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

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

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                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

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