You are given two strings word1 and word2. Merge the strings by adding letters in alternating order, starting with word1. If a string is longer than the other, append the additional letters onto the end of the merged string.

Return the merged string.

Merge Strings Alternatively

Example 1:

Input: word1 = "abc", word2 = "pqr"
Output: "apbqcr"
Explanation: The merged string will be merged as so:
word1:  a   b   c
word2:    p   q   r
merged: a p b q c r

Example 2:

Input: word1 = "ab", word2 = "pqrs"
Output: "apbqrs"
Explanation: Notice that as word2 is longer, "rs" is appended to the end.
word1:  a   b 
word2:    p   q   r   s
merged: a p b q   r   s

Example 3:

Input: word1 = "abcd", word2 = "pq"
Output: "apbqcd"
Explanation: Notice that as word1 is longer, "cd" is appended to the end.
word1:  a   b   c   d
word2:    p   q 
merged: a p b q c   d

Python Solution : (python3 , python) #Two pointer approach 

class Solution:
    def mergeAlternately(self, word1: str, word2: str) -> str:
        merge = ""
        i = 0 
        j = 0 
        while(i<len(word1) or j <len(word2)):
            if(i<len(word1)):
                merge += word1[i]
                i += 1 
            if(j <len(word2)):
                merge += word2[j]
                j += 1 
        return merge

Comments

Post a Comment

Popular posts from this blog

1050. Actors and Directors Who Cooperated At Least Three Times || Leetcode sql solution || Simple and Easy Approach || ✅✅✅✅💯💯💯🔥🔥🔥🔥

-
  Actors and Directors Who Cooperated At Least Three Times || Leetcode sql solution || Simple and Easy Approach || ✅✅✅✅💯💯💯🔥🔥🔥🔥 Link For The Question :   Actors and Directors Who Cooperated At Least Three Times SQL SOLUTION : USING   GROUP BY: select actor_id , director_id from ( select actor_id , director_id , count ( timestamp ) as cooperated from ActorDirector group by actor_id , director_id ) as table1 where cooperated >= 3 ; GROUP WITH HAVING CLAUSE: select actor_id , director_id from ActorDirector group by actor_id , director_id Having count ( timestamp ) >= 3 ;
Read more

Power of Millets || Happy And Healthy Life

-
Image
Foundation Day Meeting : By   D. Anjaneyulu Reddy : Power of millets: When food is right no medicine is needed. When food is wrong no medicine works 1. Cosmetologist : due to unbalanced diet, regularly going to cosmetologist 2. Harmonal imbalance : hair loss 3. 1 gm of sugar / 1 litre of blood : It is good for health. 4. Staple food : It is eaten regularly , large part of person's diet. 5. Don't eat icecream as it increase the sugar content in blood. Embrace meaning : include , hug
Read more

c program to clear nth bit of the number || Bitwise Operators || Problem Solving ||

-
#include <stdio.h> int main() {     int num, n, newNum;     printf("Enter any number: ");     scanf("%d", &num);     printf("Enter nth bit to clear (0-31): ");     scanf("%d", &n);     newNum = num & (~(1 << n));     printf("Number before clearing %d bit: %d (in decimal)\n", n, num);     printf("Number after clearing %d bit: %d (in decimal)\n", n, newNum);     return 0; } Logic :-  num&(~(1<<n));
Read more