i and j, to point to the end of s and t respectively.i or j is greater than or equal to 0:
a. Initialize two counters skipS and skipT to 0.
b. Move i and j backwards, skipping over characters based on the number of backspaces encountered.
c. If the current characters of s and t are not backspaces and are not equal, return false.
d. If one string is exhausted before the other, ensure that the remaining characters in the other string are all backspaces.true.