c - What is the time complexity of my codes: reversing words in a string? Should I account for asymmetrical spacing? -

problem statement: write function of declaration void str_words_in_rev(char *input, int length) reverses words in string of word/(s). i've implemented 2 different solutions it, 1 of extremely complicated.

solution 1: reverse entire string using void reverse(char* string, int start, int end) function takes index of first , last letter of string or substring want reverse. after that, reverse each word in string using same reverse function.

#include<stdio.h>  void swapchar(char *a, char *b) {   char temp = *a;   *a = *b;   *b = temp; }  int length(char *string) {   if(string == null)      return -1;   int i=0;   for(;string[i++];);   return i; }  void reverse(char* string, int start, int end) {   int mid = start + (end - start + 1)/2;   for(int i=start,j=end;i<mid;)     swapchar(&string[i++], &string[j--]); }  int main() {   char str[] = "abc def ghij klmn";   int i=0, j=0, len = length(str), flag = 0;    if(str == null || !(*str))      return 0;   for(;str[j]!='\0';++j) {      if(str[j] == ' ') {        flag = 1;        reverse(str,i,j-1);        = j+1;      }   }   if(flag == 1) {   reverse(str, i, j-1);   reverse(str,0,len-2);   }   printf("%s\n", str);   return 0; } 

solution 2: shift every word , following space character (after rearranging it) towards end using pushtoend function , then shift end before end word pushed. there printf statements can commented out understand what's happening, if explanation inadequate (which apologise for).

        int pushtoend(char *str, int length) {          int len = 0;         int = 0, j = 0, k = 0, n = 0;         (i = 0; str[i++] != ' ' && <= length;)             ++len;         if (len == length)             return len;         //printf("length of first word:");         //printf("%d\n", len);         (j = - 1; j>0; --j)             swapchar(&str[j], &str[j - 1]);         ++len;         //printf("after rearranging first word:");         //printf("%s\n", str);         //printf("pushing first word toward end\n");         (; k + 2 * len <= length; k = k + len) {             (int l = k; l<k + len; ++l) {                 swapchar(&str[l], &str[l + len]);             }             //printf("---%s---\n", str);         }         //printf("as blocks:");         //printf("%s\n", str);         //printf("k value: %d\n", k);         (; k + len<length; ++k) {             (n = k + len; n>k; --n) {                 //printf("n value: %d\n", n);                 swapchar(&str[n], &str[n - 1]);             }         }         //printf("one one:");         //printf("%s\n", str);         return len;     }      void str_words_in_rev1(char *input, int length){     int len = 0;     (int end = length; end>0; end = end - len) {         //printf("end:%d\n", end);         len = pushtoend(input, end);     } } 

once you've understood problem statement , solutions i've implemented, here questions:

what time complexity of solutions? think first solution has o(n) complexity , second 1 o(n^2) i'm not sure.

neither solutions account asymmetrical spacing. words reversed , spacing reversed well. it's imperceptible in cases there symmetrical spacing/multi-symmetrical spacing. problem statement unclear on this, i'd views on whether should accounted for.

solution1 o(n) time complexity. as,
solution2 o(kn) k number of words in string, in worst case words single characters o(n^2). , moving word word towards end , swapping other words towards front, o(kn).