Reverse Prefix of Word (LeetCode 2000) | Full Solution with stack data structure

[@IshaZaka]

Hi Nikhil, plz make for hard challegnes as well

[@AliRaza-gh3hp]

Brother You are doing awesome work for students like us please keep going. can you please do the problem 6. Zizag Conversion on leetcode. I will be thankful.

[@unemployedcse3514]

awesome ❤

[@akankshasonkar6846]

Nice solution Nikhil :)

[@rdrahuldhiman19]

I hope your eye heals quickly

[@nikoo28]

thanks a lot for the good wishes...the eye infection caused me some delay to publish videos. Thanks for being a viewer of my channel and your patience :)

[@arupgope2676]

I solve it using 2 pointer

[@hemanthcse1]

We can achieve this without using extra space like stack
fun reversePrefixSolution3(word: String, ch: Char): String {
val firstOccurrence = word.indexOf(ch)
if (firstOccurrence == -1){
return word
}
val result = CharArray(word.length)
for (i in 0..firstOccurrence){
result[i] = word[firstOccurrence-i]
}
for (i in (firstOccurrence+1) until word.length){
result[i] = word[i]
}
return String(result)
}
correct me if i am wrong

[@nikoo28]

When you use the charArray it does take up extra O(n) space. So the time complexity remains the same

[@anuragrawat4350]

int index = word.indexOf(ch)
if(index != -1){
return new StringBuilder(word.substring(0,index+1).reverse).toString() + word(substring(index+1,word.length()));
}
return word;

[@RameshGaridapuri]

// Online C compiler to run C program online
#include
#include
void swap(char *p, char *q)
{
int temp;
while ( p < q)
{
temp = *p ;
*p++ = *q;
*q-- = temp;
}

}
char * fun_ub(char arr[], char p , int len)
{
int i = 0 ;
while ( arr[i]!= '\0')
{
if ( arr[i] == 'd')
{
swap( &arr[0], &arr[i]);
break;
}
i++;
}
return arr;
}
int main() {
char arr[] = "abcdefd";
char a = 'd';
int stl = strlen(arr)-1;
char *p = fun_ub(arr, a , stl);
printf("%s", p);
}